Q141 / 150 · MHT CET 2020 (20 Oct Shift 1) · Physics

Two different radioactive elements with half-lives ' $\mathrm{T}_{1}$ ' and ' $\mathrm{T}_{2}$ ' have undecayed atoms ' $\mathrm{N}_{1}{ }^{\prime}$ and ${ }^{\prime} \mathrm{N}_{2}{ }^{\prime}$ respectively, present at a given instant. The ratio of their activities at this instant is

  • A
    $\frac{\mathrm{T}_{1} \mathrm{~T}_{2}}{\mathrm{~N}_{1} \mathrm{~N}_{2}}$
  • B
    $\frac{\mathrm{N}_{1} \mathrm{~N}_{2}}{\mathrm{~T}_{1} \mathrm{~T}_{2}}$
  • C
    $\frac{\mathrm{N}_{1} \mathrm{~T}_{1}}{\mathrm{~N}_{2} \mathrm{~T}_{2}}$
  • Correct answer
    D
    $\frac{\mathrm{N}_{1} \mathrm{~T}_{2}}{\mathrm{~N}_{2} \mathrm{~T}_{1}}$

Solution

Activity, $\mathrm{R}\left(=\frac{\mathrm{dN}}{\mathrm{dt}}\right)=\lambda \mathrm{N}$ Decay constant $\lambda=\frac{\log _{e} 2}{\mathrm{~T}}$ $\therefore$ Activity $R=\frac{\left(\log _{\mathrm{e}} 2\right) \mathrm{N}}{\mathrm{T}}$ $\therefore \mathrm{R}_{1}=\frac{\left(\log _{e} 2\right) \mathrm{N}_{1}}{\mathrm{~T}_{1}}, \mathrm{R}_{2}=\frac{\left(\log _{e} 2\right) \mathrm{N}_{2}}{\mathrm{~T}_{2}}$ For two elements $\frac{R_{1}}{R_{2}}=\frac{N_{1}}{T_{1}} \times \frac{T_{2}}{N_{2}}=\left(\frac{N_{1}}{N_{2}}\right)=\left(\frac{T_{2}}{T_{1}}\right)$

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

Relevant Questions