Question

The half-life of a radioactive substance is 30 minutes. The time (in minutes) taken between 40% decay of the same radioactive substance is :

• A. $15$
• B. $45$
• C. $22$
• D. $30$

Answer 1

Answer: (C)

We know that, $N=N_0{e}^{-\lambda t}$

Here, remaining amount:

$N=N_0-\left(\frac{40}{100}\right)N_0=\left(\frac{3}{5}\right)N_0$

$\frac{N_0}{N}=\frac{5}{3}$ and decay constant $\lambda =\frac{ln2}{t_{1/2}}=\frac{ln2}{30}$

Thus, $\frac{N_0}{N}=e^{\lambda t}$

Taking $ln$,

$ln\left(N_0/N\right)=\lambda t$

$t=\frac{ln\left(N_0/N\right)}{\lambda }=\frac{ln\left(5/3\right)}{\left(ln2\right)/30}$

$\therefore$ $t=\frac{0.51}{0.693/30}=22min$