Question

If $\alpha =2N_0\lambda ,$ the number of nuclei of A after one half-life of A becomes:

• A. zero
• B. $2N_0$
• C. 1.5 $N_0$
• D. 0.5 $N_0$

Answer 1

The correct option is C 1.5 $N_0$

The number of nuclei of A at time 't' is $N=\frac{1}{\lambda }[\alpha -(\alpha -\lambda N_0)c^{-\lambda t}]$
But $\alpha =2N_0\lambda$ and $t=t_{1/2}$
Substitute values in the above expression.
$N=\frac{1}{\lambda }[2N_0\lambda -(2N_0\lambda -\lambda N_0\left)e^{-\lambda t}\right]=2N_0-N_0{e}^{-\lambda t}=2N_0-N_0{e}^{-0.693}=1.5N_0$.