.

EMISSION AND ABSORPTION OF ENERGY

ATOMS AND MOLECULES

QUESTIONS FOR THE CLASS ON THE TOPIC:

1. Thermal radiation. Its main characteristics: radiation flux Ф, energy luminosity (intensity) R, spectral density of energy luminosity r λ; absorption coefficient α, monochromatic absorption coefficient α λ. Absolutely black body. Kirchhoff's law.

2. Thermal radiation spectra of a.ch.t. (schedule). The quantum nature of thermal radiation (Planck’s hypothesis; there is no need to remember the formula for ε λ). Dependence of the spectrum of a.ch.t. on temperature (graph). Wine's Law. Stefan-Boltzmann law for a.ch.t. (without output) and for other bodies.

3. The structure of the electronic shells of atoms. Energy levels. Emission of energy during transitions between energy levels. Bohr's formula ( for frequency and for wavelength). Spectra of atoms. Spectrum of a hydrogen atom. Spectral series. General concept of the spectra of molecules and condensed matter (liquids, solids). The concept of spectral analysis and its use in medicine.

4. Luminescence. Types of luminescence. Fluorescence and phosphorescence. The role of metastable levels. Luminescence spectra. Stokes rule. Luminescent analysis and its use in medicine.

5. Law of light absorption (Bouguer’s law; conclusion). Transmittance τ and optical density D. Determination of the concentration of solutions by light absorption.

Laboratory work: “recording the absorption spectrum and determining the concentration of the solution using a photoelectrocolorimeter.”

LITERATURE:

Mandatory: A.N. Remizov. “Medical and biological physics”, M., “Higher School”, 1996, ch. 27, §§ 1–3; Chapter 29, §§ 1,2

• additional: Emission and absorption of energy by atoms and molecules, lecture, risograph, ed. department, 2002

BASIC DEFINITIONS AND FORMULAS

1. Thermal radiation

All bodies, even without any external influence, emit electromagnetic waves. The source of energy for this radiation is the thermal motion of the particles that make up the body, which is why it is called thermal radiation. At high temperatures (about 1000 K or more), this radiation falls partially in the range of visible light; at lower temperatures, infrared rays are emitted, and at very low temperatures, radio waves are emitted.

Radiation flux F - This radiation power emitted by the source, or radiation energy emitted per unit time: Ф = Р = ; flow unit - watt.

Energetic luminosity R - This flux of radiation that is emitted from a unit surface of the body: ; unit of energetic luminosity – W.m –2 .

Spectral density of energetic luminosity r λ - This the ratio of the energetic luminosity of a body within a small wavelength interval (ΔR λ ) to the value of this interval Δ λ:

Dimension r λ – W.m - 3

Absolutely black body (a.b.t.) called t ate whichfully absorbs incident radiation. There are no such bodies in nature, but a good model of an a.ch.t. is a small hole in a closed cavity.

The ability of bodies to absorb incident radiation characterizes absorption coefficient α , that is ratio of absorbed to incident radiation flux: .

Monochromatic absorption coefficient is the value of the absorption coefficient measured in a narrow spectral range around a certain value λ.

Kirchhoff's law: at constant temperature, the ratio of the spectral density of energetic luminosity at a certain wavelength to the monochromatic absorption coefficient at the same wavelength the same for all bodies and is equal to the spectral density of the energy luminosity of the a.b.t. at this wavelength:

(sometimes r λ A.Ch.T denote ε λ)

A completely black body absorbs and emits radiation all wavelengths, That's why spectrum of a.h.t. always solid. Type of this spectrum depends on body temperature. As the temperature rises, firstly, the energetic luminosity increases significantly; Secondly, wavelength corresponding to maximum radiation(λ max ) , shifts towards shorter wavelengths :, where b ≈ 29090 µm.K -1 ( Wien's law).

Stefan-Boltzmann law: energetic luminosity of a.h.t. proportional to the fourth power of body temperature on the Kelvin scale: R = σT 4

2. Emission of energy by atoms and molecules

As is known, in the electron shell of an atom, the energy of an electron can only take on strictly defined values ​​characteristic of a given atom. In other words they say that electron can only be located on certainenergy levels. When an electron is at a given energy level, it does not change its energy, that is, it does not absorb or emit light. When moving from one level to another the energy of the electron changes, and at the same time absorbed or emittedquantum of light (photon).The energy of a quantum is equal to the difference in the energies of the levels between which the transition occurs: E QUANTUM = hν = E n – E m where n and m are level numbers (Bohr formula).

Electron transitions between different levelsoccur with different probabilities. In some cases, the transition probability is very close to zero; the corresponding spectral lines are not observed under normal conditions. Such transitions are called prohibited.

In many cases, the energy of an electron may not be converted into quantum energy, but rather converted into the energy of thermal motion of atoms or molecules. Such transitions are called non-radiative.

