Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

I. E. Irodov Solutions for Physics Class 11 & Class 12

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Q.246. Using Wien's formula, demonstrate that
 (a) the most probable radiation frequency Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev 
 (b) the maximum spectral density of thermal radiation 

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRevIrodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev
 (c) the radiosity Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Ans. a ) The most probable radiation frequency Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev is the frequency for which
Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

The maximum frequency is the root other than ω = 0 of this equation. It is

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRevwhere x0 is the solution of the transcendental equation

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

(b) The maximum spectral density is the density corresponding to most probable frequency. It is

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

where x0 is defined above, 

(c) The radiosity is

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

 

Q.247. The temperature of one of the two heated black bodies is T1 = 2500 K. Find the temperature of the other body if the wavelength corresponding to its maximum emissive capacity exceeds by Δλ = 0.50 gm the wavelength corresponding to the maximum emissive capacity of the first black body.

Ans.  For the first black body

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Then

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Hence

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

 

Q.248. The radiosity of a black body is Me  =3.0 W/cm2. Find the wavelength corresponding to the maximum emissive capacity of that body.

Ans. From the radiosity we get the temperature of the black body. It is

 

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Hence the wavelength corresponding to the maximum emissive capacity of the body is

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

 

Q.249. The spectral composition of solar radiation is much the same as that of a black body whose maximum emission corresponds to the wavelength 0.48μm. Find the mass lost by the Sun every second due to radiation. Evaluate the time interval during which the mass of the Sun diminishes by 1 per cent. 

Ans. The black body temperature of the sun maybe taken as 

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Thus the radiosity is

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Energy lost by sun is

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

This corresponds to a mass loss of

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

The sun loses 1 % of its mass in

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

 

Q.250. Find the temperature of totally ionized hydrogen plasma of density p = 0.10 g/cm3  at which the thermal radiation pressure is equal to the gas kinetic pressure of the particles of plasma. Take into account that the thermal radiation pressure p = u/3, where u is the space density of radiation energy, and at high temperatures all substances obey the equation of state of an ideal gas. 

Ans. For an ideal gas p = n k T where n = number density of the particles and Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev i.s Boltzman constant In a fully ionized hydrogen plasma, both H ions (protons) and electrons contribute to pressure but since the mass of electrons is quite small Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev only protons contribute to mass density. Thus

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

and Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Where Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev is the proton or hydrogen mass.

Equating this to thermal radiation pressure

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Then  Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev molecular weight of hydrogen = 2 x10-3 kg .

Thus Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

 

Q.251. A copper ball of diameter d = 1.2 cm was placed in an evacuated vessel whose walls are kept at the absolute zero temperature. The initial temperature of the ball is T = 300 K. Assuming the surface of the ball to be absolutely black, find how soon its temperature decreases η = 2.0 times. 

Ans. In time dt after the instant t when the temperature of the ball is T, it loses 

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Joules of energy. As & result its temperature falls by - dT and

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Where ρ = density of copper, C = its sp.heat

Thus  Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

or Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

 

Q.252. There are two cavities (Fig. 5.39) with small holes of equal diameters d = 1.0 cm and perfectly reflecting outer surfaces. The 

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

distance between the holes is l = 10 cm. A constant temperature T1 = 1700 K is maintained in cavity 1. Calculate the steady-state temperature inside cavity 2. Instruction. Take into account that a black body radiation obeys the cosine emission law. 

Ans. Taking account of cosine low of emission we write for the energy radiated per second by the hole in cavity # 1 as

 

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

where A is an constant, dΩ is an element of solid angle around some direciton defined by the symbol Ω . Integrating over the whole forward hemisphere we get

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

We find A by equating this to the quantity Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev is stefan- Boltzman constant and d is the diameter of th hole.

Then Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Now energy reaching 2 from 1 is Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

where Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev is the solid angle subtended by the hole of 2 at 1. {We are assuming

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

This must equal Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

which is the energy emitted by 2. Thus equating

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

or  Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Substituting we get T2 = 0.380 k K = 380 K .

