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 r, 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

Answer: (C)

Let $N_0$ be the number of atoms of $x$ at time $t=0$ . Then at $t=4h,$ ie, (two half-lives) $N_x=N_0{\left(\frac{1}{2}\right)}^2$ $N_x=\frac{N_0}{4}$ $\therefore$ $N_y=N_0-\frac{N_0}{4}=\frac{3N_0}{4}$ $\therefore$ $\frac{N_x}{N_y}=\frac{1}{3}$ Now, at $t=6h,$ ie, (three half-lives) $N_x=N_0{\left(\frac{1}{2}\right)}^3=\frac{N_0}{8}$ and $N_x=N_0-N_x$ $=N_0-\frac{N_0}{8}$ $=\frac{7N_0}{8}$ or $\frac{N_x}{N_y}=\frac{1}{7}$ The given ratio lies between $\frac{1}{3}$ and $\frac{1}{7}$ Therefore, t lies between 4 h and 6 h.