MHT CET · Physics · Nuclear Physics
For a substance, fraction of its initial quantify \(\left(\mathrm{N}_0\right)\) which will disintegrate in its average life time is about \((\mathrm{e}=2.71)\)
- A \(\left(\frac{1}{3}\right) \mathrm{N}_0\)
- B \(\left(\frac{1}{2}\right) \mathrm{N}_0\)
- C \(\left(\frac{2}{3}\right) \mathrm{N}_0\)
- D \((0.9) \mathrm{N}_0\)
Answer & Solution
Correct Answer
(C) \(\left(\frac{2}{3}\right) \mathrm{N}_0\)
Step-by-step Solution
Detailed explanation
For a radioactive substance, \(\mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda t}\)
Given the mean life, \(\mathrm{t}=\frac{1}{\lambda}\)
\(\therefore \mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda \times \frac{1}{\lambda}}=\frac{\mathrm{N}_0}{\mathrm{e}}=\frac{\mathrm{N}_0}{2.71}=0.37 \mathrm{~N}_0\)
\(0.37 \mathrm{~N}_0\) is the amount of radioactive substance left after time \(\tau\).
\(\therefore\) The amount disintegrated in time \(\tau\)
\(=\mathrm{N}_0-0.37 \mathrm{~N}_0=0.63 \mathrm{~N}_0=\frac{2}{3} \mathrm{~N}_0\)
Given the mean life, \(\mathrm{t}=\frac{1}{\lambda}\)
\(\therefore \mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda \times \frac{1}{\lambda}}=\frac{\mathrm{N}_0}{\mathrm{e}}=\frac{\mathrm{N}_0}{2.71}=0.37 \mathrm{~N}_0\)
\(0.37 \mathrm{~N}_0\) is the amount of radioactive substance left after time \(\tau\).
\(\therefore\) The amount disintegrated in time \(\tau\)
\(=\mathrm{N}_0-0.37 \mathrm{~N}_0=0.63 \mathrm{~N}_0=\frac{2}{3} \mathrm{~N}_0\)
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