In addition to the transition probability, the brightness of spectral lines is directly proportional to the number of atoms of the emitting substance. This dependence underlies quantitative spectral analysis.
3. Luminescence

Luminescence call any not thermal radiation. The energy sources for this radiation can be different; accordingly, they speak of different types of luminescence. The most important of them are: chemiluminescence– glow that occurs during certain chemical reactions; bioluminescence– this is chemiluminescence in living organisms; cathodoluminescence – glow under the influence of a flow of electrons, which is used in television picture tubes, cathode ray tubes, gas light lamps, etc.; electroluminescence– glow that occurs in an electric field (most often in semiconductors). The most interesting type of luminescence is photoluminescence. This is a process in which atoms or molecules absorb light (or UV radiation) in one wavelength range and emit it in another (for example, they absorb blue rays and emit yellow ones). In this case, the substance absorbs quanta with relatively high energy hν 0 (with a short wavelength). Then the electron may not immediately return to the ground level, but first go to the intermediate level, and then to the ground level (there may be several intermediate levels). In most cases, some transitions are non-radiative, that is, the electron energy is converted into the energy of thermal motion. Therefore, the energy of quanta emitted during luminescence will be less than the energy of the absorbed quantum. The wavelengths of the emitted light must be greater than the wavelength of the absorbed light. If we formulate the above in general form, we get law Stokes : the luminescence spectrum is shifted towards longer waves relative to the spectrum of radiation causing luminescence.

There are two types of luminescent substances. In some, the glow stops almost instantly after the exciting light is turned off. This short-term the glow is called fluorescence.

In substances of another type, after turning off the exciting light, the glow fades gradually(according to exponential law). This long-term the glow is called phosphorescence. The reason for the long glow is that the atoms or molecules of such substances contain metastable levels.Metastable This energy level is called in which electrons can remain much longer than at normal levels. Therefore, the duration of phosphorescence can be minutes, hours and even days.
4. Law of light absorption (Bouguer's law)

When a radiation flux passes through a substance, it loses part of its energy (the absorbed energy turns into heat). The law of light absorption is called Bouguer's law: Ф = Ф 0 ∙ e – κ λ · L ,

where Ф 0 is the incident flow, Ф is the flow passing through a layer of substance with thickness L; the coefficient κ λ is called natural absorption rate ( its magnitude depends on the wavelength) . For practical calculations, they prefer to use decimal logarithms instead of natural logarithms. Then Bouguer’s law takes the form: Ф = Ф 0 ∙ 10 – k λ ∙ L ,

where kλ – decimal absorption rate.

Transmittance name the quantity

Optical density D - this is the quantity defined by the equality: . We can say it another way: optical density D is a quantity that is in the exponent in the formula of Bouguer’s law: D = k λ ∙ L

For solutions of most substances optical density is directly proportional to the concentration of the solute:D = χ λ ∙ C∙ L ;

coefficient χ λ is called molar absorption rate(if the concentration is given in moles) or specific absorption rate(if the concentration is indicated in grams). From the last formula we get: Ф = Ф 0 ∙10 - χ λ ∙ C ∙ L(law Bugera–Bera)

These formulas underlie the most common in clinical and biochemical laboratories method for determining the concentrations of dissolved substances by light absorption.

TEACHING TYPE PROBLEMS WITH SOLUTIONS

(In the future, for brevity, we will simply write “training tasks”)

Learning Objective #1

An electric heater (radiator) emits a stream of infrared rays of 500 W. The surface area of ​​the radiator is 3300 cm2. Find the energy emitted by the radiator in 1 hour and the energetic luminosity of the radiator.

Given: Find

Ф = 500 W W and R

t = 1 hour = 3600 s

S = 3300 cm 2 = 0.33 m 2

Solution:

Radiation flux Ф is the radiation power or energy emitted per unit time: . From here

W = F t = 500 W 3600 s = 18 10 5 J = 1800 kJ

Learning Objective #2

At what wavelength is the thermal radiation of human skin maximum (that is, r λ = max)? The skin temperature on exposed parts of the body (face, hands) is approximately 30 o C.

Given: Find:

Т = 30 о С = 303 К λ max

Solution:

We substitute the data into the Wien formula: ,

that is, almost all the radiation lies in the IR range of the spectrum.

Learning Objective #3

The electron is at an energy level with an energy of 4.7.10 –19 J

When irradiated with light with a wavelength of 600 nm, it moved to a higher energy level. Find the energy of this level.

Solution:

Learning Objective #4

The decimal water absorption rate for sunlight is 0.09 m–1. What fraction of the radiation will reach the depth L = 100 m?

Given Find:

k = 0.09 m – 1

Solution:

Let's write down Bouguer's law: . The fraction of radiation reaching depth L is, obviously,

that is, one billionth of sunlight will reach a depth of 100 m.
Learning Objective #5

Light passes sequentially through two filters. The first has an optical density D 1 = 0.6; the second has D 2 = 0.4. What percentage of the radiation flux will pass through this system?

