Stefan’s Law and Spectrum of Black Body

Stefan’s Law

Stefan’s Law states that the rate of emission of radiant energy unit area of a perfectly black body is directly proportional to the fourth power of its absolute temperature.

$$ \text {i.e}\: E = \sigma T^4 \dots (1) $$

where \(\sigma \) is a constant called Stefan’s constant.

When a black body is at absolute temperature T and is surrounded by another black body at absolute temperature \(T_0\), then the amount of heat lost by the former black body per unit time is given by

$$ E = \sigma (T^4 – T_0^4) $$

Proof:

Let us consider a cylinder enclosure ABCD of uniform cross-section filled with diffuse radiations of density u at uniform temperature T. If V be the volume of enclosure.

$$ U = uV \dots (i) $$

Let a small amount of heat dQ flow the enclosure from outside and piston moves out so that volume changes by by dV. The temperature and the energy density u changes. If dV is the change in internal energy of radiation and dW is the external work done, then by 1^st law of thermodynamic

\begin{align*} dQ &= dU + dW \\ dQ &= d(uV) + PdV [\text {From equation (1)}]\\ \text {or,}\: dQ = Vdu + udV + PdV \\ \text {or,}\:dQ &= Vdu + u dV + \frac 13udV \: \: [\text {Since} P = \frac 13 u] \\ &= vdu + \frac 43udV (2) ]\\ \text {Now, from the}\: (3)\: \text {law of thermodynamics} \\ dS &= \frac {dQ}{T} \dots (3) \\ \therefore dS &= \frac VTdu + \frac 43 \frac uT dV \dots (4) \\ \end{align*}

The function of two independent variables u and V.

\begin{align*} S &= f(u, V) \\ \ dS &= \frac {\delta S}{\delta u}du + \frac {\delta S}{\delta V}dV \dots (5) \\ \text {Comparing}\: (4)\: \text {and} \: (5)\: \text {we get} \\ \frac {\delta S}{\delta u} &= \frac VT\\ \frac {\delta S}{\delta V} &= \frac {4u}{3T}\\ \text {As dS is a perfect differential, we have}\\ \frac {\delta S}{\delta u\delta V} &= \frac {\delta ^2 S}{\delta V\delta u} \\ \text {or,}\: \frac {\delta }{du}\left (\frac {\delta S}{\delta V} \right ) &= \frac {\delta }{\delta V} \left (\frac {\delta S}{\delta u}\right ) \\ \text {or,}\: \frac {\delta }{\delta u} \left (\frac {4u}{3T}\right ) &= \frac {\delta }{\delta V}\left (\frac VT\right ) \\ \text {As dS is a perfect differential, }\\ \text {we have } \\ \end{align*}

\begin{align*} \frac {\delta ^2S}{\delta u\delta V} &= \frac {\delta ^2}{\delta V \delta u} \\ \frac {\delta }{\delta V}\left (\frac {\delta S}{\delta u} \right ) &= \frac {\delta }{\delta V}\left (\frac {\delta S}{\delta u }\right ) \\ \frac {\delta }{\delta u} \left (\frac {4u}{3T}\right ) &= \frac {\delta }{\delta V} \left (\frac VT \right ) \dots (6) \\ \end{align*}

Differentiating equation (6) with respect to u, we get

\begin{align*} \frac 43\frac 1T - \frac 43\frac {u}{T^2} \frac {\delta T}{\delta u} &= \frac 1T \\ \frac 13\frac 1T &= \frac 34\frac {u}{T^2}\frac {\delta T}{\delta u} \\ \text {or,}\: \frac {\delta u}{u} &= \frac {4\delta T}{T} \dots (7)\end{align*}

Integrating we have \(\log u = 4\log T + \log A \) where \(\log A\) be constant of integration

$$ u = AT^4 \dots (8) $$

Total rate of emission of radiant energy per unit area is related to the energy density by the following relation

\begin{align*} E &= \frac 14 uc \dots (9)\\ E &= \frac 14 AT^4 \dots (10) \\ \therefore E &= \sigma T^4 \\ \end{align*}

where \(\sigma =\frac 14 Ac\) is constant. The value of \(\sigma \: \text {is}\: 5.672 \times 10^{-8}\: JM^{-2}s^{-1}K^{-4} \)

Distribution of Energy in the Spectrum of a Black Body

Lummer and Pringshein studied the distribution of energy among the radiation emitted by a black body at various temperatures.

The radiation from the black body passes through the slit S₁ and falls on the reflector M₁. The reflected parallel beam of falls on the rock salt or fluorspar prism AB placed on the turn table of the spectrometer. The emergent ray is focused by the reflector M₂ on the bolometer placed behind the slit S₂. The turn table is rotated slowly so that the different parts of radiation spectrum fall on the bolometer and corresponding deflection in the galvanometer is noticed. Then the curves between the intensity and the wavelength are plotted at different temperatures.

Results

1. The energy is not uniformly distributed in the radiation spectrum of black body.
2. At a given temperature the intensity of radiation increases with the increase in wavelength becomes maximum and then decreases with further increase in wavelength.
3. As increase in temperature causes decrease in the wavelength for which the energy emitted is maximum \(\lambda _m\).
4. An increase in temperature causes an increase in energy emission for all wavelengths.
5. The area of the curve at a particular temperature for the desired range of wavelength increases as temperature increases.

Quantum Theory of Radiation

According to Planck’s hypothesis, “A black body radiation chamber is filled up not only with radiation; but also with simple harmonic oscillators of the molecular dimensions, and the exchange of energy between the matter and radiation does not take place continuously but discontinuously and discretly as an integral multiple of small unit energy called quantum or photon.

Planck assumed that the energy of a photon is proportional to the frequency of radiation, \(\text {i.e.}\: E \propto \nu \).
$$ \text {or,}\: E = h\nu $$

where \(h \) is a universal constant the Planck’s constant, its value being equal to \(6.62 \times 10^{-34}\) joule sec and is dimensionally equal to the product of linear momentum and distance or the product of energy and time.

Bibliography

S.S. Singhal, J.P. Agarwal, Satya Prakash. heat and thermodynamics and statistical physics. pragati prakashan, 2010.

—. Heat and Thermodynamics and Statistical Physics. Pragati Prakashan, 2010.

Vatsyayan, Dr. Rakesh Ranjan. Refresher Course in Physics. kathmandu: Surya Book Traders, 2015.