Question

If a radioactive substance reduces to $\frac{1}{16}$ of its original mass in 40 days, what is its half-life.

• A. 10 days
• B. 20 days
• C. 40 days
• D. None of these

Answer 1

Answer: (A)

10 days

Answer 2

$\frac{N}{N_{0}} = \left( \frac{1}{2} \right)^{n}$ ⇒ $\frac{1}{16} = \left( \frac{1}{2} \right)^{4} = \left( \frac{1}{2} \right)^{n}$ ⇒ n = 4.

Also $n = \frac{t}{T_{1/2}}$ ⇒ $T_{1/2} = \frac{40}{4} = 10days$