"Thermal radiation of bodies and the photoelectric effect" Introduction. Characteristics of thermal radiation What type of radiation heats the substance

18.1. Find the temperature T of the furnace if it is known that the radiation from a hole in it with an area S = 6.1 cm 2 has a power N = 34.6 W. The radiation is considered to be close to the radiation of an absolutely black body.

18.2. What is the N radiation power of the Sun? The radiation of the Sun is considered close to the radiation of a completely black body. The temperature of the surface of the Sun T = 5800 K.

18.3. What energy luminosity R" E has hardened lead? The ratio of the energy luminosities of lead and a black body for a given temperature k =0.6.

18.4. Radiation power of a black body N = 34 kW. Find temperature T this body, if it is known that its surface S\u003d 0.6 m 2.

18.5. The radiation power of a hot metal surface is N = 0.67 kW. Surface temperature T = 2500K, its area S = 10 cm2. What radiation power N would this surface have if it were absolutely black? Find the ratio k of the energy luminosities of this surface and a completely black body at a given temperature.

18.6. Diameter of a tungsten filament in a light bulb d= 0.3 mm, length of the spiral l = 5 cm. U 127 V current flows through the bulb I \u003d 0.31 A. Find the temperature T spirals. Assume that after equilibrium is established, all the heat released in the filament is lost as a result of radiation. The ratio of the energy luminosities of tungsten and a black body for a given temperature is k = 0.31.

18.7. The temperature of a tungsten filament in a 25-watt light bulb is T = 2450 K. The ratio of its luminosity to the luminosity of a black body at a given temperature k = 0.3 . Find the area S of the radiating surface of the spiral.

18.8. Find the solar constant K, i.e., the amount of radiant energy sent by the Sun per unit time through a unit area perpendicular to the sun's rays and located at the same distance from it as the Earth. The temperature of the Sun's surface is T = 5800K. The radiation of the Sun is considered close to the radiation of a completely black body.

18.9. Assuming that the atmosphere absorbs 10% of radiant energy,. sent by the Sun, find the radiation power N received from the Sun by a horizontal section of the Earth with an area S= 0.5 ha. The height of the Sun above the horizon is φ = 30°. The radiation of the Sun is considered close to the radiation of a completely black body.

18.10. Knowing the value of the solar constant for the Earth (see problem 18.8), find the value of the solar constant for Mars.

18.11. What energy luminosity R e does an absolutely black body have if the maximum spectral density of its energy luminosity falls at a wavelength λ = 484 nm?

18.12. Radiation power of a completely black body N = 10 kW Find the area S of the radiating surface of the body if the maximum spectral density of its energy luminosity falls on the wavelength λ = 700 nm.

18.13. In what regions of the spectrum are the wavelengths corresponding to the maximum spectral density of energy luminosity, if the light source is: a) a spiral of an electric light bulb (T = 3000 K); b) the surface of the Sun (T = 6000 K); c) an atomic bomb, in which at the moment of explosion the temperature develops T = 10 7 K? The radiation is considered to be close to the radiation of an absolutely black body.

18.14. The figure shows the dependence curve of the spectral density of the energy luminosity of a black body r λ on the wavelength λ at a certain temperature. To what temperature T does this curve apply? What percentage of the emitted energy is in the visible spectrum at this temperature?

18.15. When heating a completely black body, the wavelength λ, which accounts for the maximum spectral density of energy luminosity, changed from 690 to 500 nm. How many times did the energy permeability of the body increase in this case?

18.16. What wavelength λ accounts for the maximum spectral density of the energy luminosity of a completely black body with a temperature equal to the temperature t = 37° human body, i.e. T = 310K?

18.17. The temperature T of an absolutely black body changed when heated from 1000 to 3000 K. How many times did its energy luminosity R e increase? How much has the wavelength λ changed, which accounts for the maximum spectral density of energy luminosity? How many times has its maximum spectral density of energy luminosity r λ increased ?

18.18. An absolutely black body has a temperature T 1 = 2900 K. As a result of the cooling of the body, the wavelength, which accounts for the maximum spectral density of energy luminosity, has changed by Δλ = 9 μm. To what temperature T2 has the body cooled?

18.19. The surface of the body is heated to a temperature T = 1000K. Then one half of this surface is heated by ΔT = 100K, the other half is cooled by ΔT = 100K. How many times will the energy luminosity change R uh the surface of this body?

18.20. What power N must be supplied to a blackened metal ball of radius r = 2 cm to keep the temperature ΔT = 27K above the temperature environment? Ambient temperature T = 293 K. Assume that heat is lost only due to radiation.

18.21. The blackened ball cools down from temperature T 1 = 300 K to T 2 = 293 K. How much has the wavelength λ changed , corresponding to the maximum spectral density of its energy luminosity?

18.22. By how much will the mass of the Sun decrease in a year due to radiation? How long will it take for the mass of the sun to halve? Sun surface temperature T= 5800K. Radiation from the Sun is assumed to be constant.

Polarization

1) The magnetic rotation of the plane of polarization is determined by the following formula. 4

2) Determine the thickness of the quartz plate, for which the angle of rotation of the plane of polarization is 180. The specific rotation in quartz for a given wavelength is 0.52 rad/mm. 3

3) Plane polarized light, whose wavelength in vacuum is 600 nm, falls on a plate of Icelandic spar, perpendicular to its optical axis. The refractive indices for ordinary and extraordinary rays are 1.66 and 1.49, respectively. Determine the wavelength of an ordinary ray in a crystal. 3

4) Some substance was placed in the longitudinal magnetic field of a solenoid located between two polarizers. The length of the tube with the substance l. Find the Verdet constant if, at a field strength H, the angle of rotation of the polarization plane for one direction of the field and for the opposite direction of the field. 4

5) Monochromatic plane-polarized light with a circular frequency passes through a substance along a homogeneous magnetic field with a strength H. Find the difference in the refractive indices for the right and left circularly polarized components of the light beam if the Verdet constant is equal to V. 1

6) Find the angle between the main planes of the polarizer and analyzer if the intensity of natural light passing through the polarizer and analyzer is reduced by a factor of 4. 45

7) Linearly polarized light of intensity I0 falls on the analyzer, the vector E0 of which makes an angle of 30 with the transmission plane. What fraction of the incident light does the analyzer let through? 0.75

8) If we pass natural light through two polarizers, the main planes of which form an angle, then the intensity of this light is I \u003d 1/2 * Iest * cos ^ 2 (a). What is the intensity of the plane polarized light that comes out of the first polarizer? 1

9) Natural light passes through two polarizers, the main planes of which form an angle a between themselves. What is the intensity of the plane polarized light that comes out of the second polarizer? 4

10) The angle between the main planes of the polarizer and the analyzer is 60. Determine the change in the intensity of the light passing through them if the angle between the main planes becomes 45. 2

11) A beam of natural light falls on a system of 6 polarizers, the transmission plane of each of which is rotated by an angle of 30 relative to the transmission plane of the previous polarizer. What part of the light flux passes through this system? 12

12) A quartz plate 2 mm thick, cut perpendicular to the optical axis of the crystal, rotates the plane of polarization of monochromatic light of a certain wavelength by an angle of 30. Determine the thickness of the quartz plate placed between parallel nicols so that this monochromatic light is extinguished. 3

13) Natural light passes through a polarizer and an analyzer set so that the angle between their principal planes is phi. Both the polarizer and the analyzer absorb and reflect 8% of the light incident on them. It turned out that the intensity of the beam that emerged from the analyzer is equal to 9% of the intensity of natural light incident on the polarizer. 62

14) When adding two linearly polarized light waves oscillating in perpendicular directions with a phase shift ... 3

15) In what cases, when light passes through the analyzer, is the Malus law applicable? 2

16) What types of waves have the property of polarization? 3

17) What type of waves are electromagnetic? 2

18) Determine the intensity of the reflected light if the oscillations of the light vector of the incident light are perpendicular to the plane of incidence. 1

19) Light falls on the interface between two media with refractive indices n1 and n2, respectively. Denote the angle of incidence by a and let n1>n2. Total reflection of light occurs when ... 2

20) Determine the intensity of the reflected light, in which the oscillations of the light vector lie in the plane of incidence. 5

21) A crystal plate that creates a phase difference between ordinary and extraordinary rays is placed between two polarizers. The angle between the plane of transmission of the polarizers and the optical axis of the plate is 45. In this case, the intensity of the light transmitted through the polarizer will be maximum under the following conditions ... 1

22) Which statements about partially polarized light are true? 3

23) Which statements about plane polarized light are true? 3

24) Two polarizers are placed in the path of the natural light beam, the axes of the polarizers are oriented in parallel. How are the vectors E and B oriented in the beam of light coming out of the second polarizer? 1

25) Which of the following statements are true only for plane polarized electromagnetic waves? 3

26) Which of the following statements are true for both plane polarized electromagnetic waves and non-polarized ones? 4

27) Determine the path difference for a quarter-wave plate cut parallel to the optical axis? 1

28) What is the difference between the refractive indices of ordinary and extraordinary rays in the direction perpendicular to the optical axis in the case of deformation. 1

29) A parallel beam of light falls normally on a 50 mm thick plate of Icelandic spar cut parallel to the optical axis. Taking the refractive indices of Icelandic spar for ordinary and extraordinary rays, respectively, 1.66 and 1.49, determine the difference in the path of these rays passing through this plate. 1

