Need of Quantum Theory ( Black body radiation )

Ultraviolet catastrophe:

An ideal blackbody is an object which absorbs all radiation falling on it. When it is heated it emits radiation of all frequencies.

According to classical theory, A blackbody can exchange energy between electromagnetic radiation [ black body and radiation fields] and the enclouser [ wall of cavity, electron , atom, molecules ] in any amount. The amount of energy exchange is continuous. According to Rayleigh Jeans law, the energy density of radiation inside black body at temperature and frequency \(\nu\) is given by.

$$u(\nu, T)= \frac{8\pi\nu^2}{c^3}\times\dotsm(1)$$

Where,

= Mean energy of each oscillator

C= Speed of light

According to classical thermodynamics the average energy at temperature T is given by equipartition theorem of energy.

$$=\frac{\int_0^\infty E\cdot e^{\frac{-E}{KT}}}{\int_0^\infty e^{\frac{-E}{KT}}dE}dE\dotsm(2)$$

Where,

\(e^{\frac{-E}{KT}}\)= Boltzmann factor

\(\int dE\)= Because energy exchange is continuous.

Solving equation (2), we get

$$ = KT\dotsm(3)$$

Where, k= Boltzmann constant

$$\therefore u(\nu, T)=\frac{8\pi\nu^2}{C^3} KT\dotsm(4)$$

Energy density of radiation inside black body within frequency interval \(\nu\) and \(\nu+d\nu\) is given by

$$u(\nu, T)d\nu= \frac{8\pi\nu^2}{c^3} KT. d\nu$$

Integrating above equation,

$$\int_0^\infty u(\nu, T)d\nu=\frac{8\pi KT}{C^3}\int_0^\infty \nu^2 d\nu$$

$$=\frac{8\pi KT}{C^3}\biggl[\frac{\nu^3}{3}\biggr]_0^\infty$$

$$i.e \int_0^\infty u(\nu, T))d\nu=\frac{8\pi KT}{C^3}\biggl[\frac{\infty^3-0^3}{3}\biggr]=\infty\dotsm(5)$$

From equation (5) the total energy density come out to be infinity. i.e energy of black body at temperature is infinity, which is known as ultraviolet catastrophe.

The total energy of black body at a given temperature must be measurable and finite. Reyleigh Jeans law ( Classical Theory ) couldn't explain the measurable value of total energy of black body.

Resolution of ultraviolet catatrhophe:

1. According to Plank's theorem the energy exchange between electromagnetic radiation and black body [ enclouser ] is discrete.

2. Minimum energy exchange at frequency \(\nu\) is

$$E=h\nu$$

Where, h= Planck's constant

3. The energy exchange is integral multiple of minimum energy (i.e \(h\nu\))

$$\therefore\;\; E_n= nh\nu\dotsm(6)$$

Where, n=0.1,2,3,4,...,\(\infty\)

The energy density per unit frequencies according to Planck's theory f radiation.

$$u(\nu, T)= \frac{8\pi\nu^2}{C^3}\times <E>\dotsm(7)$$

Where,

$$<E>=\frac{\sum_{n=0}^\infty E_n e^{\frac{-E_n}{KT}}}{\sum_{n=0}^\infty e^{\frac{-E_n}{KT}}}$$

$$=\frac{\sum_{n=0}^\infty nh\nu e^{\frac{-E_n h\nu}{KT}}}{\sum_{n=0}^\infty e^{\frac{-n h\nu}{KT}}}$$

$$=\frac{h\nu\biggl[ 0\cdot e^0+1\cdot e^{\frac{-1 h\nu}{KT}}+ 2\cdot e^{\frac{-2h\nu}{KT}}+ 3\cdot e^{\frac{- 3h\nu}{KT}}+\dotsm \biggr]}{\biggl[e^0+ e^{\frac{-h\nu}{KT}}+e^{\frac{-2h\nu}{KT}}+e^{\frac{-3h\nu}{KT}}+\dotsm\biggr]}$$

$$=\frac{h\nu\biggl[ e^{\frac{-h\nu}{KT}}+2e^{\frac{-2h\nu}{KT}}+3e^{\frac{-3h\nu}{KT}}+\dotsm\biggr]}{\biggl[1+e^{\frac{-h\nu}{KT}}+e^{\frac{-2h\nu}{KT}}+e^{\frac{-3h\nu}{KT}}+\dotsm\biggr]}$$

