Question

A radioactive element X converts into another stable element Y. Half-life of X is 2 h. Initially only X is present. After time t, the ratio of atoms of X and Y is found to be 1 : 4, then t in hour is

• A. 2
• B. 4
• C. Between 4 and 6
• D. 6

Answer 1

The correct option is C

Between 4 and 6

Let $N_0$ be the number of atoms of X at time t = 0.
Then at t = 4h (two half-lives)
$N_x=\frac{N_0}{4}$ and $N_y=\frac{3N_0}{4}$
$\therefore \frac{N_x}{N_y}=\frac{1}{3}$
and at t = 6h (three half-lives)
$N_x=\frac{N_0}{8}$ and $N_y=\frac{7N_0}{8}$
or $\frac{N_x}{N_y}=\frac{1}{7}$
The given ratio $\frac{1}{4}$ lies between $\frac{1}{3}$ and $\frac{1}{7}$
Therefore, t lies between 4h and 6h