Let \(N_\beta\) be the number of \(\beta\) particles emitted by 1 gram of \(^{24}\mathrm{Na}\) radioactive nuclei (half life = 15 hrs) in 7.5 hours, \(N_\beta\) is close to (Avogadro number \(= 6.023\times10^{23}\)/g. mole) :
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- \(6.2\times10^{21}\)
- \(7.5\times10^{21}\)
- \(1.25\times10^{22}\)
- \(1.75\times10^{22}\)
The Correct Option is B
Solution and Explanation
Elapsed time \(t = 7.5\) h \(= \frac{T_{1/2}}{2}\), so \(N = N_0 \left(\frac12\right)^{1/2} = 0.7071\,N_0\).
Nuclei decayed: \(N_0 - N = 0.2929\,N_0 = 0.2929 \times 2.51\times10^{22} \approx 7.35\times10^{21}\).
Each decay emits one \(\beta\) particle, so \(N_\beta \approx 7.35\times10^{21}\), closest to \(7.5\times10^{21}\).
Answer: (2) \(7.5\times10^{21}\)
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