Given: Find:

D 1 = 0.6 (in %%)

Solution:

We start the solution with a drawing of this system

SF-1 SF-2

Find Ф 1: Ф 1 = Ф 0 10 – D 1

Similarly, the flux passing through the second light filter is equal to:

Ф 2 = Ф 1 10 – D 2 = Ф 0 10 – D 1 10 – D 2 = Ф 0 10 – (D 1 + D 2)

The result obtained has general significance: if light passes sequentially through a system of several objects,the total optical density will be equal to the sum of the optical densities of these objects .

Under the conditions of our problem, a flow of F 2 = 100%∙10 – (0.6 + 0.4) = 100%∙10 – 1 = 10% will pass through a system of two light filters

Learning Objective #6

According to the Bouguer-Baer law, it is possible, in particular, to determine the concentration of DNA. In the visible region, solutions of nucleic acids are transparent, but they strongly absorb in the UV part of the spectrum; The absorption maximum lies around 260 nm. It is obvious that it is precisely in this region of the spectrum that the absorption of radiation must be measured; in this case, the sensitivity and accuracy of measurement will be the best.

Conditions of the problem: When measuring the absorption of UV rays with a wavelength of 260 nm by a DNA solution, the transmitted radiation flux was attenuated by 15%. The path length of the beam in the cuvette with solution “x” is 2 cm. The molar absorption index (decimal) for DNA at a wavelength of 260 nm is 1.3.10 5 mol – 1.cm 2 Find the concentration of DNA in the solution.

Given:

Ф 0 = 100%; F = 100% – 15% = 85% Find: With DNA

x = 2 cm; λ = 260 nm

χ 260 = 1.3.10 5 mol –1 .cm 2

Solution:

(we “flipped” the fraction to get rid of the negative exponent). . Now let's logarithm: , and ; we substitute:

0.07 and C = 2.7.10 – 7 mol/cm3

Pay attention to the high sensitivity of the method!

TASKS FOR INDEPENDENT SOLUTION
When solving problems, take the values ​​of the constants:

b = 2900 µm.K; σ = 5.7.10 – 8 W.K 4; h = 6.6.10 – 34 J.s; c = 3.10 8 m.s –1

1. What is the energetic luminosity of the surface of the human body if the maximum radiation occurs at a wavelength of 9.67 microns? The skin can be considered an absolutely black body.

2. Two light bulbs have exactly the same design, except that in one the filament is made of pure tungsten (α = 0.3), and in the other it is coated with platinum black (α = 0.93). Which light bulb has more radiation flux? How many times?

3. In what areas of the spectrum do the wavelengths corresponding to the maximum spectral density of energy luminosity lie if the source of radiation is: a) the spiral of an electric light bulb (T = 2,300 K); b) the surface of the Sun (T = 5,800 K); c) the surface of the fireball of a nuclear explosion at the moment when its temperature is about 30,000 K? The difference in the properties of these radiation sources from the a.ch.t. neglect.

4. A red-hot metal body, the surface of which is 2.10 - 3 m 2, at a surface temperature of 1000 K emits a flux of 45.6. Tue What is the absorption coefficient of the surface of this body?

5. The light bulb has a power of 100 W. The surface area of ​​the filament is 0.5.10 - 4 m 2. The temperature of the filament is 2,400 K. What is the absorption coefficient of the filament surface?

6. At a skin temperature of 27 0 C, 0.454 W is emitted from each square centimeter of the body surface. Is it possible (with an accuracy of no worse than 2%) to consider the skin to be an absolutely black body?

7. In the spectrum of a blue star, the maximum emission corresponds to a wavelength of 0.3 microns. What is the surface temperature of this star?

8. What energy does a body with a surface of 4,000 cm 2 radiate in one hour?

at a temperature of 400 K, if the absorption coefficient of the body is 0.6?

9. Plate (A) has a surface area of ​​400 cm 2 ; its absorption coefficient is 0.4. Another plate (B) with an area of ​​200 cm 2 has an absorption coefficient of 0.2. The temperature of the plates is the same. Which plate emits more energy and by how much?

10 – 16. Qualitative spectral analysis. Based on the absorption spectrum of one of the organic compounds, the spectra of which

are shown in the figure, determine which functional groups are part of this substance, Use the table data:

Group; connection type	Absorbed wavelengths, microns	Group, connection type	Absorbed wavelengths, µm
-HE	2,66 – 2,98	-NH 4	7,0 – 7,4
-NH	2,94 – 3,0	-SH	7,76
 CH	3,3	-CF	8,3
-N  N	4,67	-NH 2	8,9
-C=N	5,94	-NO	12,3
-N=N	6,35	-SO 2	19,2
-CN 2	6,77	-C=O	23,9

10 – graph a); 11 – graph b); 12 – graph c); 13 – graph d);

14 – graph d); 15 – graph f); 16 – graph g).

Pay attention to what value on your graph is plotted on the vertical axis!

17. Light passes sequentially through two light filters with transmittance coefficients of 0.2 and 0.5. What percentage of radiation will come out of such a system?

18. Light passes sequentially through two filters with optical densities of 0.7 and 0.4. What percentage of radiation will pass through such a system?