30) A linearly polarized light beam is incident on a polarizer rotating around the beam axis at an angular velocity of 27 rad/s. The energy flux in the incident beam is 4 mW. Find the light energy passing through the polarizer in one revolution. 2

31) A beam of polarized light (lambda = 589nm) falls on a plate of Icelandic spar. Find the wavelength of an ordinary ray in a crystal if its refractive index is 1.66. 355

32) A linearly polarized light beam falls on a polarizer whose transmission plane rotates around the beam axis with an angular velocity w. Find the light energy W passing through the polarizer in one revolution if the energy flux in the incident beam is phi. 1

33) A beam of plane polarized light (lambla = 640nm) falls on a plate of Icelandic spar perpendicular to its optical axis. Find the wavelengths of the ordinary and extraordinary rays in a crystal if the refractive index of Icelandic spar for the ordinary and extraordinary rays is 1.66 and 1.49. 1

34) Plane polarized light falls on an analyzer rotating around the beam axis at an angular velocity of 21 rad/s. Find the light energy passing through the analyzer in one revolution. The intensity of polarized light is 4 W. 4

35) Determine the difference between the refractive index of the ordinary and extraordinary rays of a substance, if the smallest thickness of a half-wave crystalline plate made of this substance for lambda0 \u003d 560 nm is 28 microns. 0.01

36) Plane polarized light, with a wavelength of lambda \u003d 589 nm in vacuum, falls on a crystal plate perpendicular to its optical axis. Finding nm (in modulus) is the difference in wavelengths in a crystal, if the refractive index of the ordinary and extraordinary rays in it is 1.66 and 1.49, respectively. 40

37) Determine the smallest thickness of the crystal plate in half a wave for lambda = 589 nm, if the difference between the refractive indices of ordinary and extraordinary rays for a given wavelength is 0.17. 1.73

38) A parallel beam of light falls normally on a 50 mm thick Icelandic spar plate cut parallel to the optical axis. Taking the refractive indices of the ordinary and extraordinary rays to be 1.66 and 1.49, respectively, determine the difference in the path of the rays that have passed through the plate. 8.5

39) Determine the path difference for a half-wave plate cut parallel to the optical axis? 2

40) A linearly polarized light beam is incident on a polarizer, the transmission plane of which rotates around the beam axis with an angular velocity of 20. Find the light energy W passing through the polarizer in one revolution if the power of the incident beam is 3 W. 4

41) A beam of natural light falls on a glass prism with an angle at the base of 32 (see figure). Determine the refractive index of the glass if the reflected beam is plane polarized. 2

42) Determine at what angle to the horizon the Sun should be in order for the rays reflected from the surface of the lake (n=1.33) to be maximally polarized. 2

43) Natural light falls on glass with a refractive index of n=1.73. Determine the angle of refraction, to the nearest degree, at which the light reflected from the glass is completely polarized. thirty

44) Find the refractive index n of glass if, when light is reflected from it, the reflected beam is completely polarized at an angle of refraction of 35. 1.43

45) Find the angle of full polarization when light is reflected from glass, the refractive index of which is n \u003d 1.57 57.5

46) A beam of light reflected from a dielectric with a refractive index n is completely polarized when the reflected beam forms an angle of 90 with the refracted beam. At what angle of incidence is complete polarization of the reflected light achieved? 3

47) A beam of light falls on the surface of the water (n=1.33). Determine the angle of refraction to the nearest degree if the reflected beam is completely polarized. 37

48) In what case is Brewster's law inaccurate? 4

49) A natural beam of light falls on the surface of a glass plate with refractive indices n1 = 1.52, placed in a liquid. The reflected beam makes an angle of 100 with the incident beam and is completely polarized. Determine the refractive index of the liquid. 1.27

50) Determine the speed of propagation of light in glass if, when light falls from air onto glass, the angle of incidence corresponding to the full polarization of the reflected beam is 58. 1

51) Angle of total internal reflection at the "glass-air" interface 42. Find the angle of incidence of a light beam from the air onto the glass surface, at which the beam is completely polarized to within a degree. 56

52) Determine the refractive index of the medium to the second decimal place, upon reflection from which at an angle of 57 the light will be completely polarized. 1.54

53) Find the refractive index of glass if, when light is reflected from it, the reflected beam is completely polarized at an angle of refraction of 35. 1.43

54) A beam of natural light falls on a glass prism, as shown in the figure. Angle at the base of the prism 30. Determine the refractive index of the glass if the reflected beam is plane polarized. 1.73

55) Determine at what angle to the horizon the Sun should be so that the rays reflected from the surface of the lake (n = 1.33) are maximally polarized. 37

56) A beam of natural light falls on a glass prism with an angle at the base a (see figure). Refractive index of glass n=1.28. Find the angle a to the nearest degree if the reflected beam is plane polarized. 38

57) Determine the refractive index of glass if, when light is reflected from it, the reflected beam is completely polarized at the angle of refraction. 4

58) A beam of plane polarized light falls on the surface of water at the Brewster angle. Its plane of polarization makes an angle of 45 with the plane of incidence. Find the reflection coefficient. 3

59) Determine the refractive index of glass if, when light is reflected from it, the reflected beam is completely polarized at an angle of incidence of 55. 4

60) The degree of polarization of partially polarized light is 0.2. Determine the ratio of the maximum intensity of light transmitted by the analyzer to the minimum. 1.5

61) What are Imax, Imin, P for plane polarized light, where ... 1

62) Determine the degree of polarization of partially polarized light if the amplitude of the light vector corresponding to the maximum light intensity is twice the amplitude corresponding to the minimum intensity. 0.6

63) Determine the degree of polarization of partially polarized light if the amplitude of the light vector corresponding to the maximum light intensity is three times the amplitude corresponding to the maximum intensity. 1

64) The degree of polarization of partially polarized light is 0.75. Determine the ratio of the maximum intensity of light transmitted by the analyzer to the minimum. 1

65) Determine the degree of polarization P of light, which is a mixture of natural light with plane polarized light, if the intensity of polarized light is 3 times the intensity of natural light. 3

66) Determine the degree of polarization P of light, which is a mixture of natural light with plane polarized light, if the intensity of polarized light is 4 times the intensity of natural light. 2

67) Natural light falls at the Brewster angle on the surface of the water. In this case, part of the incident light is reflected. Find the degree of polarization of the refracted light. 1

68) Natural light falls at the Brewster angle on the glass surface (n=1.5). Determine the reflection coefficient in percent. 7

69) Natural light falls at the Brewster angle on the glass surface (n=1.6). Determine the reflection coefficient in percent using Fresnel formulas. 10

70) Determine, using the Fresnel formulas, the reflection coefficient of natural light at normal incidence on the glass surface (n=1.50). 3

71) The reflectance of natural light at normal incidence on the surface of a glass plate is 4%. What is the refractive index of the plate? 3

72) Degree of polarization of partially polarized light P=0.25. Find the ratio of the intensity of the polarized component of this light to the intensity of the natural component. 0.33

73) Determine the degree of polarization P of light, which is a mixture of natural light with plane polarized light, if the intensity of polarized light is equal to the intensity of natural light. 4

74) Degree of polarization of partially polarized light P=0.75. Find the ratio of the intensity of the polarized component of this light to the intensity of the natural component. 3

75) Determine the degree of polarization P of light, which is a mixture of natural light with plane polarized light, if the intensity of polarized light is half the intensity of natural light. 0.33

76) A narrow beam of natural light passes through a gas of optically isotropic molecules. Find the degree of polarization of light scattered at an angle a to the beam. 1

3) The gray body is... 2

5) In fig. graphs of the dependence of the spectral density of the energy luminosity of a black body on the wavelength of radiation at different temperatures T1 and T2 are given, and T1>

Quantum mechanics

quantum mechanics

8) A particle with charge Q and rest mass m0 accelerates in electric field, passing the potential difference U. Can the de Broglie wavelength of a particle be less than its Compton wavelength. (Maybe if QU>0.41m0*c^2)

10) Determine at what numerical value of the speed the de Broglie wavelength for an electron is equal to its Compton wavelength. (2.12e8. lambda(c)=2pi*h/m0*c; lambda=2pi*h*sqrt(1-v^2/c^2)/m0*v; lambda(c)=lambda; 1/ c=sqrt(1-v^2/c^2)/v;v^2=c^2*(1-v^2/c^2);v^2=c^2-v^2;v =c/sqrt(2);v=2.12e8 m/s)

<=x<=1. Используя условие нормировки, определите нормировочный множитель. (A=sqrt(2/l))

>Dpr)

32) The uncertainty relation for energy and time means that (the lifetime of the state of the system (particle) and the uncertainty of the energy of this state of relations >=h)

35) Which of the following ratios is not the Heisenberg ratio. (VEV(x)>=h)

quantum mechanics

1) The kinetic energy of a moving electron is 0.6 MeV. Determine the de Broglie wavelength of an electron. (1.44 pm; 0.6 MeV = 9.613*10^-14 J; lambda=2pi*h/(sqrt(2mT))=1.44 pm)

