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  

 

Answer 1

Stefan Boltzmann Law relates the temperature of the blackbody to the amount of the power it emits per unit area. The law states that;

 

“The total energy emitted/radiated per unit surface area of a blackbody across all wavelengths per unit time is directly proportional to the fourth power of the black body’s thermodynamic temperature. ”

Here, lm becomes half, the Temperature doubles. 

T1/T2=1/2 put in eqn

⇒ ε = (σT)^{^4} 

^{e1/e2 = (T1/T2)^4 resiprocal values to get e2}

 

^{⇒ e2 = (T2/T1)^4 . e1 = (2)^4 . 16}

 

^{= 16.16 = 256 J m-2 s-1}

Wein’s displacement law is, lm.T = b. 

T- temperature

 

i.e. Tμ [1/lm. )\( \)

_{Now from Stefan Boltzmann Law, e = (sT)^{^4}}