19. To protect against the light radiation of a nuclear explosion, you need glasses that attenuate the light by at least a million times. The glass from which they want to make such glasses has an optical density of 3 with a thickness of 1 mm. What thickness of glass should be taken to achieve the required result?

20 To protect the eyes when working with a laser, it is required that a radiation flux not exceeding 0.0001% of the flux generated by the laser can enter the eye. What optical density should glasses have to ensure safety?

General assignment for problems 21 – 28 (quantitative analysis):

The figure shows the absorption spectra of colored solutions of some substances. In addition, the problems indicate the values ​​of D (the optical density of the solution at the wavelength corresponding to the maximum absorption of light) and X(cuvette thickness). Find the concentration of the solution.

Pay attention to the units in which the absorption rate is indicated on your graph.

21. Graph a). D = 0.8 x = 2 cm

22. Graph b). D = 1.2 x = 1 cm

... 23. Graph c). D = 0.5 x = 4 cm

24. Graph d). D = 0.25 x = 2 cm

25 Schedule d). D = 0.4 x = 3 cm

26. Graph e) D = 0.9 x = 1 cm

27. Graph g). D = 0.2 x = 2 cm

§ 4 Energy luminosity. Stefan-Boltzmann law.

Wien's displacement law

RE(integrated energy luminosity) - energy luminosity determines the amount of energy emitted from a unit surface per unit time over the entire frequency range from 0 to ∞ at a given temperature T.

Connection energetic luminosity and emissivity

[ R E ] = J/(m 2 s) = W/m 2

Law of J. Stefan (Austrian scientist) and L. Boltzmann (German scientist)

Where

σ = 5.67·10 -8 W/(m 2 · K 4) - Steph-on-Boltzmann constant.

The energetic luminosity of a black body is proportional to the fourth power of thermodynamic temperature.

Stefan-Boltzmann law, defining the dependenceREon temperature does not provide an answer regarding the spectral composition of black body radiation. From experimental dependence curvesrλ ,T from λ at different T it follows that the energy distribution in the spectrum of an absolutely black body is uneven. All curves have a maximum, which, with increasing T shifts towards shorter wavelengths. Area limited by the dependence curverλ ,T from λ, is equal RE(this follows from the geometric meaning of the integral) and is proportional T 4 .

Wien's displacement law (1864 - 1928): Length, waves (λ max), which accounts for the maximum emissivity of the a.ch.t. at a given temperature, inversely proportional to temperature T.

b= 2.9·10 -3 m·K - Wien's constant.

The Wien shift occurs because as temperature increases, the maximum emissivity shifts toward shorter wavelengths.

§ 5 Rayleigh-Jeans formula, Wien formula and ultraviolet catastrophe

The Stefan-Boltzmann law allows us to determine the energetic luminosityREa.ch.t. according to its temperature. Wien's displacement law relates body temperature to the wavelength at which maximum emissivity occurs. But neither one nor the other law solves the main problem of how great the radiation emission ability is for each λ in the spectrum of the a.ch.t. at a temperature T. To do this, you need to establish a functional dependencerλ ,T from λ and T.

Based on the idea of ​​the continuous nature of the emission of electromagnetic waves in the law of uniform distribution of energies over degrees of freedom, two formulas were obtained for the emissivity of the AC:

• Wine formula

Where A, b = const.

• Rayleigh-Jeans formula

k =1.38·10 -23 J/K - Boltzmann's constant.

Experimental testing has shown that for a given temperature, Wien's formula is correct for short waves and gives sharp discrepancies with experiment in the region of long waves. The Rayleigh-Jeans formula turned out to be true for long waves and not applicable for short ones.

The study of thermal radiation using the Rayleigh-Jeans formula showed that, within the framework of classical physics, it is impossible to solve the question of the function characterizing the emissivity of the AC. This unsuccessful attempt to explain the laws of radiation of a.ch.t. Using the apparatus of classical physics, it was called the “ultraviolet catastrophe.”

If you try to calculateREusing the Rayleigh-Jeans formula, then

• “ ultraviolet disaster”

§6 Quantum hypothesis and Planck's formula.

In 1900, M. Planck (a German scientist) put forward a hypothesis according to which the emission and absorption of energy does not occur continuously, but in certain small portions - quanta, and the energy of a quantum is proportional to the frequency of oscillations (Planck's formula):

h = 6.625·10 -34 J·s - Planck’s constant or

Where

Since radiation occurs in portions, the energy of the oscillator (oscillating atom, electron) E takes only values ​​that are multiples of an integer number of elementary portions of energy, that is, only discrete values

E = n E o = nhν .

PHOTOELECTRIC EFFECT

The influence of light on the course of electrical processes was first studied by Hertz in 1887. He conducted experiments with an electric discharger and discovered that when irradiated with ultraviolet radiation, the discharge occurs at a significantly lower voltage.

