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)

10

Answer 2

By using $N = N_{0}\left( \frac{1}{2} \right)^{t/T_{1/2}}$

⇒ $\frac{N}{N_{0}} = \frac{1}{16} = \left( \frac{1}{2} \right)^{40/T_{1/2}}$

⇒ $T_{1/2} = 10$days.