Question

Certain radio-active substance reduces to 25% of its value in 16 days. Its half-life is.

• A. 32 days
• B. 8 days
• C. 64 days
• D. 28 days

Answer 1

Answer: (D)

28 days

Answer 2

$N = N_{0}\left( \frac{1}{2} \right)^{\frac{t}{T}} \Rightarrow \frac{N_{0}}{4} = N_{0}\left( \frac{1}{2} \right)^{\frac{16}{T}} \Rightarrow T = 8$days.