Discussion Forum

An active nucleus decays to $(\frac{1}{3})$ rd in 20 h . tHe fraction of original activity remaining after 80 h is 

Answer:

Explanation:

 Decay equation , $N=N_{0}exp(-\lambda t)$

 Given,  $N_{1}= N_{0}/3, N_{2}=  ?$

  $t_{1}=20 h$  , $t_{2}= 20 h$

 So,   $N_{1}=N_{0}exp(-\lambda t_{1})$

$\Rightarrow$        $\frac{1}{3}=e(-20 \lambda)\Rightarrow e^{20 \lambda}=3$

   Now,  $N_{2}=N_{0}exp(-\lambda t_{2})$  $\Rightarrow$  $\frac{N_{0}}{N_{2}}= e^{80 \lambda}$

 $\Rightarrow$     $\frac{N_{0}}{N_{2}}=(e^{20 \lambda})^{4}=(3)^{4}$

 $\Rightarrow$    $N_{2}= \frac{1}{(3)^{4}}N_{0}=   \frac{N_{0}}{81}$