Question

A radioactive substance with decay constant of $0.5\, s^{-1}$ is being produced at a constant rate of 50 nuclei per second. If there are no nuclei present initially, the time (in second) after which 25 nuclei will be present is

• A. $1$
• B. $2ln\left(\frac{4}{3}\right)$
• C. $ln2$
• D. $ln\left(\frac{4}{3}\right)$

Answer 1

Answer: (B)

$2ln\left(\frac{4}{3}\right)$

Answer 2

Let N be the number of nuclei at any time t. Then $\frac{dN}{dt} = 50-\lambda N$ or $\frac{dN}{50-\lambda N} = dt$ Integrate both sides, we get $\int^{N}_{0} \frac{dN}{50-\lambda N} = \int^{t}_{0} dt\quad\Rightarrow\quad- \frac{1}{\lambda}\left[ln\left(50-\lambda N\right)\right]^{N}_{0} = t$ $ln \left(\frac{50-\lambda N}{50}\right) = -\lambda t$ $\frac{50-\lambda N}{50} = e^{-\lambda t} ; 1-\frac{\lambda N}{50} = e^{-\lambda t} N = \frac{50}{\lambda} \left(1-e^{-\lambda t}\right)$ As $N = 25$ and $\lambda= 0.5 \,s^{-1}$ $25 = \frac{50}{0.5}\left(1-e^{-0.5t}\right), \frac{1}{4} = 1-e^{-0.5t}$ $e^{-0.5t} = \frac{3}{4} \therefore t = 2ln \left(\frac{4}{3}\right)$