Answer to Question #197954 in Electric Circuits for Plea

Question

A hot black body emits the energy at the rate of 16 J m^-2 s^- 1 and its most intense radiation corresponds to 20,000 Å. When the temperature of this body is further increased and its most intense radiation corresponds to 10,000 Å, then find the value of energy radiated in Jm^-2 s^-1

Answer 1

We know according to Wein's displacement law,

$\lambda_m.T = b$

i.e. if $\lambda_m$ becomes half temperature doubles.

Hence,

$\dfrac{\lambda_{m1}}{\lambda_{m2}} = \dfrac{20000}{10000} = \dfrac{T_2}{T_1}$

$T_2 = 2T_1$

Also, $e = \sigma T^4$

Therefore,

$\dfrac{e_1}{e_2} = (\dfrac{T_1}{T_2})^4$

$e_2 = (\dfrac{T_2}{T_1})^4e_1$

$e_2 = (\dfrac{2T_1}{T_1})^4e_1$

$e_2= 16\times 16$

$e_2 = 256Jm^{-2}s^{-1}$