Question

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

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

Answer 1

Answer: (C)

between 4 and 6

Answer 2

Let No 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 $t=6\, h$ , ie, (three half-lives) $N_{x}=N_{0} \left(\frac{1}{2}\right)^{3} = \frac{N_{0}}{8}$ and $N_{y} =N_{0} -N_{x}$ $=N_{0} -\frac{N_{0}}{8}$ $= \frac{7N_{0}}{8}$ or $\frac{N_{x}}{N_{y}} = \frac{1}{7}$ The given ratio lines between $\frac{1}{3}$ and $\frac{1}{7}$ Therefore, $t$ lies between $4\, h$ and $6\, h$ .