2) Find the de Broglie wavelength for a proton with a kinetic energy of 100 eV. (2.86 rm. phi=h/sqrt(2m*E(k))=2.86 pm)

3) The kinetic energy of the neutron is 1 keV. Determine the de Broglie wavelength. (0.91 pm. 1keV=1600*10^-19 J. lambda=2pi*h/sqrt(2m*T))=0.91pm)

4) a) Is it possible to represent the De Broglie wave as a wave packet? b) How will the group velocity of the wave packet U and the particle velocity V be related in this case? (no, u=v)

5) Find the ratio of the Compton wavelength of the proton to the De Broglie wavelength for a proton moving at a speed of 3*10^6 m/s. (0.01. lambda(c)=2pi*h/mc=h/mc; lambda=2pi*h/sqrt(2m*T); lambda(c)/phi=0.01)

6) The kinetic energies of two electrons are 3 keV and 4 keV, respectively. Determine the ratio of the corresponding De Broglie lengths. (1.15. lambda=2pi*h/sqrt(2mT); phi1/phi2=1.15)

7) Calculate the de Broglie wavelength of a 0.2 kg ball flying at a speed of 15 m/s. (2.2*10^-34; lambda=h/mv=2.2*10^-34)

8) A particle with charge Q and rest mass m0 accelerates in an electric field, having passed a potential difference U. Can the de Broglie wavelength of a particle be less than its Compton wavelength. (Maybe if QU>0.41m0*c^2)

9) Determine what accelerating potential difference a proton must pass in order for the de Broglie wavelength for it to be 1 nm. (0.822 mV. lambda=2pi*h/sqrt(2m0*T); lambda^2*2m0*T=4*pi^2*h^2; T=2*pi^2*h^2/lambda^2 *m0=2.39e-19; T=eU; U=T/e=2pi^2*h^2/lambda^2*m0*e=0.822 mV)

10) Determine at what numerical value of the speed the de Broglie wavelength for an electron is equal to its Compton wavelength. (2.12e8. lambda(c)=2pi*h/m0*c; lambda=2pi*h*sqrt(1-v^2/c^2)/m0*v; lambda(c)=lambda; 1/ c=sqrt(1-v^2/c^2)/v;v^2=c^2*(1-v^2/c^2);v^2=c^2-v^2;v =c/sqrt(2);v=2.12e8 m/s)

11) Determine the minimum probable energy for a quantum particle located in an infinitely deep potential well of width a. (E=h^2/8ma^2)

12) A particle of mass m is in a one-dimensional potential rectangular well with infinitely high walls. Find the number dN of energy levels in the energy interval (E, E+dE) if the levels are very dense. (dN=l/pi*n*sqrt(m/2E)dE)

13) A quantum particle is located in an infinitely deep potential well of width L. At what points of the electron location on the first (n=1) energy level is the function maximum. (x=L/2)

14) A quantum particle is in an infinitely deep potential well of width a. At what points of the third energy level the particle cannot be. (a, b, d, e)

15) The particle is in an infinitely deep hole. At what energy level is its energy defined as 2h^2/ml^2? (4)

16) The wave function psi(x)=Asin(2pi*x/l) is defined only in the region 0<=x<=1. Используя условие нормировки, определите норировочный множитель. (A=sqrt(2/l))

17) The particle is in the ground state (n=1) in a one-dimensional infinite deep potential well of lambda width with absolutely impenetrable walls (0

18) The particle is in a one-dimensional rectangular potential well with infinitely high walls. Find the quantum number of the energy level of the particle if the energy intervals to the neighboring levels (upper and lower) are related as n:1, where n=1.4. (2.)

19) Determine the wavelength of the photon emitted during the transition of an electron in a one-dimensional rectangular potential well with infinitely high walls of width 1 from state 2 to the state with the lowest energy. (lambda=8cml^2/3h.)

20) The electron hits a potential barrier of finite height. At what energy value of an electron will it not pass through a potential barrier of height U0. (no correct answers)

21) Complete the definition: The tunnel effect is a phenomenon in which a quantum particle passes through a potential barrier at (E

22) The coefficient of transparency of the potential barrier - (the ratio of the flux density of passing particles to the flux density of incident particles)

23) What will be the transparency coefficient of the potential barrier if its width is doubled? (D^2)

24) A particle of mass m falls on a rectangular potential barrier, and its energy E >Dpr)

25) A proton and an electron, having the same energy, move in the positive direction of the X axis and meet a rectangular potential barrier on their way. Determine how many times it is necessary to narrow the potential barrier so that the probability of its passage by a proton is the same as for an electron. (42.8)

26) The rectangular potential barrier has a width of 0.3 nm. Determine the energy difference at which the probability of an electron passing through the barrier is 0.8. (5.13)

27) An electron with an energy of 25 eV meets on its way a low potential step with a height of 9 eV. Determine the refractive index of de Broglie waves at the step boundary. (0.8)

28) A proton with an energy of 100 eV changes the de Broglie wavelength by 1% when passing through a potential step. Determine the height of the potential barrier. (2)

29) The uncertainty relation for position and momentum means that (it is possible to simultaneously measure the coordinates and momentum of a particle only with a certain accuracy, and the product of the uncertainties of position and momentum must be at least h / 2)

30) Estimate the uncertainty of the speed of an electron in a hydrogen atom, assuming the size of the hydrogen atom is 0.10 nm. (1.16*10^6)

31) The uncertainty relation for position and momentum means that (it is possible to simultaneously measure the coordinates and momentum of a particle only with a certain accuracy, and the product of the uncertainties of the position and momentum must be at least h/2)

32) The uncertainty relation for energy and time means that (the lifetime of the state of the system (particle) and the uncertainty of the energy of this state of relations >=h)

33) The uncertainty relation follows from (wave properties of microparticles)

34) The average kinetic energy of an electron in an atom is 10 eV. What is the order of the smallest error with which the coordinate of an electron in an atom can be calculated. (10^-10)

35) Which of the following ratios is not the Heisenberg ratio. (VEV(x)>=h)

36) The uncertainty relation for the position and momentum of a particle means that (it is possible to simultaneously measure the coordinates and momentum of a particle only with a certain accuracy, and the uncertainties of the position and momentum must be at least h/2)

37) Choose the INCORRECT statement (with n=1 an atom can only be in the first energy level for a very small amount of time n=1)

38) Determine the ratio of the uncertainties of the speed of an electron and a dust particle with a mass of 10 ^ -12 kg, if their coordinates are set with an accuracy of 10 ^ -5 m. (1.1 * 10 ^ 18)

39) Determine the speed of an electron in the third orbit of a hydrogen atom. (v=e^2/(12*pi*E0*h))

40) Derive the relationship between the radius of a circular electron orbit and the de Broglie wavelength, where n is the number of the stationary orbit. (2pi*r=n*lambda)

41) Determine the energy of a photon emitted during the transition of an electron in a hydrogen atom from the third energy level to the second. (1.89 eV)

42) Determine the speed of the electron in the third Bohr orbit of the hydrogen atom. (0.731 mm/s)

43) Using Bohr's theory for hydrogen, determine the speed of an electron in an excited state at n=2. (1.14 mm/s)

44) Determine the period of revolution of an electron located in a hydrogen atom in a stationary state (0.15 * 10^-15)

45) An electron is knocked out of a hydrogen atom in a stationary state by a photon whose energy is 17.7. Determine the speed of an electron outside the atom. (1.2 mm/s)

46) Determine the maximum and minimum photon energies in the visible series of the hydrogen spectrum (Bolmer series). (5/36hR, 1/4hR)

47) Calculate for the hydrogen atom the radius of the second Bohr orbit and the speed of the electron on it. (2.12*10^-10, 1.09*10^6)

48) Using Bohr's theory, determine the orbital magnetic moment of an electron moving along the third orbit of a hydrogen atom. (2.8*10^-23)

49) Determine the binding energy of an electron in the ground state for the He+ ion. (54.5)

50) Based on the fact that the ionization energy of the hydrogen atom is 13.6 eV, determine the first excitation potential of this atom. (10.2)

51) An electron is knocked out of a hydrogen atom, which is in the ground state, by a photon of energy e. Determine the speed of an electron outside the atom. (sqrt(2(E-Ei)/m))

52) What is the maximum speed that electrons must have in order to transfer a hydrogen atom from the first state to the third by impact. (2.06)

53) Determine the energy of a photon emitted during the transition of an electron in a hydrogen atom from the third energy level to the second. (1.89)

54) What orbit from the main one will an electron in a hydrogen atom go to when a photon with an energy of 1.93 * 10^-18 J is absorbed. (3)

55) As a result of the absorption of a photon, the electron in the hydrogen atom moved from the first Bohr orbit to the second. What is the frequency of this photon? (2.5*10^15)

56) An electron in a hydrogen atom passes from one energy level to another. What transitions correspond to the absorption of energy. (1,2,5)

57) Determine the minimum electron velocity required for the ionization of a hydrogen atom if the ionization potential of a hydrogen atom is 13.6. (2.2*10^6)

58) At what temperature do mercury atoms have translational kinetic energy sufficient for ionization? The ionization potential of the mercury atom is 10.4 V. The molar mass of mercury is 200.5 g/mol, the universal gas constant is 8.31. (8*10^4)

59) The binding energy of an electron in the ground state of the He atom is 24.6 eV. Find the energy required to remove both electrons from this atom. (79)

