A radioactive substance decays to $\frac{1}{32}^{\text {th }}$ of its initial activity in 30 days. Its half life period will be:
Answer & explanation
Correct answer: option 1
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]
$∴\frac{1}{32}=\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$
$⇒\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}=\left(\frac{1}{2}\right)^5$
$⇒\frac{t}{T_{1/2}}=5⇒T_{1/2}=\frac{30}{5}=6days$