Question

The radioactive sources $A$ and $B$ of half lives of $2\text{hr}$ and $4\text{hr}$ respectively, initially contain the same number of radioactive atoms. At the end of $2\text{hours}$, their rates of disintegration are in the ratio

• A. $4:1$
• B. $2:1$
• C. $\sqrt{2}:1$
• D. $1:1$

Answer 1

The correct option is C $\sqrt{2}:1$

Rate of disintegration of a Radio active sample
$\frac{dN}{dt}=\lambda N_0{e}^{-\lambda t}$
where $\lambda =\text{Decay}\text{constant}=\frac{\ln 2}{t_{1/2}}$
For radio active source $A$ :
${\left[\frac{dN}{dt}\right]}_A=-\left[\frac{\ln 2}{2}\right]N_0{e}^{-\left[\frac{\ln 2}{2}\right]\left[2\right]}$
${\left[\frac{dN}{dt}\right]}_A=-\left[\frac{N_0\ln 2}{4}\right]...\left(1\right)$
For radio-active source $B$ :
${\left[\frac{dN}{dt}\right]}_B=-\left[\frac{\ln 2}{4}\right]N_0{e}^{-\left[\frac{\ln 2}{4}\right]\left[2\right]}$
${\left[\frac{dN}{dt}\right]}_B=-\left[\frac{N_0\ln 2}{4}\right]\left[\frac{1}{\sqrt{2}}\right]...\left(2\right)$
From (1) and (2) :
$\frac{{\left[\frac{dN}{dt}\right]}_A}{{\left[\frac{dN}{dt}\right]}_B}=\frac{\sqrt{2}}{1}$
Hence option (C) is correct.