60) With what minimum kinetic energy must a hydrogen atom move in order for one of them to be able to emit a photon in an inelastic head-on collision with another hydrogen atom at rest. It is assumed that both atoms are in the ground state before the collision. (20.4)

61) Determine the first excitation potential of the hydrogen atom, where R is the Rydberg constant. (3Rhc/4e)

62) Find the difference between the wavelengths of the head lines of the Lyman series for atoms of light and heavy hydrogen. (33 pm)

1) Choose the correct statement regarding the way electromagnetic waves are emitted. 4

2) Absolutely black and gray bodies, having the same surface area, are heated to the same temperature. Compare the fluxes of thermal radiation of these bodies Ф0 (black) and Ф (gray). 2

3) The gray body is... 2

4) The characteristics of thermal radiation are given below. Which of them is called the spectral density of energy luminosity? 3

5) In fig. graphs of the dependence of the spectral density of the energy luminosity of a blackbody on the wavelength of radiation at different temperatures T1 and T2, and T1>T2 are shown. Which of the figures correctly takes into account the laws of thermal radiation? 1

6) Determine how many times it is necessary to reduce the thermodynamic temperature of a black body so that its energy luminosity R is weakened by 39 times? 3

7) A completely black body is ... 1

8) Can the absorption capacity of a gray body depend on a) radiation frequency b) temperature? 3

9) When studying stars A and stars B, the ratio of the masses lost by them per unit time (delta)mA=2(delta)mB and their radii Ra=2.5Rb was established. The maximum radiation energy of the star B corresponds to the wave lambda B = 0.55 μm. Which wave corresponds to the maximum radiation energy of the star A? 1

10) Choose the correct statement. (absolutely white body) 2

11) Find the wavelength of lambda0 light corresponding to the red border of the photoelectric effect for lithium. (Work function A=2.4 eV). Planck's constant h=6.62*10^-34 J*s. 1

12) Find the wavelength of lambda0 light corresponding to the red border of the photoelectric effect for sodium. (Work function A=2.3 eV). Planck's constant h=6.62*10^-34 J*s. 1

13) Find the wavelength of lambda0 light corresponding to the red border of the photoelectric effect for potassium. (Work function A=2.0 eV). Planck's constant h=6.62*10^-34 J*s. 3

14) Find the wavelength of lambda0 light corresponding to the red border of the photoelectric effect for cesium. (Work function A=1.9 eV). Planck's constant h=6.62*10^-34 J*s. 653

15) The wavelength of light corresponding to the red border of the photoelectric effect for some metal lambda0. Find the minimum energy of a photon that causes the photoelectric effect. 1

16) The wavelength of light corresponding to the red border of the photoelectric effect for some metal lambda0. Find the work function A of an electron from a metal. 1

17) The wavelength of light corresponding to the red border of the photoelectric effect for some metal is equal to lambda0. Find the maximum kinetic energy W of electrons ejected from the metal by light with a wavelength of lambda. 1

18) Find the retarding potential difference U for electrons ejected when a certain substance is illuminated by light with a wavelength of lambda, where A is the work function for this substance. 1

19) Photons with energy e eject electrons from the metal with work function A. Find the maximum momentum p transferred to the surface of the metal when each electron is emitted. 3

20) A vacuum photocell consists of a central cathode (tungsten ball) and an anode (the inner surface of the bulb silvered from the inside). The contact potential difference between the electrodes U0 accelerates the outgoing electrons. The photocell is illuminated by light with a wavelength of lambda. What speed v will the electrons get when they reach the anode if no potential difference is applied between the cathode and the anode? 4

21) In fig. Graphs of the dependence of the maximum energy of photoelectrons on the energy of photons incident on the photocathode are shown. In which case does the photocell cathode material have the lowest work function? 1

22) Einstein's equation for the many photon photoelectric effect has the form. 1

23) Determine the maximum speed of electrons emitted from the cathode if U=3V. 1

24) External photoelectric effect - ... 1

25) Internal photoelectric effect - ... 2

26) Valve photoelectric effect - ... 1) consists ... 3

27) Determine the speed of photoelectrons pulled out from the surface of silver by ultraviolet rays (lambda \u003d 0.15 microns, m \u003d 9.1 * 10^-31 kg), if the work function is 4.74 eV. 3

28) Determine the "red border" of the photoelectric effect for silver if the work function is 4.74 eV. 2

29) The red border of the photoelectric effect for metal (lambda0) is 550 nm. Find the minimum value of the photon energy (Emin) that causes the photoelectric effect. 1

30) The work function of an electron from the surface of one metal is A1=1 eV, and from the other - A2=2 eV. Will the photoelectric effect be observed in these metals if the photon energy of the radiation incident on them is 4.8 * 10^-19 J? 3

31) Valve photoelectric effect is ... 1) occurrence ... 1

32) The figure shows the current-voltage characteristic of the photoelectric effect. Determine which curve corresponds to the high illumination of the cathode, at the same frequency of light. 1

33) Determine the maximum speed Vmax of photoelectrons ejected from the surface of silver by ultraviolet radiation with a wavelength of 0.155 μm at a work function for silver of 4.7 eV. 1

34) Compton found that the optical difference between the wavelength of the scattered and incident radiation depends on ... 3

35) The Compton wavelength (when a photon is scattered into electrons) is equal to. 1

36) Determine the wavelength of X-ray radiation if, with Compton scattering of this radiation at an angle of 60, the wavelength of the scattered radiation turned out to be 57 pm. 5

37) A photon with a wavelength of 5 pm experienced Compton scattering at an angle of 60. Determine the change in wavelength during scattering. 2

38) What was the wavelength of X-rays, if, when this radiation is scattered by some substance at an angle of 60, the wavelength of the scattered X-rays is 4 * 10^-11 m.

39) Are the following statements true: a) scattering occurs when a photon interacts with a free electron, and the photoelectric effect occurs when interacting with bound electrons b) absorption of a photon by a free electron is impossible, since this process is in conflict with the laws of conservation of momentum and energy. 3

40) Figure 3 shows a vector diagram of Compton scattering. Which of the vectors represents the momentum of the scattered photon? 2

41) Directed monochromatic light flux Ф falls at an angle of 30 on absolutely black (A) and mirror (B) plates (Fig. 4). Compare the light pressure on plates A and B, respectively, if the plates are fixed. 3

42) Which of the following expressions is the formula experimentally obtained by Compton? 1

43) Can a free electron absorb a photon? 2

44) A photon with an energy of 0.12 MeV was scattered by an initially resting free electron. It is known that the wavelength of the scattered photon has changed by 10%. Determine the kinetic energy of the recoil electron (T). 1

45) X-ray radiation with a wavelength of 55.8 pm is scattered by a graphite tile (Compton effect). Determine the wavelength of light scattered at an angle of 60 to the direction of the incident light beam. 1

85) In Young's experiment, the hole is illuminated with monochrome light (lambda=600 nm). The distance between the holes is d=1 nm, the distance from the holes to the screen is L=3 m. Find the position of the first three light bands. 4

86) The installation for obtaining Newton's rings is illuminated by monochromatic light falling normally. Light wavelength lambda=400 nm. What is the thickness of the air wedge between the lens and the glass plate for the third bright ring in reflected light? 3

87) In Young's experiment (interference of light from two narrow slits), a thin glass plate was placed in the path of one of the interfering rays, as a result of which the central bright stripe shifted to the position originally occupied by the fifth bright stripe (not counting the central one). The beam is incident perpendicular to the surface of the plate. The refractive index of the plate n=1.5. Wavelength lambda=600 nm. What is the thickness h of the plate? 2

88) Installation for the observation of Newton's rings is illuminated by monochromatic light with a wavelength of lambda = 0.6 microns, falling normally. Observation is carried out in reflected light. The radius of curvature of the lens is R=4 m. Determine the refractive index of the liquid that fills the space between the lens and the glass plate if the radius of the third light ring is r=2.1 mm. It is known that the refractive index of a liquid is less than that of glass. 3

89) Determine the length of the segment l1, on which the same number of wavelengths of monochromatic light in vacuum fit as they fit on the cutter l2=5 mm in glass. Refractive index of glass n2=1.5. 3 http://ivandriver.blogspot.ru/2015/01/l1-l25-n15.html

90) A normally parallel beam of monochromatic light (lambda = 0.6 μm) falls on a thick glass plate covered with a very thin film, the refractive index of which is n = 1.4. At what minimum film thickness will reflected light be maximally attenuated? 3

91) What should be the allowable width of the slits d0 in Young's experiment so that the interference pattern is visible on a screen located at a distance L from the slits. Distance between slots d, wavelength lambda0. 1

92) A point source of radiation contains wavelengths in the range from lambda1=480 nm to lambda2=500 nm. Estimate the coherence length for this radiation. 1

93) Determine how many times the width of the interference fringes on the screen will change in the experiment with Fresnel mirrors if the violet light filter (0.4 microns) is replaced by red (0.7 microns). max: delta=+-m*lambda, delta=xd/l, xd/l=+-m*lambda, x=+-(ml/d)*lambda, delta x=(ml*lambda/d)-( (m-1)l*lambda/d)=l*lambda/d, delta x1/delta x2=lambda2/lambda1 = 1.75 (1)

94) In the Young installation, the distance between the slots is 1.5 mm, and the screen is located at a distance of 2 m from the slots. Determine the distance between the interference fringes on the screen if the wavelength of monochromatic light is 670 nm. 3

95) Two coherent beams (lambda = 589 nm) maximize each other at a certain point. A normal soap film was placed on the path of one of them (n=1.33). At what minimum thickness d of the soap film will these coherent rays maximally attenuate each other at some point. 3

96) Installation for obtaining Newton's rings is illuminated by monochromatic light incident along the normal to the surface of the plate. The radius of curvature of the lens R=15 m. Observation is carried out in reflected light. The distance between the fifth and twenty-fifth light Newton's rings l=9 mm. Find the lambda wavelength of monochromatic light. r=sqrt((2m-1)lambda*R/2), delta d=r2-r1=sqrt((2*m2-1)lambda*R/2)-sqrt((2*m1-1)lambda* R/2)=7sqrt(lambda*R/2)-3sqrt(lambda*R/2)=4sqrt(lambda*R/2), lambda=sqr(delta d)/8R = 675 nm.

