Question

If $20$% of a radioactive substance disappears in $1$ year, then its half life approximately equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

For radioactive decay,

$N\left(t\right)=N{e}^{-\lambda t}$ where $N\left(t\right)$ is the substance left after $t$ years of disintegration, $N$ is the amount of initial substance, and $\lambda$ is the rate of disintegration.

So, $N\left(1\right)=\frac{(100-20)}{100}N=\frac{4}{5}N=N{e}^{-\lambda }$

Then, $\lambda =\ln \frac{5}{4}$

Half life of a radioactive substance$=\frac{\ln 2}{\lambda }=\frac{\ln 2}{\ln \frac{5}{4}}\approx 3$

Then Option C is right answer.