Question

The decay constant of radioactive sample is $\lambda$. The half-life and mean-life of the sample are, respectively, given by

• A. $1/\lambda$ and $\left(ln2\right)/\lambda$
• B. $\left(ln2\right)/\lambda$ and $1/\lambda$
• C. $\lambda \left(ln2\right)$ and $1/\lambda$
• D. $\lambda /\left(ln2\right)$ and $1/\lambda$

Answer 1

The correct option is B $\left(ln2\right)/\lambda$ and $1/\lambda$

We know that, $N=N_0{e}^{-\lambda t}$
$e^{\lambda t}=\frac{N_0}{N}\Rightarrow \lambda t=ln\left(\frac{N_0}{N}\right)$

For half life, $t=t_{1/2},N=N_0/2$ so $t_{1/2}=\frac{1}{\lambda }ln2$
Mean life is $t_{avg}=\frac{{\int }_0^{\infty }Ntdt}{{\int }_0^{\infty }Ndt}$

Now, ${\int }_0^{\infty }Ntdt={\int }_0^{\infty }{N}_0{e}^{-\lambda t}tdt=N_0\left(\right[(-t\lambda )e^{-\lambda t}{]}_0^{\infty }+\frac{1}{\lambda }{\int }_0^{\infty }{e}^{-\lambda t})dt$ (integrated by parts )

$=0+\frac{N_0}{{\lambda }^2}[e^{-\lambda t}{]}_0^{\infty }=\frac{N_0}{{\lambda }^2}[0-1]=\frac{N_0}{{\lambda }^2}$

${\int }_0^{\infty }Ndt={\int }_0^{\infty }{N}_0{e}^{-\lambda t}dt=N_0\frac{1}{{\lambda }^2}[e^{-\lambda t}{]}_0^{\infty }=\frac{N_0}{\lambda }$

$t_{avg}=\frac{\frac{N_0}{{\lambda }^2}}{N_0/\lambda }=\frac{1}{\lambda }$