97) The two slots are 0.1 mm apart and 1.20 m from the screen. From a remote source, light with a wavelength of lambda=500 nm falls on the slit. How far apart are the light stripes on the screen? 2

98) Monochromatic light with a wavelength of lambda = 0.66 microns falls on the installation for obtaining Newton's rings. The radius of the fifth light ring in reflected light is 3 mm. Determine the radius of curvature of the lens. 3m or 2.5m

100) An interference pattern from two coherent light sources with a wavelength of lambda \u003d 760 nm is observed on the screen. By how many fringes will the interference pattern on the screen shift if a plastic of fused quartz d = 1 mm thick with a refractive index n = 1.46 is placed in the path of one of the rays? The beam falls on the plate normally. 2

101) An interference pattern from two coherent light sources with a wavelength of 589 nm is observed on the screen. By how many fringes will the interference pattern on the screen shift if a fused quartz plastic 0.41 mm thick with a refractive index n = 1.46 is placed in the path of one of the rays? The beam falls on the plate normally. 3

103) If you look, squinting your eyes, at the filament of an incandescent lamp, then the filament seems to be bordered by light highlights in two perpendicular directions. If the lamp filament is parallel to the observer's nose, then a series of iridescent images of the filament can be observed. Explain the reason for this phenomenon. 4

104) Light falls normally on a transparent diffraction grating of width l=7 cm. Determine the smallest wave difference that this grating can resolve in the lambda=600 nm region. Type your answer in pm to the nearest tenth. 7.98*10^-12=8.0*10^-12

105) Let the intensity of a monochromatic wave be I0. The diffraction pattern is observed using an opaque screen with a round hole on which the given wave is incident perpendicularly. Considering the hole to be equal to the first Fresnel zone, compare the intensities I1 and I2, where I1 is the light intensity behind the screen with the hole fully open, and I2 is the light intensity behind the screen with the hole half closed (in diameter). 2

106) Monochromatic light with a wavelength of 0.6 microns normally falls on a diffraction grating. The diffraction angle for the fifth maximum is 30, and the minimum wavelength difference resolvable by the grating is 0.2 nm for this maximum. Determine: 1) diffraction grating constant; 2) the length of the diffraction grating. 4

107) A parallel beam of light falls on a diaphragm with a round hole. Determine the maximum distance from the center of the hole to the screen, at which a dark spot will still be observed in the center of the diffraction pattern, if the radius of the hole is r=1 mm, the wavelength of the incident light is 0.5 μm. 2

108) Normally monochromatic light falls on a narrow slit. Its direction to the fourth dark diffraction band is 30. Determine the total number of diffraction peaks. 4

109) A normally monochromatic wave of lambda length falls on a diffraction grating with a period of d \u003d 2.8 * lambda. What is the highest order of the diffraction maximum given by the grating? Determine the total number of maxima? 1

110) Light with a wavelength of 750 nm passes through a slit with a width of D \u003d 20 microns. What is the width of the central maximum on the screen located at a distance L=20 cm from the slit? 4

111) A beam of light from a discharge tube normally falls on a diffraction grating. What should be the constant d of the diffraction grating so that the maxima of the lines lambda1=656.3 nm and lambda2=410.2 nm coincide in the direction phi=41. 1

112) Using a diffraction grating with a period of 0.01 mm, the first diffraction maximum was obtained at a distance of 2.8 cm from the central maximum and at a distance of 1.4 m from the grating. Find the length of the light wave. 4

113) A point source of light with a wavelength of 0.6 microns is located at a distance a = 110 cm in front of a diaphragm with a round hole with a radius of 0.8 mm. Find the distance b from the diaphragm to the observation point for which the number of Fresnel zones in the hole is k=2. 3

114) A point light source (lambda = 0.5 μm) is located at a distance a = 1 m in front of a diaphragm with a round hole of diameter d = 2 mm. Determine the distance b (m) from the diaphragm to the observation point if the hole opens three Fresnel zones. 2 http://studyport.ru/images/stories/tasks/Physics/difraktsija-sveta/1.gif

116) Normally monochromatic light with a wavelength of 550 nm falls on a diffraction grating with a length l \u003d 15 mm, containing N \u003d 3000 strokes. Find: 1) the number of maxima observed in the spectrum of the diffraction grating 2) the angle corresponding to the last maximum. 2

117) How does the pattern of the diffraction spectrum change as the screen moves away from the grating? 2

118) A parallel beam of monochromatic light with a wavelength of 0.5 microns normally falls on a screen with a round hole of radius r \u003d 1.5 mm. The observation point is located on the axis of the hole at a distance of 1.5 m from it. Determine: 1) the number of Fresnel zones that fit in the hole 2) a dark or light ring is observed in the center of the diffraction pattern if a screen is placed at the observation point. r=sqrt(bm*lambda), m=r^2/b*lambda=3 - odd, light ring. 2

119) A plane wave falls normally on a diaphragm with a round hole. Determine the radius of the fourth Fresnel zone if the radius of the second Fresnel zone = 2 mm. 4

120) The angular dispersion of the diffraction grating in the spectrum of the first order dphi / dlambda \u003d 2.02 * 10 ^ 5 rad / m. Find the linear dispersion D of the diffraction grating if the focal length of the lens projecting the spectrum onto the screen is F=40 cm. 3

The emission of electromagnetic waves by matter occurs due to

intraatomic and intramolecular processes. Energy sources and, therefore, the type of glow can be different: a TV screen, a fluorescent lamp, an incandescent lamp, a rotting tree, a firefly, etc.

Of all the variety of electromagnetic radiation, visible or not visible to the human eye, one can be distinguished, which is inherent in all bodies. This is the radiation of heated bodies, or thermal radiation.

thermal radiation is characteristic of all bodies at absolute temperature T>0, and its source is the internal energy of radiating bodies, or rather, the energy of the chaotic thermal motion of their atoms and molecules. Depending on the body temperature, the intensity of radiation and the spectral composition change, therefore, thermal radiation is not always perceived by the eye as a glow.

Let us consider some basic characteristics of thermal radiation. The average radiation power for a time much longer than the period of light oscillations is taken as radiation flux F. In SI it is expressed in watts(W).

The flux of radiation emitted by 1 m 2 of the surface is called energy luminosityR e. It is expressed in watts per square meter (W/m2).

A heated body emits electromagnetic waves of various wavelengths. We single out a small interval of wavelengths from λ up to λ + Δλ . The energy luminosity corresponding to this interval is proportional to the width of the interval:

Where - spectral density of the energy luminosity of the body, equal to the ratio of the energy luminosity of a narrow section of the spectrum to the width of this section, W / m 3.

The dependence of the spectral density of energy luminosity on the wavelength is called body radiation spectrum.

Integrating (13), we obtain an expression for the energy luminosity of the body:

The ability of a body to absorb radiation energy is characterized by absorption coefficient, equal to the ratio of the radiation flux absorbed by a given body to the radiation flux that fell on it:

α = Fpogl /Fpad (15)

Since the absorption coefficient depends on the wavelength, then (15) is written for monochromatic radiation fluxes, and then this ratio determines monochromatic absorption coefficient:

αλ = Фpl (λ) / Фpad (λ)

It follows from (15) that the absorption coefficients can take values ​​from 0 to 1. Black bodies absorb radiation especially well: black paper, fabrics, velvet, soot, platinum black, etc.; poorly absorb bodies with a white surface and mirrors.

A body whose absorption coefficient is equal to unity for all wavelengths (frequencies) is called black. It absorbs all radiation falling on it at any temperature.

There are no black bodies in nature, this concept is a physical abstraction. The model of a black body is a small hole in a closed opaque cavity. The beam that fell into this hole, reflected many times from the walls, will be almost completely absorbed. In the future, this particular model will be taken as a black body (Fig. 26).

A body whose absorption coefficient is less than unity and does not depend on the wavelength of light falling on it is called grey.

There are no gray bodies in nature, however, some bodies emit and absorb as gray in a certain range of wavelengths. Thus, for example, the human body is sometimes considered gray, having an absorption coefficient of approximately 0.9 for the infrared region of the spectrum.

