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JEE Mains · Physics · STD 12 - 13. Nuclei
Let \(N_{\beta}\) be the number of \(\beta \) particles emitted by \(1\) gram of \(Na^{24}\) radioactive nucler (half life \(= 15\, hrs\)) in \(7.5\, hours\), \(N_{\beta}\) is close to (Avogadro number \(= 6.023\times10^{23}\,/g.\, mole\))
- A \(6.2\times10^{21}\)
- B \(7.5\times10^{21}\)
- C \(1.25\times10^{22}\)
- D \(1.75\times10^{22}\)
Answer & Solution
Correct Answer
(B) \(7.5\times10^{21}\)
Step-by-step Solution
Detailed explanation
We know that \(N_{\beta}=N_{0}\left(1-e^{-\lambda t}\right)\) \(N_{\beta}=\frac{6.023 \times 10^{23}}{24} \cdot\left[1-e^{\left(-\frac{\ln ^{2} }{15} \times 75\right)}\right]\) on solving we get, \(\mathrm{N}_{\beta}=7.4 \times 10^{21}\)
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