Discussion Forum



1)

A radioactive nucleus A with a half-life T, decays into a nucleus B. At t=0, there is no nucleus B. After sometimes t. the ratio of the number of B to that of A is 0.3. Then, t is given by


A) $t=T\frac{log1.3}{\log_{e}{2}}$

B) $t=T\log_{}{1.3}$

C) $t=\frac{T}{\log_{}{1.3}}$

D) $t=\frac{T\log_{e}{2}}{2\log_{}{1.3}}$

Answer:

Option A

Explanation:

Decay scheme is,

                 18102019770_tools.PNG

Given        $\frac{N_{B}}{N_{A}}=0.3=\frac{3}{10}$

                 $\Rightarrow\frac{N_{B}}{N_{A}}=\frac{30}{100}$

  So,    N0  =100+30=130 atoms

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

   We have , $100= 130e^{-\lambda t}$

           $\Rightarrow     \frac{1}{13}=e^{-\lambda t}\Rightarrow log1 .3=\lambda t$

    If T is half-life, then $t=\frac{\log_{e}{2}}{T}$

                 $\Rightarrow \log_{}{1.3}=\frac{\log_{e}{2}}{T}.t$

                               $\therefore     t=\frac{T.\log_{}{(1.3)}}{\log_{e}{2}}$