The quantitative relationship between radiation and absorption was established by G. Kirchhoff in 1859: at the same temperature, the ratio of the spectral density of energy luminosity to the monochromatic absorption coefficient is the same for any bodies, including black ones ( Kirchhoff's law):

where is the spectral density of the energy luminosity of a black body (subscripts in brackets mean bodies1 , 2 etc.).

Kirchhoff's law can also be written in the following form:

The ratio of the spectral density of the energy luminosity of any body to its corresponding monochromatic absorption coefficient is equal to the spectral density of the energy luminosity of a black body at the same temperature.

From (17) we find one more expression:

Since for any body (non-black)< 1, то, как следует из (18), спектральная плотность энергетической светимости любо­го тела меньше спектральной плотности энергетической свети­мости черного тела при той же температуре. Черное тело при про­чих равных условиях является наиболее интенсивным источником thermal radiation.

From (18) it can be seen that if the body does not absorb any radiation (= 0), then it does not emit it (= 0).

Black body radiation has a continuous spectrum. Graphs of the emission spectra for different temperatures are shown in Fig. 27.

A number of conclusions can be drawn from these experimental curves.

There is a maximum spectral density of energy luminosity, which shifts towards short waves with increasing temperature.

Based on (14) the energy luminosity of the black body can be found as the area bounded by the curve and the x-axis.

From fig. 27 shows that the energy luminosity increases as the black body heats up.

For a long time, they could not theoretically obtain the dependence of the spectral density of the energy luminosity of a black body on the wavelength and temperature, which would correspond to the experiment. In 1900, this was done by M. Planck.

In classical physics, the emission and absorption of radiation by a body was considered as a continuous wave process. Planck came to the conclusion that it is precisely these basic provisions that do not allow obtaining the correct dependence. He put forward a hypothesis, from which it followed that the black body radiates and absorbs energy not continuously, but in certain discrete portions - quanta.

For the energy luminosity of a black body, we get:

where is the Boltzmann constant.

This Stefan-Boltzmann law: the energy luminosity of a black body is proportional to the fourth power of its thermodynamic temperature.

Wien's Displacement Law:

where - wavelength, which accounts for the maximum spectral density of the energy luminosity of the black body, b = 0.28978.10 -2 m.K - Wien's constant. This law is also valid for gray bodies.

The manifestation of Wien's law is known from everyday observations. At room temperature, the thermal radiation of bodies is mainly in the infrared region and is not perceived by the human eye, and at a very high temperature - white with a blue tint, the feeling of body heating increases.

The laws of Stefan-Boltzmann and Wien allow, by registering the radiation of bodies, to determine their temperatures (optical pyrometry).

The most powerful source of thermal radiation is the Sun.

The weakening of radiation by the atmosphere is accompanied by a change in its spectral composition. On fig. 28 shows the spectrum of solar radiation at the boundary of the Earth's atmosphere (curve 1) and on the Earth's surface (curve 2) at the highest position of the Sun. Curve 1 is close to the spectrum of a black body, its maximum corresponds to a wavelength of 470 nm, which, according to Wien's law, makes it possible to determine the temperature of the solar surface - about 6100 K. Curve 2 has several absorption lines, its maximum is located near 555 nm. The intensity of direct solar radiation is measured actinometer.

Its principle of operation is based on the use of heating of the blackened surfaces of bodies, which comes from solar radiation.

Dosed solar radiation is used as sun therapy (heliotherapy), as well as a means of hardening the body. For therapeutic purposes, artificial sources of thermal radiation are used: incandescent lamps ( sollux) and infrared emitters infrared), mounted in a special reflector on a tripod. Infrared emitters are designed like household electric heaters with a round reflector. The spiral of the heating element is heated by current to a temperature of about 400-500 °C. Electromagnetic radiation occupying the spectral region between the red border of visible light (λ=0.76 µm) and shortwave radio emission [λ=(1-2) mm] is called infrared (IR). The infrared region of the spectrum is usually conditionally divided into near (from 0.74 to 2.5 microns), middle (2.5 - 50 microns) and far (50-2000 microns).

The spectrum of infrared radiation, as well as the spectrum of visible and ultraviolet radiation, may consist of separate lines, bands, or be continuous, depending on the nature of the infrared source.

radiation (Fig. 29).

Excited atoms or ions emit ruled infrared spectra. Excited molecules emit striped infrared spectra due to their oscillations and rotations. Vibrational and vibrational-rotational spectra are located mainly in the middle, and purely rotational - in the far infrared region.

Heated solids and liquids emit a continuous infrared spectrum. If we substitute the limits of IR radiation in Wien's displacement law, we will obtain temperatures of 3800-1.5 K, respectively. This means that all liquid and solid bodies under normal conditions (at ordinary temperatures) are practically not only sources of IR radiation, but and have a maximum emission in the IR region of the spectrum. The deviation of real bodies from gray ones does not change the nature of the conclusion.

A heated solid body radiates in a very wide range of wavelengths. At low temperatures (below 800 K), the radiation of a heated solid body is located almost entirely in the infrared region, and such a body appears dark. As the temperature rises, the proportion of radiation in the visible region increases, and the body first appears dark red, then red, yellow, and finally, at high temperatures (above 5000 K), white; in this case, both the total energy of radiation and the energy of infrared radiation increase.

PROPERTIES of infrared radiation:

optical properties- many substances that are transparent in the visible region are opaque in some areas of infrared radiation and vice versa. For example: a few cm layer of water is opaque, while black paper is transparent in the far IR region.

At a low temperature, the energy luminosity of bodies is small. Therefore, not all bodies can be used as sources IR radiation. In this regard, along with thermal sources of infrared radiation, high-pressure mercury lamps and lasers are also used, which, unlike other sources, do not give a continuous spectrum. The Sun is a powerful source of IR radiation, about 50% of its radiation lies in the IR region of the spectrum.

Methods detection and measurement IR radiation is based on the conversion of infrared radiation energy into other types of energy that can be measured by conventional methods. They are divided mainly into two groups: thermal and photovoltaic. An example of a thermal receiver is a thermoelement, the heating of which causes an electric current. Photoelectric receivers include photocells and photoresistors.

It is also possible to detect and register infrared radiation with photographic plates and photographic films with a special coating.

The therapeutic use of infrared radiation is based on its thermal effect. The greatest effect is achieved by short-wave infrared radiation, close to visible light. For treatment, special lamps are used.

Infrared radiation penetrates the body to a depth of about 20 mm, so the surface layers are heated to a greater extent. The therapeutic effect is precisely due to the emerging temperature gradient, which activates the activity of the thermoregulatory system. Increased blood supply to the irradiated site leads to favorable therapeutic consequences.

Pros and cons of IR radiation:

Infrared rays for the treatment of diseases have been used since ancient times, when doctors used burning coals, hearths, heated iron, sand, salt, clay, etc. to cure frostbite, ulcers, bruises, bruises, etc. Hippocrates described how they were used to treat wounds, ulcers, cold injuries, etc.

It has been proven that infrared rays simultaneously have analgesic (due to hyperemia caused by infrared rays), antispasmodic, anti-inflammatory, stimulating, distracting effects; improve blood circulation; surgery performed with IR radiation is easier to tolerate and cell regeneration occurs faster.

IR radiation is used to prevent the development of fibrosis and pneumosclerosis in the lung tissue (to enhance regeneration in the affected organ).

Magnetic laser therapy is carried out in the infrared spectrum of radiation for the treatment of liver pathology (for example, in order to correct the toxic effect of chemotherapy drugs in the treatment of tuberculosis).

2. - On bright sunny days, on the water, in the highlands, on the snow, there may be an excess of IR radiation. And although the effects of UV sound more ominous, excess IR for the eyes is just as undesirable. The energy of these rays is absorbed by the cornea and lens and converted into heat. An excess of this completely imperceptible heat can lead to irreversible damage. Unlike UV, IR radiation passes through glass lenses perfectly. In special glasses for pilots, climbers, skiers, the factor of increased infrared radiation must be taken into account. Radiation with a wavelength of 1-1.9 microns especially heats the lens and aqueous humor. This causes various disturbances, the main of which is photophobia(photophobia) - a hypersensitive condition of the eye, when normal light exposure gives rise to painful sensations. Photophobia often does not depend on the extent of the damage: with a small damage to the eye, the patient may feel severely affected.

Electromagnetic radiation occupying the spectral region between the violet border of visible light (λ=400 nm) and the long-wavelength part of X-ray radiation (λ=10 nm) is called ultraviolet (UV).

In the wavelength range below 200 nm, UV radiation is strongly absorbed by all bodies, including thin layers of air, so it is not of particular interest to medicine. The rest of the UV spectrum is conditionally divided into three regions (see § 24.9): A (400-315 nm-), B (315-280 nm-erythema) and C (280-200 nm-bactericidal).

Incandescent solids at high temperatures emit a significant amount of UV radiation. However, the maximum spectral density of energy luminosity, in accordance with Wien's displacement law, even for the longest wavelength of the UV range (0.4 μm) falls at 7000 K. In practice, this means that under normal conditions, the thermal radiation of bodies cannot serve as an effective source of powerful UV radiation. radiation. The most powerful source of thermal UV radiation is the Sun, 9% radiation of which at the boundary of the earth's atmosphere falls in the UV range.