Put, \(\frac{-h\nu}{KT}=x\)

$$<E>= \frac{h\nu[e^x+2e^{2x}+3e^{3x}+\dotsm]}{[1+e^x+e^{2x}+e^{3x}+\dotsm]}$$

$$=h\nu \frac{d}{dx}[log(1+e^x+e^{2x}+e^{3x}+\dotsm]$$

$$=h\nu\frac{d}{dx}\biggl[ log(\frac{1}{1-e^x})\biggr]$$

$$= h\nu\frac{d}{dx}\biggl[log(1)-log(1-e^x)\biggr]$$

$$=h\nu\biggl[0-\frac{1}{(1-e^x)}\times (0-e^x)\biggr]$$

$$=h\nu\biggl(\frac{e^x}{1-e^x}\biggr)$$

$$=\frac{h\nu}{e^{-x}-1}$$

$$\therefore\;\; <E>= \frac{h\nu}{e^{\frac{h\nu}{KT}}-1}\dotsm(8)$$

Using equation (8) in (7)

$$u(\nu , T)=\frac{8\pi\nu^2}{C^3}\times\frac{h\nu}{e^{\frac{h\nu}{KT}}-1}$$

$$=\frac{8\pi\nu^3}{C^3}\times \frac{1}{e^{\frac{h\nu}{KT}}-1}\dotsm(9)$$

Equation (9) gives the energy density pper uni frequency of black body radiation. Also known as Planck's formula for energy density.

The total energy density in the interval of frequency \(\nu\) to \(\nu+d\nu\) is given by

$$\int_0^\infty u( \nu, T) d\nu=\int_0^\infty \frac{8\pi h\nu^3}{C^3}\cdot \frac{d\nu}{e^{\frac{h\nu}{KT}}-1}$$

$$=\frac{8\pi h}{C^3}\int_0^\infty \frac{\nu^3 d\nu}{e^{\frac{h\nu}{KT}}-1}$$

Put, \(\frac{h\nu}{KT}=x\)

\(\Rightarrow d\nu= dx\cdot \frac{KT}{h}\nu^3= \frac{x^3 (KT)^3}{h^3}\)

When,

\( \nu=0\Rightarrow x=0\)

\(\nu=\infty\Rightarrow x=\infty\)

Or, total energy density

$$=\frac{8\pi h}{C^3}\int_0^\infty \frac{x^3 K^3 T^3}{h^3}\cdot dx\frac{KT}{h}\times \frac{1}{e^x-1}$$

$$=\frac{8\pi h}{C^3}\int_0^\infty \frac{x^3 K^3 T^3}{h^3}\cdot \frac{dx KT}{h}\times \frac{1}{e^x-1}$$

$$=\frac{8\pi h}{C^3}\frac{K^4 T^4}{h^4}\int_0^\infty \frac{x^3}{e^x-1}$$

Using Standared integral

$$\int_0T\infty \frac{x^3}{e^x-1}dx= \frac{\pi^4}{15}$$

Total energy density = \(\frac{8\pi K^4 T^4}{C^3 h^3}\times \frac{\pi^4}{15}\)

$$=\frac{4}{c}\sigma\cdot T^4$$

Where, \(\sigma=\frac{2\pi^5 K^4}{15 c^2 h^3}= 5.67\times 10^{-18}\; Nm^{-2} k^{-4}\) is Stefan- Boltzmann constant.

the energy comes out to be finite and measurable at given temperature.

Thus Planck's theory of radiation avoid Ultravoilet catastrophe.

Reference:

1. Mathews, P.M and K Venkatesan. A Text Book of Quantum Mechanics. New Delhi: Tata McGraw Hill Publishing Co. Ltd, 1997.
2. Merzbacher, E. Quantum Mechanics . New York: John Wiley, 1969.
3. Prakash, S and S Salauja. Quantum Mechanics. Kedar Nath Ram Nath Publishing Co, 2002.
4. Singh, S.P, M.K and K Singh. Quantum Mechanics. Chand & Company Ltd., 2002.