In 1889-1895. A.G. Stoletov studied the effect of light on metals using the following scheme. Two electrodes: cathode K made of the metal under study and anode A (in Stoletov’s scheme - a metal mesh that transmits light) in a vacuum tube are connected to the battery so that with the help of resistance R you can change the value and sign of the voltage applied to them. When the zinc cathode was irradiated, a current flowed in the circuit, recorded by a milliammeter. By irradiating the cathode with light of various wavelengths, Stoletov established the following basic principles:

• Ultraviolet radiation has the most powerful effect;
• When exposed to light, negative charges are released from the cathode;
• The strength of the current generated by light is directly proportional to its intensity.

Lenard and Thomson in 1898 measured the specific charge ( e/ m), particles being torn out, and it turned out that it is equal to the specific charge of an electron, therefore, electrons are ejected from the cathode.

§ 2 External photoelectric effect. Three laws of external photoelectric effect

The external photoelectric effect is the emission of electrons by a substance under the influence of light. Electrons emitted from a substance during the external photoelectric effect are called photoelectrons, and the current they generate is called photocurrent.

Using Stoletov’s scheme, the following dependence of the photocurrent onapplied voltage at a constant luminous flux F(that is, the current-voltage characteristic was obtained):

At some voltageUNphotocurrent reaches saturationI n - all electrons emitted by the cathode reach the anode, hence the saturation currentI n determined by the number of electrons emitted by the cathode per unit time under the influence of light. The number of released photoelectrons is proportional to the number of light quanta incident on the cathode surface. And the number of light quanta is determined by the luminous flux F, incident on the cathode. Number of photonsN, falling over timet to the surface is determined by the formula:

Where W- radiation energy received by the surface during time Δt,

Photon energy,

F e -luminous flux (radiation power).

1st law of external photoelectric effect (Stoletov’s law):

At a fixed frequency of incident light, the saturation photocurrent is proportional to the incident light flux:

Ius~ Ф, ν =const

Uh - holding voltage- the voltage at which not a single electron can reach the anode. Consequently, the law of conservation of energy in this case can be written: the energy of the emitted electrons is equal to the stopping energy of the electric field

therefore, we can find the maximum speed of emitted photoelectronsVmax

2nd law of photoelectric effect : maximum initial speedVmaxphoto-electrons does not depend on the intensity of the incident light (from F), and is determined only by its frequency ν

3rd law of photoelectric effect : for each substance there is "red border" photo effect, that is, the minimum frequency ν kp, depending on the chemical nature of the substance and the state of its surface, at which the external photoelectric effect is still possible.

The second and third laws of the photoelectric effect cannot be explained using the wave nature of light (or the classical electromagnetic theory of light). According to this theory, the ejection of conduction electrons from a metal is the result of their “swinging” by the electromagnetic field of a light wave. With increasing light intensity ( F) the energy transferred by the electron of the metal must increase, therefore, it must increaseVmax, and this contradicts the 2nd law of the photoelectric effect.

Since, according to the wave theory, the energy transmitted by the electromagnetic field is proportional to the intensity of light ( F), then any light; frequency, but with a sufficiently high intensity, it would have to pull electrons out of the metal, that is, the red limit of the photoelectric effect would not exist, which contradicts the 3rd law of the photoelectric effect. The external photoelectric effect is inertialess. But the wave theory cannot explain its inertialessness.

§ 3 Einstein's equation for the external photoelectric effect.

Work function

In 1905, A. Einstein explained the photoelectric effect based on quantum concepts. According to Einstein, light is not only emitted by quanta in accordance with Planck's hypothesis, but spreads in space and is absorbed by matter in separate portions - quanta with energy E 0 = hv. Quanta of electromagnetic radiation are called photons.

Einstein's equation (law of conservation of energy for external photo-effect):

Incident photon energy hv is spent on ejecting an electron from the metal, that is, on the work function And out, and to communicate kinetic energy to the emitted photoelectron.

The minimum energy that must be imparted to an electron in order to remove it from a solid into a vacuum is called work function.

Since the Ferm energy to E Fdepends on temperature and E F, also changes with temperature changes, then, consequently, And out depends on temperature.

In addition, the work function is very sensitive to surface cleanliness. Applying a film to the surface ( Ca, SG, Va) on WAnd outdecreases from 4.5 eV for pureW up to 1.5 ÷ 2 eV for impurityW.

Einstein's equation allows us to explain in c e three laws of external photoeffect,

1st law: each quantum is absorbed by only one electron. Therefore, the number of ejected photoelectrons should be proportional to the intensity ( F) Sveta

2nd law: Vmax~ ν, etc. And out does not depend on F, thenVmax does not depend on F

3rd law: As ν decreases, it decreasesVmax and for ν = ν 0 Vmax = 0, therefore,hν 0 = And out, therefore, i.e. There is a minimum frequency from which the external photoelectric effect is possible.

Thermal radiation of bodies is electromagnetic radiation arising from that part of the internal energy of the body, which is associated with the thermal motion of its particles.

The main characteristics of thermal radiation of bodies heated to a temperature T are:

1. Energy luminosityR (T ) -the amount of energy emitted per unit time from a unit surface of a body, over the entire wavelength range. Depends on the temperature, nature and condition of the surface of the radiating body. In the SI system R ( T ) has a dimension [W/m2].