Under laboratory conditions, electric discharges in gases and metal vapors are used as sources of UV radiation. Such radiation is no longer thermal and has a line spectrum.

Measurement UV radiation is mainly carried out by photovoltaic receivers. Indicators are luminescent substances and photographic plates.

UV radiation is necessary for the operation of ultraviolet microscopes, luminescent microscopes, and for luminescent analysis. The main use of UV radiation in medicine is associated with its specific biological effects, which are caused by photochemical processes.

Ultraviolet rays have the highest energy, therefore, when they are absorbed, significant changes occur in the electronic structure of atoms and molecules. The absorbed energy of ultraviolet rays can migrate and be used to break weak bonds in protein molecules.

Short-wave ultraviolet rays cause denaturation of protein polymers, which precipitate, losing their biological activity.

A special effect of ultraviolet rays was noted on DNA molecules: DNA duplication and cell division are disturbed, oxidative destruction of protein structures occurs, which leads to cell death. The irradiated cell first loses the ability to divide, and then, after dividing two or three times, dies.

Vitamin-forming action of ultraviolet rays is also important. Pro-vitamins found in the skin are converted into vitamin D under the influence of medium-wave ultraviolet radiation. .

Ultraviolet rays penetrate only 0.1 mm, but carry more energy than other electromagnetic waves in the visible and infrared spectrum.

Protein breakdown products cause vasodilation, skin edema, migration of leukocytes with irritation of skin receptors, internal organs with the development of neuroreflex reactions. Protein degradation products are carried through the bloodstream, causing a humoral effect.

In cosmetology, ultraviolet irradiation is widely used in solariums to obtain an even, beautiful tan. In solariums, unlike natural conditions, filters are used that absorb short-wave and medium-wave rays. Irradiation in solariums begins with a minimum time - one minute, and then gradually the duration of insolation increases. An overdose of ultraviolet rays leads to premature aging, a decrease in skin elasticity, the development of skin and oncological diseases.

All modern protective skin care creams contain complexes that provide ultraviolet protection.

Deficiency of ultraviolet rays leads to beriberi, decreased immunity, weak functioning of the nervous system, and the appearance of mental instability.

Ultraviolet radiation has a significant effect on phosphorus-calcium metabolism, stimulates the formation of vitamin D and improves all metabolic processes.

Ultraviolet rays are useful, moreover, necessary for humans, if only because vitamin D is formed in the body when irradiated in the range of 280-320 nm. However, this is common knowledge. Less often you can find references to the fact that ultraviolet radiation in reasonable doses helps the body suppress colds, infectious and allergic diseases, enhances metabolic processes and improves blood formation. And also increases resistance to many harmful substances, including lead, mercury, cadmium, benzene, carbon tetrachloride and carbon disulfide.

However, UV is not for everyone. It is contraindicated in active forms of tuberculosis, with severe atherosclerosis, hypertension II and III degree, kidney disease and some other diseases. If in doubt, consult your doctor. To receive a prophylactic dose of ultraviolet radiation, you need to be in the fresh air for a sufficient time, not caring especially about whether sunlight hits the skin or does not.

However, in order to get a good tan, it is not at all necessary to climb into the heat, under direct rays. Against. Sunbathing in the shade - you see, there is something in this ... It is quite enough if a significant part of the celestial sphere is not fenced off from you, say, by houses or a dense forest. Ideal conditions are the shade of a lone tree on a clear day. Or the shade from a large umbrella (or a small awning) on ​​a sunny beach. Sunbathe for health!

The human body has a certain temperature due to

thermoregulation, an essential part of which is the heat exchange of the body with the environment. Let us consider some features of such heat transfer, assuming that the ambient temperature is lower than the temperature of the human body.

Heat exchange occurs through heat conduction, convection, evaporation and radiation (absorption).

It is difficult or even impossible to accurately indicate the distribution of the amount of heat given off between the listed processes, since it depends on many factors: the state of the body (temperature, emotional state, mobility, etc.), the state of the environment (temperature, humidity, air movement, etc.). . p.), clothing (material, shape, color, thickness).

However, it is possible to make approximate and average estimates for people who do not have much physical activity and live in a temperate climate.

Since the thermal conductivity of air is small, this type of heat transfer is very insignificant. More significant is convection, it can be not only ordinary, natural, but also forced, in which air blows over a heated body. Clothing plays an important role in reducing convection. In temperate climates, 15-20% of human heat transfer is carried out by convection.

Evaporation occurs from the surface of the skin and lungs, with about 30% of heat loss taking place.

The largest share of heat loss (about 50%) is due to radiation to the external environment from open parts of the body and clothing. The main part of this radiation belongs to the infrared range with a wavelength of 4 to 50 microns.

The maximum spectral density of the energy luminosity of the body

a person, in accordance with Wien's law, falls on a wavelength of approximately 9.5 microns at a skin surface temperature of 32 degrees C.

Due to the strong temperature dependence of the energy luminosity (fourth power of thermodynamic temperature), even a slight increase in surface temperature can cause such a change in the radiated power, which is reliably recorded by the instruments.

In healthy people, the distribution of temperature at various points on the surface of the body is quite characteristic. However, inflammatory processes, tumors can change the local temperature.

The temperature of the veins depends on the state of circulation, as well as on cooling or heating of the extremities. Thus, registration of radiation from different parts of the human body surface and determination of their temperature is a diagnostic method. Such a method, called thermography, is increasingly being used in clinical practice.

Thermography is absolutely harmless and in the future can become a method of mass preventive examination of our population.

The determination of the difference in body surface temperature during thermography is mainly carried out two methods. In one case, liquid crystal indicators are used, the optical properties of which are very sensitive to small temperature changes. By placing these indicators on the patient's body, it is possible to visually determine the local temperature difference by changing their color. Another method, more common, is the technical method, which is based on the use thermal imagers. A thermal imager is a technical system similar to a television that is able to perceive infrared radiation coming from the body, convert this radiation into the optical range and reproduce the image of the body on the screen. Body parts with different temperatures are shown on the screen in different colors.

Thermal radiation of bodies

The main questions of the topic:

1. Characteristics of thermal radiation.

2. Laws of thermal radiation (Kirchhoff's law, Stefan-Boltzmann's law, Wien's law); Planck's formula.

3. Physical foundations of thermography (thermal imaging).

4. Body heat transfer.

Any body at temperatures above absolute zero (0 K) is a source of electromagnetic radiation, which is called thermal radiation. It arises due to the internal energy of the body.

The range of wavelengths of electromagnetic waves (spectral range) emitted by a heated body is very wide. In the theory of thermal radiation, it is often believed that here the wavelength varies from 0 to ¥.

The distribution of the energy of thermal radiation of a body over wavelengths depends on its temperature. At room temperature, almost all energy is concentrated in the infrared region of the electromagnetic wave scale. At a high temperature (1000°C) a significant part of the energy is emitted in the visible range as well.

Characteristics of thermal radiation

1. Flux (power) of radiation Ф(sometimes referred to as R) is the energy emitted in 1 second from the entire surface of the heated body in all directions in space and in the entire spectral range:

, in SI . (1)

2. Energy luminosity R- energy radiated for 1 sec from 1 m 2 of the body surface in all directions in space and in the entire spectral range. If S is the surface area of ​​the body

, , in SI , (2)

It's obvious that .

3. Spectral density of energy luminosity r λ- energy radiated in 1 sec from 1m 2 of the body surface in all directions at wavelength λ in a single spectral range , →

Rice. 1

The dependence of r l on l is called spectrum thermal radiation of the body at a given temperature (at T= const). The spectrum gives the distribution of the energy emitted by the body over wavelengths. It is shown in fig. 1.

It can be shown that the energy luminosity R is equal to the area of ​​the figure bounded by the spectrum and the axis (Fig. 1).

4. The ability of a heated body to absorb the energy of external radiation is determined by monochromatic absorption coefficient a l,

those. a l the ratio of the radiation flux with a wavelength l, absorbed by the body, to the radiation flux of the same wavelength, incident on the body. From (3.) it follows that and l - dimensionless quantity and .

By type of addiction A from l all bodies are divided into 3 groups:

1). Completely black bodies:

A= 1 at all wavelengths at any temperatures (Fig. 3, 1 ), i.e. A black body completely absorbs all radiation incident on it. There are no “absolutely black” bodies in nature; a closed opaque cavity with a small hole can serve as a model for such a body (Fig. 2). The beam that fell into this hole, after multiple reflections from the walls, will be almost completely absorbed.

The sun is close to an absolutely black body, its T = 6000 K.

2). gray bodies: their absorption coefficient A < 1 и одинаков на всех длинах волн при любых температурах (рис. 3, 2 ). For example, a human body can be considered a gray body in problems of heat exchange with the environment.

3). All other bodies:

for them, the absorption coefficient A< 1 и зависит от длины волны, т.е. A l = f(l), this dependence is the absorption spectrum of the body (Fig. 3 , 3 ).

Finally, there is another way to characterize electromagnetic radiation - by specifying its temperature. Strictly speaking, this method is suitable only for the so-called black-body or thermal radiation. In physics, an absolutely black body is an object that absorbs all radiation falling on it. However, ideal absorbing properties do not prevent the body from emitting radiation itself. On the contrary, for such an idealized body it is possible to accurately calculate the shape of the radiation spectrum. This is the so-called Planck curve, the shape of which is determined by a single parameter - temperature. The famous hump of this curve shows that a heated body radiates little at both very long and very short wavelengths. The maximum radiation falls on a well-defined wavelength, the value of which is directly proportional to the temperature.