 

 

Q.253. A cavity of volume V = 1.0 1 is filled with thermal radiation at a temperature T = 1000 K. Find:
 (a) the heat capacity Cv;
 (b) the entropy S of that radiation. 

Ans. (a ) The total internal eneigy of the cavity is 

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Hence  Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

(b) From first law

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

so Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Hence Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

 

Q.254. Assuming the spectral distribution of thermal radiation energy to obey Wien's formula Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev where a = 7.64 ps•K , find for a temperature T = 2000 K the most probable
 (a) radiation frequency;
 (b) radiation wavelength. 

Ans. We are given

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

(a) then Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

so  Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

(b) We determine the spectral distribution in wavelength.

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

But Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

soIrodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

(we have put a minus sign before dλ to subsume just this fact dλ is -ve where dω is +ve.)

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

This is maximum when

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

or  Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

 

Q.255. Using Planck's formula, derive the approximate expressions for the space spectral density uω of radiation
 (a) in the range where hω << kT (Rayleigh-Jeans formula);
 (b) in the range where Nω >> kT (Wien's formula). 

Ans. From Planek’s formula 

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

(a) In a range Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev (long wavelength or high temperature).

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRevfor small x.

(b) In the range  Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev (high frequency or low temperature) :

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

andIrodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

 

Q.256. Transform Planck's formula for space spectral density uω. of radiation from the variable ω to the variables v (linear frequency) and λ (wavelength). 

Ans. We write

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Then

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Also Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

 

 

 

Q.257. Using Planck's formula, find the power radiated by a unit area of a black body within a narrow wavelength interval Δλ = = 1.0 nm close to the maximum of spectral radiation density at a temperature T = 3000 K of the body. 

Ans. We write the required power in terms of the radiosity by considering only the energy radiated in the given range. Then from the previous problem

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

But Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

so  Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Using the data

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

and  Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

 

 

Q.258. Fig. 5.40 shows the plot of the function y (x) representing a fraction of the total power of thermal radiation falling within 

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

the spectral interval from 0 to x. Here Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev is the wavelength corresponding to the maximum of spectral radiation density). Using this plot, find:
 (a) the wavelength which divides the radiation spectrum into two equal (in terms of energy) parts at the temperature 3700 K;
 (b) the fraction of the total radiation power falling within the visible range of the spectrum (0.40-0.76 Rm) at the temperature 5000 K; 
 (c) how many times the power radiated at wavelengths exceeding 0.76 Jim will increase if the temperature rises from 3000 to 5000 K. 

Ans. (a) From the curve of the function y(x) we see that y = 0-5 when x = 1.41

ThusIrodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

(b) At 5000 K

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

So the visible range (0.40 to 0.70) ^im corresponds to a range (0.69 to 1.31) of x

 From the curve

y (0.69 ) = 0.07 

y( 1-31) = 0.44

so the fraction is 0.37

(c) The value of x corresponding to 0.76 are

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

The requisite fraction is then

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

 

Q.259. Making use of Planck's formula, derive the expressions determining the number of photons per 1 cm of a cavity at a temperature T in the spectral intervals Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Ans. We use the formulaIrodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Then the number of photons in the spectral interval (ω, ω + dω) is

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

usingIrodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

we get  Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

 

Q.260. An isotropic point source emits light with wavelength λ = 589 nm. The radiation power of the source is P = 10 W. Find:
 (a) the mean density of the flow of photons at a distance r = = 2.0 m from the source;
 (b) the distance between the source and the point at which the mean concentration of photons is equal to n = 100 cm -3

Ans. (a) The mean density of the flow of photons at a distance r is

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

(b) If n(r) is the mean concentration (number per unit volume) of photons at a distance r form the source, then, since all photons are moving outwards with a velocity c, there is an outward flux of cn which is balanced by the flux from the source. In steady state, the two are equal and so

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

so Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

Irodov Solutions: Thermal Radiation. Quantum Nature of Light- 1 Notes | EduRev

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