2. Spectral density of energetic luminosityr (  ,T) =dW/ d - the amount of energy emitted by a unit surface of a body per unit time in a unit wavelength interval (near the wavelength in question ). Those. this quantity is numerically equal to the energy ratio dW, emitted from a unit area per unit time in a narrow range of wavelengths from  before  +d , to the width of this interval. It depends on the body temperature, wavelength, and also on the nature and condition of the surface of the emitting body. In the SI system r( , T) has a dimension [W/m 3 ].

Energetic luminosity R(T) related to the spectral density of energetic luminosity r( , T) in the following way:

(1) [W/m2]

3. All bodies not only emit, but also absorb electromagnetic waves incident on their surface. To determine the absorption capacity of bodies in relation to electromagnetic waves of a certain wavelength, the concept is introduced monochromatic absorption coefficient-the ratio of the magnitude of the energy of a monochromatic wave absorbed by the surface of a body to the magnitude of the energy of the incident monochromatic wave:

The monochromatic absorption coefficient is a dimensionless quantity that depends on temperature and wavelength. It shows what fraction of the energy of an incident monochromatic wave is absorbed by the surface of the body. Value  ( , T) can take values ​​from 0 to 1.

Radiation in an adiabatically closed system (not exchanging heat with the external environment) is called equilibrium. If you create a small hole in the wall of the cavity, the equilibrium state will change slightly and the radiation emerging from the cavity will correspond to the equilibrium radiation.

If a beam is directed into such a hole, then after repeated reflections and absorption on the walls of the cavity, it will not be able to come back out. This means that for such a hole the absorption coefficient ( , T) = 1.

The considered closed cavity with a small hole serves as one of the models absolutely black body.

Absolutely black bodyis a body that absorbs all radiation incident on it, regardless of the direction of the incident radiation, its spectral composition and polarization (without reflecting or transmitting anything).

For a completely black body, the spectral luminosity density is some universal function of wavelength and temperature f( , T) and does not depend on its nature.

All bodies in nature partially reflect radiation incident on their surface and therefore are not classified as absolute black bodies. If the monochromatic absorption coefficient of a body is the same for all wavelengths and lessunits(( , T) = Т =const<1),then such a body is called gray. The monochromatic absorption coefficient of a gray body depends only on the temperature of the body, its nature and the state of its surface.

Kirchhoff showed that for all bodies, regardless of their nature, the ratio of the spectral density of energy luminosity to the monochromatic absorption coefficient is the same universal function of wavelength and temperature f( , T) , the same as the spectral density of the energy luminosity of a completely black body :

Equation (3) represents Kirchhoff's law.

Kirchhoff's law can be formulated this way: for all bodies of the system that are in thermodynamic equilibrium, the ratio of the spectral density of energy luminosity to the coefficient monochromatic absorption does not depend on the nature of the body, is the same function for all bodies, depending on the wavelength and temperature T.

From the above and formula (3) it is clear that at a given temperature those gray bodies that have a large absorption coefficient emit more strongly, and absolutely black bodies emit the most strongly. Since for an absolutely black body(  , T)=1, then from formula (3) it follows that the universal function f( , T) represents the spectral luminosity density of a black body

Examples of problem solving. Example 1. The maximum spectral density of the solar energy luminosity occurs at wavelength = 0.48 microns

Example 1. The maximum spectral density of the solar energy luminosity occurs at wavelength = 0.48 microns. Assuming that the Sun radiates as a black body, determine: 1) the temperature of its surface; 2) the power emitted by its surface.

According to Wien's displacement law, the desired temperature of the solar surface is:

where b= is Wien's constant.

Power emitted by the surface of the Sun:

where is the energetic luminosity of the black body (Sun), is the surface area of ​​the Sun, is the radius of the Sun.

According to the Stefan-Boltzmann law:

where = W/ is the Stefan-Boltzmann constant.

Let's substitute the written expressions into formula (2) and find the required power emitted by the surface of the Sun:

Calculating, we get: T=6.04 kK; P=W.

Example 2. Determine the wavelength, mass and momentum of a photon with energy = 1 MeV.

The photon energy is related to the wavelength of light by the relation: ,

where h is Planck’s constant, c is the speed of light in vacuum. From here.

Substituting the numerical values, we get: m.

Let's determine the photon mass using Einstein's formula. Photon mass = kg.

Photon momentum = kg m/s.

Example 3. The sodium cathode of a vacuum photocell is illuminated with monochromatic light with a wavelength of 40 nm. Determine the delay voltage at which the photocurrent stops. The “red limit” of the photoelectric effect for sodium = 584 nm.

