WEIN'S DISPLACEMENT LAW

At ordinary temperatures (below about 600°C) the thermal radiation emitted by a body is not visible, most of it is concentrated in wavelengths much longer than those of visible light.

Figure shows how the energy of a black body radiation varies with temperature and wavelength. As the temperature of the black body increases, two distinct behaviours are observed. The first effect is that the peak of the distribution shifts to shorter wavelengths. This shift if found to obey the following relationship called Wein's displacement law

$\lambda$_maxT = b

Here b is a constant called Wein's constant. The value of this constant in SI unit is 2.898 x 10^–3 m–K. Thus,

${\lambda }_{\max }\propto \frac{1}{T}$

Here $\lambda$_max is the wavelength corresponding to the maximum spectral emissive power e$\lambda$.

The second effect is that the total amount of energy the black body emits per unit area per unit time (= $\sigma$T⁴) increases with fourth power of absolute temperature T. This is also known as the emissive power. We know

e = ${\int }_0^{\infty }{e}_{\lambda }d\lambda$ = Area under e$\lambda$–$\lambda$ graph

= $\sigma$T⁴

or Area $\propto$ T⁴

A² = (2)⁴A₁ = 16A₁

Thus, if the temperature of the black body is made two fold, $\lambda$_max remains half while the area becomes 16 times.

Illustration 1. Estimate the surface temperature of sun. Given for solar radiations, $\lambda$_m = 4753 A°

Solution: From Wien's displacement law

$\lambda$_mT = b

$\therefore$ T = $\frac{b}{{\lambda }_m}=\frac{(2.89\times {10}^{-3}m-k)}{(4753\times {10}^{-10}m)}$= 6097°K.