Discussion Forum

For a radioactive material , its activity A and rate of change of its activity R are defined as $A=-\frac{dN}{dt}$  and    $A=-\frac{dA}{dt}$ , where N(t)  is the number of nuclei at time  t. Two radioactive source P(mean life τ )  and Q (mean life 2τ)  have the same activity at t=0. Their rate of change of activities at t=2τ is R_p and R_Q, respectively. If  $\frac{R_{P}}{R_{Q}}=\frac{n}{e}$, then the value of n is 

Answer:

Explanation:

For  initial numbers are N₁   and N₂.

      $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\tau_{2}}{\tau_{1}}=\frac{2\tau}{\tau}$

                              $=2=\frac{T_{2}}{T_{1}}$               (T= Half life)

                                     $A= \frac{-dN}{dt}=\lambda N$

 initial activity is same

$\therefore$               $\lambda_{1}N_{1}=\lambda_{2}N_{2}$          ........(i)

Activity at time t,

                                               $A=\lambda N=\lambda N_{0}e^{-\lambda t}$

                                     $A_{1}=\lambda_{1}N_{1}e^{-\lambda_{1}t}$

$\Rightarrow$            $R_{1}=-\frac{dA_{1}}{dt}=\lambda_{1}^{2}N_{1}e^{-\lambda_{1}t}$

similarly,                     $R_{2}=\lambda^{2}_{2}N_{2}e^{-\lambda_{2}t}$

After t= 2τ                        

                       $\lambda_{1}t=\frac{1}{\tau_{1}}(t)=\frac{1}{\tau}(2\tau)=2$

                          $\lambda_{2}t=\frac{1}{\tau_{2}}(t)=1$

                                      $\frac{1}{2\tau}(2\tau)=1$

                      $\frac{R_{P}}{R_{Q}}=\frac{\lambda^{2}_{1}N_{1}e^{-\lambda_{1}t}}{\lambda^{2}_{2}N_{2}e^{-\lambda_{2}t}}$

   $\frac{R_{P}}{R_{Q}}=\frac{\lambda_{1}}{\lambda_{2}}\left(\frac{e^{-2}}{e^{-1}}\right)=\frac{2}{e}$