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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
Options:
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
Solution:
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Verified Answer
The correct answer is:
$\frac{\mathrm{N}_{1} \mathrm{~T}_{2}}{\mathrm{~N}_{2} \mathrm{~T}_{1}}$
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 $\frac{R_{1}}{R_{2}}=\frac{N_{1}}{T_{1}} \times \frac{T_{2}}{N_{2}}=\left(\frac{N_{1}}{N_{2}}\right)=\left(\frac{T_{2}}{T_{1}}\right)$
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 $\frac{R_{1}}{R_{2}}=\frac{N_{1}}{T_{1}} \times \frac{T_{2}}{N_{2}}=\left(\frac{N_{1}}{N_{2}}\right)=\left(\frac{T_{2}}{T_{1}}\right)$
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