Question

If $N_t=N_0{e}^{-\lambda t}$, then number of disintegrated atoms between $t_1$ to $t_2$ $(t_2>t_1)$ will be

• A. $N_0[e^{\lambda t_2}-e^{\lambda t_1}]$
• B. $N_0[e^{-\lambda t_2}-e^{-\lambda t_1}]$
• C. $N_0[e^{-\lambda t_1}-e^{-\lambda t_2}]$
• D. None

Answer 1

The correct option is C $N_0[e^{-\lambda t_1}-e^{-\lambda t_2}]$

The number of nuclei at time $t_1$,
$N_1=N_0{e}^{-\lambda t_1}$
The number of nuclei at time $t_2$,
$N_2=N_0{e}^{-\lambda t_2}$
Hence the number of disintegrated atoms between $t_1$ and $t_2$
= $N_1-N_2$
= $N_0[e^{-\lambda t_1}-e^{-\lambda t_2}]$