KCET · Physics · Nuclear Physics
Decay constants of two radio-active samples A and \(\mathrm{B}\) are \(15 \mathrm{x}\) and \(3 \mathrm{x}\) respectively. They have equal number of initial nuclei. The ratio of the number of nuclei left in \(\mathrm{A}\) and \(\mathrm{B}\) after a time \(\frac{1}{6 \mathrm{x}}\) is
- A \(\mathrm{e}^{-2}\)
- B \(\mathrm{e}\)
- C \(\mathrm{e}^{2}\)
- D \(\mathrm{e}^{-1}\)
Answer & Solution
Correct Answer
(A) \(\mathrm{e}^{-2}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{N}_{1}=\mathrm{N}_{01} \mathrm{e}^{-\lambda_{1} \mathrm{t}}=\mathrm{N}_{0} \mathrm{e}^{-15 \mathrm{x} \times \frac{1}{6 \mathrm{x}}}=\mathrm{N}_{0} \mathrm{e}^{-5 / 2}\)
and \(\mathrm{N}_{2}=\mathrm{N}_{02} \mathrm{e}^{-\lambda_{2} \mathrm{t}}=\mathrm{N}_{0} \mathrm{e}^{-3 \mathrm{x} \times \frac{1}{6 \mathrm{x}}}=\mathrm{N}_{0} \mathrm{e}^{-1 / 2}\)
\(\therefore \quad \frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\mathrm{e}^{-2}\)
and \(\mathrm{N}_{2}=\mathrm{N}_{02} \mathrm{e}^{-\lambda_{2} \mathrm{t}}=\mathrm{N}_{0} \mathrm{e}^{-3 \mathrm{x} \times \frac{1}{6 \mathrm{x}}}=\mathrm{N}_{0} \mathrm{e}^{-1 / 2}\)
\(\therefore \quad \frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\mathrm{e}^{-2}\)
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