Question

Nuclei of a radioactive element $A$ are being produced at a constant rate $\alpha$. The element $A$ has a decay constant $\lambda$. At $t=0$, there are $N_0$ nuclei of element $A$,.then the number of nuclei of $A$ at any time $t$ is. If $\alpha =2\lambda N_0$, the number of nuclei of $A$ after one half-life of $A$ and also the limiting value of $N$ as $t\to \infty$ is :

• A. $\left[\right(a)N=\frac{1}{\lambda }[\alpha +(\alpha -\lambda N_0)e^{-\lambda t}],(b)\frac{3N_0}{2},2N_0]$
• B. $\left[\right(a)N=\frac{1}{\lambda }[\alpha -(\alpha -\lambda N_0)e^{-\lambda t}],(b)\frac{3N_0}{2},2N_0]$
• C. $\left[\right(a)N=\frac{1}{\lambda }[\alpha -(\alpha -\lambda N_0)e^{-\lambda t}],(b)\frac{5N_0}{2},2N_0]$
• D. $\left[\right(a)N=\frac{1}{\lambda }[\alpha -(\alpha -\lambda N_0)e^{-2\lambda t}],(b)\frac{3N_0}{2},2N_0]$

Answer 1

Answer: (B)

$\left[\right(a)N=\frac{1}{\lambda }[\alpha -(\alpha -\lambda N_0)e^{-\lambda t}],(b)\frac{3N_0}{2},2N_0]$

Rate of production of the radioactive element $A$ $\frac{d{N}_A}{dt}=\alpha$
Rate of decay of the radioactive element $A$ $-\frac{d{N}_A}{dt}=-\lambda N_A$
$\therefore$ Net rate $\frac{d{N}_A}{dt}=\alpha -\lambda N_A$
To calculate the number $N$ of nuclei of $A$ at time $t$, integrating the above expression
${\int }_{N_0}^N\frac{d{N}_A}{\alpha -\lambda N_A}={\int }_0^{t}dt\Rightarrow -\frac{1}{\lambda }\ln \left(\frac{\alpha -\lambda N}{\alpha -\lambda N_0}\right)=t\Rightarrow N=\frac{1}{\lambda }[\alpha -(\alpha -\lambda N_0)e^{-\lambda t}]$
If $\alpha =2N_0\lambda$, then $\Rightarrow N=\frac{1}{\lambda }[2N_0\alpha -(2N_0\alpha -\lambda N_0)e^{-\lambda t}]=N_0(2-e^{-\lambda t})$
for $t=t_{\frac{1}{2}}=\frac{\ln 2}{\lambda }$, we have
$N=N_0(2-e^{-\ln 2})=N_0(2-\frac{1}{2})=\frac{3}{2}{N}_0$
For $t\to \infty ,e^{-\lambda t}\to 0$ and, hence $N=2N_0$