When specifying this temperature, one must keep in mind that this is not a property of the radiation itself, but only the temperature of an idealized absolutely black body, which has a maximum radiation at a given wavelength. If there is reason to believe that the radiation is emitted by a heated body, then by finding the maximum in its spectrum, one can approximately determine the temperature of the source. For example, the surface temperature of the Sun is 6,000 degrees. This just corresponds to the middle of the visible radiation range. This is hardly accidental - most likely, the eye during evolution has adapted to make the most efficient use of sunlight.

Temperature ambiguity

The point of the spectrum, which accounts for the maximum of black-body radiation, depends on which axis we are plotting on. If the wavelength in meters is evenly plotted along the abscissa axis, then the maximum will fall on

λ max = b/T= (2.9 10 -3 m· TO)/T ,

Where b= 2.9 10 -3 m· TO. This is the so-called Wien's displacement law. If we plot the same spectrum, plotting the radiation frequency uniformly on the y-axis, the location of the maximum is calculated by the formula:

ν max = (α k/h) · T= (5.9 10 10 Hz/TO) · T ,

where α = 2.8, k= 1.4 10 –23 J/TO is the Boltzmann constant, h is Planck's constant.

Everything would be fine, but as it turns out λ max and v max correspond to different points of the spectrum. This becomes obvious if we calculate the wavelength corresponding to ν max, then you get:

λ" max = With/ν max = (ch/α k)/T= (5.1 10 -3 m K) / T .

Thus, the maximum of the spectrum, determined by frequency, in λ" max/ν max = 1,8 times different in wavelength (and hence in frequency) from the maximum of the same spectrum determined from wavelengths. In other words, the frequency and wavelength of the maximum black-body radiation do not correspond to each other: λ max ≠ With/ν max .

In the visible range, it is customary to indicate the maximum of the thermal radiation spectrum along the wavelength. In the spectrum of the Sun, as already mentioned, it falls in the visible range. However, in terms of frequency, the maximum solar radiation lies in the near infrared range.

And here is the maximum of cosmic microwave radiation with a temperature of 2.7 TO it is customary to indicate by frequency - 160 MHz, which corresponds to a wavelength of 1.9 mm. Meanwhile, in the graph by wavelengths, the maximum of the cosmic microwave background falls on 1.1 mm.

All this shows that temperature must be used with great care to describe electromagnetic radiation. It can be used only in the case of radiation close in spectrum to thermal radiation, or for a very rough (up to an order of magnitude) characteristic of the range. For example, visible radiation corresponds to a temperature of thousands of degrees, X-ray - millions, microwave - about 1 kelvin.

Passing the radiation of a body through a device that decomposes it into a spectrum, one can judge the presence of waves of one or another length in the radiation, as well as evaluate the distribution of energy over parts of the spectrum. Such spectra are called emission spectra. It turns out that vapors and gases (especially monatomic ones), when heated or during an electric discharge, give (at low pressures, when the interaction of atoms is practically imperceptible) line spectra, consisting of relatively narrow "lines", i.e., narrow frequency intervals, where the radiation intensity is significant. So, hydrogen gives five lines in the visible part of the spectrum, sodium - one (yellow) line. When high-resolution spectral equipment is used, a complex structure is found in a number of lines. With an increase in pressure, when the interaction of atoms with each other affects, as well as with a complex structure of molecules, broader lines are obtained, turning into relatively broad bands of a complex structure (striped spectra). Such striped spectra, in particular, are observed in liquids. Finally, when heated, solids give practically continuous spectra, but the intensity distribution over the spectrum is different for different bodies.

The spectral composition of the radiation also depends on the temperature of the bodies. The higher the temperature, the more (ceteris paribus) the high frequencies predominate. So, as the temperature of the filament of an incandescent lamp increases, with changes in the current flowing through it, the color of the filament changes: at first, the filament glows faintly with red light, then the visible radiation becomes more intense and short-wavelength - the yellow-green part of the spectrum predominates. But, as will become clear later, in this case, too, most of the radiated energy corresponds to the invisible infrared range.

If radiation with a continuous spectrum is passed through a layer of matter, then partial absorption occurs, as a result of which lines with a minimum intensity are obtained on the continuous spectrum of radiation. In the visible part of the spectrum, they appear in contrast as dark bands (or lines); such spectra are called absorption spectra. Thus, the solar spectrum, cut by a system of thin dark lines (Fraunhofer lines), is an absorption spectrum; it takes place in the atmosphere of the sun.

The study of the spectra shows that with a change in body temperature, not only the emission of light changes, but also its absorption. In this case, it was found that well-radiating bodies also have a large absorption (Prevost), and the absorbed frequencies coincide with those emitted (Kirchhoff). Here, phenomena associated with frequency conversion (luminescence, Compton effect, Raman scattering), which usually play an insignificant role, are not taken into account.

Of particular interest to physicists of the XIX century. caused the radiation of heated bodies. The fact is that during an electric discharge, during certain chemical reactions (chemiluminescence), during ordinary luminescence, a continuous expenditure of energy is required, due to which radiation occurs, i.e., the process is nonequilibrium.

The radiation of a heated body, under certain conditions, can be in equilibrium, since the radiated energy can be absorbed. In the 19th century thermodynamics was developed only for equilibrium processes; therefore, one could hope for the creation of only a theory of the radiation of a heated body.

So, let's imagine a body that has a cavity inside with mirrored (that is, completely reflecting radiation of any frequency) walls. Let two arbitrary bodies are placed in this cavity, giving a continuous spectrum of radiation; their temperature may initially be different. They will exchange radiation energy until an equilibrium state is established: the energy absorbed per unit time by an element of the surface of each body will be equal to the energy emitted by the same element. In this case, the entire cavity will be filled with radiation of various frequencies. According to the Russian physicist B. B. Golitsyn, this radiation should be assigned the same temperature that will be established for radiating bodies after reaching an equilibrium state.

For a quantitative description, we introduce the distribution function e(ν, T), called emissivity body. Work edv, Where dv- an infinitely small frequency interval (near the frequency ν), gives the energy emitted by a unit surface of the body per unit time in the frequency interval (ν, ν+ dv).

Further we will call absorption capacity body function A(ν,T), which determines the ratio of the energy absorbed by an element of the surface of the body to the energy incident on it, contained in the frequency interval (v, ν + dv).

In the same way, one can define reflectivityr(ν , T) as the ratio of the reflected energy in the frequency range (ν, v+dν) to the incident energy.

Idealized mirror walls have a reflectivity equal to unity in the entire frequency range - from the smallest to arbitrarily large.

Let us assume that a state of equilibrium has come, while the first body per unit time radiates power from each unit of the surface

If radiation comes to this single surface from the cavity, Described by the function Ɛ(v, T) dv, then the part of the energy determined by the product a 1 (v, T) Ɛ(v, T) dv, will be absorbed, the rest of the radiation will be reflected. At the same time, a unit surface of the second body radiates power e 2 (v, T) dv, but absorbed a 2 (v, T)Ɛ(v, T) dv.

It follows that at equilibrium the following condition is satisfied:

It can be represented as

(11.1)

This entry emphasizes that the ratio of the emissivity of any body to its absorption capacity at a given temperature in a certain narrow frequency range is a constant value for all bodies. This constant is equal to the emissivity of the so-called black body(i.e., bodies with an absorbance equal to unity in the entire conceivable frequency range).

This black body turns out to be the cavity we are considering. Therefore, if a very small hole is made in the wall of a body with a cavity that does not noticeably disturb thermal equilibrium, then a weak radiation flux from this hole will be characteristic of black body radiation. At the same time, it is clear that radiation entering the cavity through such an opening has a negligibly small probability of escaping back, i.e., the cavity has complete absorption, as it should be for a black body. It can be shown that our reasoning remains valid even when the mirror walls are replaced by walls with a lower reflectivity; instead of two bodies, one can take several or one, or simply consider the radiation of the walls of the cavity itself (if they are not specular). The law expressed by formula (11.1) is called Kirchhoff's law. It follows from Kirchhoff's law that if the function Ɛ(v, T), characterizing the radiation of a black body, then the radiation of any other body could be determined by measuring its absorptive capacity.

Note that a small hole in the wall of, for example, a muffle furnace at room temperature appears black, since, absorbing all the radiation that enters the cavity, the cavity almost does not radiate, being cold. But when the walls of the furnace are heated, the hole seems to glow brightly, since the flux of "black" radiation coming out of it at a high temperature (900 K and above) is quite intense. As the temperature rises, the intensity increases and red radiation is first perceived as yellow, and then as white.

If there is, for example, a cup made of white porcelain with a dark pattern in the cavity, then the pattern will not be noticeable inside the hot furnace, since its own radiation, together with the reflected one, coincides in composition with the radiation filling the cavity. If you quickly take the cup outside, into a bright room, then at first the dark pattern glows brighter than the white background. After cooling, when the cup's own radiation becomes vanishingly small, the room-filling light again produces a dark pattern on a white background.