The half-life period of radon is 3.8 days. After how many days will only one-twentieth of the radon sample be left over?

Correct option: (D)

Step 1: Calculating the decay constant

Radioactive decay is considered as a first-order reaction. So, the half-life for radioactive decay is the same as the half-life of first-order reaction.

$\lambda =\frac{0.693}{t_{1/2}}$

$\lambda =$ decay constant

$t_{1/2}=$ half life

$\lambda =\frac{0.693}{3.8}=0.182da{y}^{-1}$

Step 2: Calculating the number of days for one-twentieth sample left

Let the initial amount of radon be $N_0$ and the amount left after $t$ days be $N$ which is equal to $\frac{N_0}{20}$

Applying the equation,

$t=\frac{2.303}{\lambda }log\frac{N_0}{N}$

$t=\frac{2.303}{0.182}log\frac{N_0}{\frac{N_0}{20}}$

$=\frac{2.303}{0.182}log\left(20\right)=16.54days$