The electric field that prevents the movement of electrons from the cathode to the anode is called reverse. The voltage at which the photocurrent stops completely is called the retardation voltage. With such a retarding voltage, none of the electrons, even those with maximum speed when leaving the cathode, can overcome the retarding field and reach the anode. In this case, the initial kinetic energy of photoelectrons () transforms into potential energy (, where e = C is the elementary charge, and is the lowest retarding voltage). According to the law of conservation of energy

We find the kinetic energy of electrons using Einstein’s equation for the external photoelectric effect:

From here (3)

The electron work function A in is determined by the red boundary of the photoelectric effect:

Substituting expression (4) into equation (3), we obtain:

Then, from equation (1).

Calculating, we get V.

Example 4. The kinetic energy of a proton is four times less than its rest energy. Calculate the de Broglie wavelength for the proton.

The de Broglie wavelength is determined by the formula: , (1)

where h is Planck’s constant and is the momentum of the particle.

According to the conditions of the problem, the kinetic energy of a proton is comparable in magnitude to its rest energy E 0 . Consequently, momentum and kinetic energy are related to each other by a relativistic relationship:

where c is the speed of light in vacuum.

Using the condition of the problem, we obtain: . Substituting the resulting expression into formula (1), we find the de Broglie wavelength:

We will find the rest energy of the electron using Einstein's formula, where m 0 is the rest mass of the electron, c is the speed of light in vacuum.

Substituting the numerical values, we get: m.

Example 5. The electron beam is accelerated in a cathode ray tube by a potential difference U=0.5 kV. Assuming that the uncertainty of the electron momentum is 0.1% of its numerical value, determine the uncertainty of the electron coordinate. Under these conditions, is the electron a quantum or classical particle?

In the direction of motion of the electron beam (X axis), the uncertainty relation has the form:

where is the uncertainty of the electron coordinate; - uncertainty of its impulse; - Planck's constant.

Having passed through the accelerating potential difference, the electron acquires kinetic energy equal to the work done by the electric field forces:

The calculation gives the value E k = 500 eV, which is much less than the rest energy of the electron (E 0 = 0.51 MeV). Consequently, under these conditions, the electron is a non-relativistic particle with momentum related to kinetic energy by the formula .

According to the conditions of the problem, the uncertainty of the impulse = 0.001 = , i.e.<< .

This means that the wave properties under these conditions are insignificant and the electron can be considered as a classical particle. From expression (1) it follows that the desired uncertainty of the electron coordinate

Having calculated, we get 8.51 nm.

Example 6. As a result of the transition from one stationary state to another, the hydrogen atom emitted a quantum with a frequency of . Find how the radius of the orbit and the speed of the electron have changed using Bohr's theory.

Radiation with a frequency corresponding to a wavelength = = 102.6 nm (c is the speed of light in vacuum), lying in the ultraviolet region. Consequently, the spectral line belongs to the Lyman series, which appears when an electron transitions to the first energy level (n=1).

We use the generalized Balmer formula to determine the number of the energy level (k) from which the transition was made: .

Let us express k from this formula:

Substituting the available data, we get k=3. Consequently, the radiation occurred as a result of the transition of an electron from the third orbit to the first.

We will find the values ​​of the radii of orbits and the velocities of electrons in these orbits from the following considerations.

An electron located in a stationary orbit in a hydrogen atom is acted upon by the Coulomb force from the nucleus.

which gives it normal acceleration. Therefore, according to the basic law of dynamics:

In addition, according to Bohr's postulate, the angular momentum of an electron in a stationary orbit must be a multiple of Planck's constant, i.e.

where n = 1, 2, 3…. – number of stationary orbit.

From equation (2) speed . Substituting this expression into equation (1), we obtain

Hence the radius of the stationary orbit of an electron in a hydrogen atom: .

Then the speed of the electron in this orbit is:

Assuming that before the radiation of the quantum the electron had the characteristics r 3, v 3, and after the radiation r 1, v 1 it is easy to obtain:

that is, the radius of the orbit decreased by 9 times, the speed of the electron increased by 3 times.

Example 7. An electron in a one-dimensional rectangular “potential well” with a width of =200 pm with infinitely high “walls” is in an excited state (n=2). Determine: 1) the probability W of detecting an electron in the middle third of the “well”; 2) points of the specified interval at which the probability density of detecting an electron is maximum and minimum.

1. Probability of detecting a particle in the interval

The excited state (n=2) corresponds to its own wave function:

Let’s substitute (2) into (1) and take into account that and:

Expressing through the cosine of the double angle using the trigonometric equality, we obtain the expression for the desired probability: = = = = = 0.195.

2. The probability density of the existence of a particle in a certain region of space is determined by the square of the modulus of its wave function. Using expression (2), we get:

The dependence of the squared modulus of the wave function of a particle on its coordinate, determined by expression (3), is shown in the figure.

Obviously, the minimum probability density w=0 corresponds to the values ​​of x for which .

That is, ,

where k = 0, 1, 2…

The probability density w reaches its maximum value within the well under the condition: . Corresponding values.

As can be seen from the graph of w= w(x) shown in the figure, in the interval

As we can see, the probability density of detecting an electron at the boundaries of a given interval is the same. Hence, , .

Example 8. Determine the amount of heat required to heat a NaCl crystal weighing m = 20 g at a temperature T 1 = 2 K. The characteristic Debye temperature for NaCl is taken equal to 320K.

