

InterviewSolution
Saved Bookmarks
1. |
Let the wavelength at which the the spectral emissive power of a black body (at a temperature T) is maximum, be denoted by `lamda_(max)` as the temperature of the body is increased by 1K, `lamda_(max)` decreases by 1 percent. The temperature T of the black body isA. 100 KB. 200 KC. 400 KD. 288 K |
Answer» `lamda_(m)T=`const. `lnlamda_(m)=lnT=C` `(dlamda_(m))/(lamda_(m))+(dT)/(T)=0" "therefore(dlamda_(m))/(lamda_(m))=(dT)/(T)` Now `(dlamda_(m))/(lamda_(m))=-1%=-(1)/(100)` (-ve sign indicates decreases) `dT=1` (given) `thereforeT=100K` |
|