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

Answer: (C)

We know that the activity of radioactive substance,
$A=A_{0}$ $e^{-\lambda t}$
$\Rightarrow \frac{A}{A_{0}}=e^{-\lambda t}=e^{-\frac{0.693}{T^{1 / 2}} t}$
$\left[\because T_{1 / 2}=\frac{0.693}{\lambda}\right]$
$\frac{1}{16}=e^{-\frac{0.693 \times 40}{T^{1 / 2}}} $
$\Rightarrow \frac{0.693 \times 40}{T_{1 / 2}}=\ln 16$
$\Rightarrow T_{1 / 2}=\frac{0.693 \times 40}{\ln 16}$
$=9.997 \approx 10$ days.