Let N_{beta} be the number of beta particles | vedclass

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(B) The initial number of nuclei $N_0$ is given by $N_0 = \frac{m}{M} 	imes N_A = \frac{1}{24} 	imes 6.023 	imes 10^{23} \approx 2.51 	imes 10^{22

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$7.5 	imes 10^{21}$

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(B) The initial number of nuclei $N_0$ is given by $N_0 = \frac{m}{M} 	imes N_A = \frac{1}{24} 	imes 6.023 	imes 10^{23} \approx 2.51 	imes 10^{22}$.The number of nuclei decayed in time $t$ is $N_{\beta} = N_0(1 - e^{-\lambda t})$.Given $t = 7.5 \, hrs$ and $T_{1/2} = 15 \, hrs$,the decay constant $\lambda = \frac{\ln 2}{T_{1/2}}$.Thus,$N_{\beta} = N_0(1 - e^{-\frac{\ln 2}{15} 	imes 7.5}) = N_0(1 - e^{-\frac{\ln 2}{2}}) = N_0(1 - 2^{-1/2})$.Using $2^{-1/2} \approx 0.707$,we get $N_{\beta} = 2.51 	imes 10^{22} 	imes (1 - 0.707) = 2.51 	imes 10^{22} 	imes 0.293 \approx 7.35 	imes 10^{21}$.Rounding to the nearest option,$N_{\beta} \approx 7.5 	imes 10^{21}$.

Let $N_{\beta}$ be the number of $\beta$ particles emitted by $1 \, g$ of $^{24}Na$ radioactive nuclei (half-life $= 15 \, hrs$) in $7.5 \, hrs$. $N_{\beta}$ is close to (Avogadro number $= 6.023 \times 10^{23} \, mol^{-1}$)

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