The amount of heat required to heat a body of mass m from temperature T 1 to temperature T 2 can be calculated by the formula:

where C is the molar heat capacity of the substance, M is the molar mass.

According to Debye's theory, at temperature the molar heat capacity of crystalline solids is given by:

Substituting expression (2) into (1) and integrating, we get:

Substituting numerical values ​​and performing calculations, we find Q = 1.22 mJ.

Example 9. Calculate the mass defect, binding energy, and specific binding energy of the nucleus.

The core mass defect is determined by the formula:

For the core: Z=5; A=11.

We will calculate the mass defect in non-systemic units – atomic mass units (a.m.u.). We take the necessary data from the table (Appendix 3):

1.00783 a.m.u., =1.00867 a.m.u., = 11.00931 a.m.u.

As a result of the calculation using formula (1), we obtain: =0.08186 a.m.u.

We will also find the nuclear binding energy in extrasystemic units (MeV), using the formula:

Proportionality coefficient = 931.4 MeV/amu, i.e.

After substituting the numerical values ​​we get:

Specific binding energy, by definition, is equal to:

Determine the atomic number and mass number of the second nucleus, give a symbolic notation of the nuclear reaction and determine its energy effect.

THERMAL RADIATION Stefan Boltzmann's law Relationship between the energy luminosity R e and the spectral density of the energy luminosity of a black body Energy luminosity of a gray body Wien's displacement law (1st law) Dependence of the maximum spectral density of the energy luminosity of a black body on temperature (2nd law) Planck's formula

THERMAL RADIATION 1. The maximum spectral density of the solar energy luminosity occurs at wavelength = 0.48 microns. Assuming that the Sun radiates as a black body, determine: 1) the temperature of its surface; 2) the power emitted by its surface. According to Wien's displacement law, Power emitted by the surface of the Sun According to Stefan Boltzmann's law,

THERMAL RADIATION 2. Determine the amount of heat lost by 50 cm 2 from the surface of molten platinum in 1 minute, if the absorption capacity of platinum A T = 0.8. The melting point of platinum is 1770 °C. The amount of heat lost by platinum is equal to the energy emitted by its hot surface. According to Stefan Boltzmann's law,

THERMAL RADIATION 3. An electric furnace consumes power P = 500 W. The temperature of its inner surface with an open small hole with a diameter of d = 5.0 cm is 700 °C. How much of the power consumption is dissipated by the walls? The total power is determined by the sum of the Power released through the hole Power dissipated by the walls According to Stefan Boltzmann's law,

THERMAL RADIATION 4 A tungsten filament is heated in a vacuum with a current of force I = 1 A to a temperature T 1 = 1000 K. At what current strength will the filament be heated to a temperature T 2 = 3000 K? The absorption coefficients of tungsten and its resistivity corresponding to temperatures T 1, T 2 are equal to: a 1 = 0.115 and a 2 = 0.334; 1 = 25, Ohm m, 2 = 96, Ohm m The power emitted is equal to the power consumed from the electrical circuit in steady state Electric power released in the conductor According to Stefan Boltzmann's law,

THERMAL RADIATION 5. In the spectrum of the Sun, the maximum spectral density of energy luminosity occurs at a wavelength of .0 = 0.47 microns. Assuming that the Sun emits as a completely black body, find the intensity of solar radiation (i.e., radiation flux density) near the Earth outside its atmosphere. Luminous intensity (radiation intensity) Luminous flux According to the laws of Stefan Boltzmann and Wien

THERMAL RADIATION 6. Wavelength 0, which accounts for the maximum energy in the black body radiation spectrum, is 0.58 microns. Determine the maximum spectral density of energy luminosity (r, T) max, calculated for the wavelength interval = 1 nm, near 0. The maximum spectral density of energy luminosity is proportional to the fifth power of temperature and is expressed by Wien’s 2nd law. Temperature T is expressed from Wien’s displacement law value C is given in SI units, in which the unit wavelength interval = 1 m. According to the conditions of the problem, it is necessary to calculate the spectral luminosity density calculated for the wavelength interval of 1 nm, so we write out the value of C in SI units and recalculate it for a given wavelength interval:

THERMAL RADIATION 7. A study of the solar radiation spectrum shows that the maximum spectral density of energy luminosity corresponds to a wavelength = 500 nm. Taking the Sun to be a black body, determine: 1) the energetic luminosity R e of the Sun; 2) energy flow F e emitted by the Sun; 3) the mass of electromagnetic waves (of all lengths) emitted by the Sun in 1 s. 1. According to the laws of Stefan Boltzmann and Wien 2. Luminous flux 3. The mass of electromagnetic waves (all lengths) emitted by the Sun during the time t = 1 s, we determine by applying the law of proportionality of mass and energy E = ms 2. The energy of electromagnetic waves emitted during time t, is equal to the product of energy flow Ф e ((radiation power) by time: E=Ф e t. Therefore, Ф e =ms 2, whence m=Ф e/s 2.