Question

A radioactive element A with half life 3 hours decays to a stable element B. After a time $ t $, the ratio of A and B atoms is 1:16 then the time $ t $ in hours is

• A. 6
• B. 12
• C. 16
• D. 20

Hint:

In problems involving radioactive decay, use the formula for half-life to find the time when the ratio of decay products is known.

Answer 1

The Correct Option is B

The relation between the amount of a radioactive element at time \( t \) and its half-life is given by the formula: \[ N(t) = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}} \] where \( N_0 \) is the initial number of atoms, \( T_{1/2} \) is the half-life, and \( N(t) \) is the number of atoms at time \( t \). The ratio \( \frac{A}{B} = 1:16 \) implies that 4 half-lives have passed.
Thus, \( t = 12 \) hours.