A radioactive substance decays to $\frac{1}{32}^{\text {th }}$ of its initial activity in 30 days. Its half life period will be:

6 days

The correct answer is Option (1) → 6 days

Formula for radioactive decay is -

$A=A_0\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$

where,

A = Remaining activity

$A_0$ = Initial activity

t = Time elapsed = 30 days

T = half time period

$\left(\frac{A}{A_0}\right)=\frac{1}{32}$ [Given]

$∴\frac{1}{32}=\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$

$⇒\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}=\left(\frac{1}{2}\right)^5$

$⇒\frac{t}{T_{1/2}}=5⇒T_{1/2}=\frac{30}{5}=6days$