Discussion Forum

Two spherical stars A and B emit blackbody radiation. The radius of A  is 400 times that of B and A emits 10⁴  times the power emitted from B. The ratio $\left(\frac{\lambda_{A}}{\lambda_{B}}\right)$   of their  wavelengths  $\lambda_{A}$  and $\lambda_{B}$  at which the peaks  occur in their respective radiation curves is 

Answer:

Explanation:

 Power,    $P=(\sigma T^{4}A)=\sigma T^{4}(4\pi R^{2})$

   or                  $P \propto T^{4}R^{2}$  .........(i)

  According to Wien's law,

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

    ($\lambda$ is the wavelength at which peak occurs)

       $\therefore$      Eq (i) will become,

              $P\propto \frac{R^{2}}{\lambda^{4}}$

    or                          $\lambda\propto \left[\frac{R^{2}}{P}\right]^{1/4}$

  $\Rightarrow$         $\frac{\lambda_{A}}{\lambda_{B}}=\left[\frac{R_{A}}{R_{B}}\right]^{1/2} \left[\frac{P_{A}}{P_{B}}\right]^{1/4}$

$=[400]^{1/2}\left[\frac{1}{10^{4}}\right]^{1/4}=2$