Q95 / 125 · BITSAT 2016 · Physics

A radioactive element X converts into another stable element Y. Half life of $\mathrm{X}$ is $2 \mathrm{hrs}$. Initially only $\mathrm{X}$ is present. After time $\mathrm{t},$ the ratio of atoms of $X$ and $Y$ is found to be $1: 4,$ then $t$ in hours is

  • A
    2
  • B
    4
  • Correct answer
    C
    between 4 and 6
  • D
    6

Solution

Let $\mathrm{N}_{0}$ be the number of atoms of $\mathrm{X}$ at time $\mathrm{t}=0$ Then at $\mathrm{t}=4$ hrs (two half lives) $\mathrm{N}_{\mathrm{x}}=\frac{\mathrm{N}_{0}}{4}$ and $\mathrm{N}_{\mathrm{y}}=\frac{3 \mathrm{~N}_{0}}{4}$ $\therefore \mathrm{N}_{\mathrm{x}} / \mathrm{N}_{\mathrm{y}}=1 / 3$ and at $\mathrm{t}=6 \mathrm{hrs}$ (three half lives) $\mathrm{N}_{\mathrm{x}}=\frac{\mathrm{N}_{0}}{8}$ and $\mathrm{N}_{\mathrm{y}}=\frac{7 \mathrm{~N}_{0}}{8}$ or $\frac{\mathrm{N}_{\mathrm{x}}}{\mathrm{N}_{\mathrm{y}}}=\frac{1}{7}$ The given ratio $\frac{1}{4}$ lies between $\frac{1}{3}$ and $\frac{1}{7}$ Therefore, t lies between $4 \mathrm{hrs}$ and $6 \mathrm{hrs}$.

Hence, the correct answer is option: 'C'.

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