Question

A radioactive substance decays to $1/{16}^{th}$ of its initial activity in $40$ days. The half-life of the radioactive substance expressed in days is:

• A. $2.5$
• B. $5$
• C. $10$
• D. $20$

Answer 1

The correct option is C $10$

Let $T$ days be the half-life period.
After one half-life period, the activity reduces to one half of its initial activity.
In 40 days, the activity reduces to $1/{16}^{th}$ of its initial activity.
$\frac{1}{16}=(\frac{1}{2}{)}^4$
This corresponds to four half life periods or $4T$ days.
$4T=40$
$T=10$
Hence, the half-life of the radioactive substance is 10 days.