A body emits maximum energy at 4253 `Å` and the same body at some other temperature emits maximum energy at 2342 `Å`. Find the ratio of the maximum energy radiated by the body in a short wavelength range.

Explanation

If `E_1` and `E_2` are the maximum energy radiated by the body in a short wavelength interval at temperature `T_1` and `T_2` then according to Wien's fifth power law, we have
`(E_1)/(E_2)=((T_1)/(T_2))^5` ..(i)
According to Wien's displacement law, we have
`lamda_(m1)=T_1=lamda_mT_2`
or `(T_1)/(T_2)=(lamda_(m2))/(lamda_(m1))` ..(ii)
Here `lamda_(m1)` and `lamda_(m2)` are the wavelengths at temperatures `T_1` and `T_2` at which the energy radiated in a small interval of wavelength is maximum
Now from Eqs. (i) and (ii) we have
`(E_1)/(E_2)=((lamda_(m2))/(lamda_(m1)))^5` or `(E_1)/(E_2)=((234xx10^-10)/(4253xx10^-10))=0.0506`