A heavy nucleus Q of half-life 20 text{ minut | vedclass

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(D) The total number of nuclei $N_0 = 1000$. Since $60 \%$ of the nuclei undergo alpha-decay,the initial number of nuclei available for alpha-decay is

Vedclass · Accepted Answer

$525$

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(D) The total number of nuclei $N_0 = 1000$. Since $60 \%$ of the nuclei undergo alpha-decay,the initial number of nuclei available for alpha-decay is $N_{0,\alpha} = 1000 	imes 0.60 = 600$.The decay constant $\lambda$ is given by $\lambda = \frac{\ln 2}{t_{1/2}} = \frac{\ln 2}{20 	ext{ min}}$.The time elapsed is $t = 1 	ext{ hour} = 60 	ext{ minutes}$.The number of nuclei remaining after time $t$ is given by $N(t) = N_{0,\alpha} e^{-\lambda t}$.Substituting the values: $N(60) = 600 	imes e^{-\left(\frac{\ln 2}{20}ight) 	imes 60} = 600 	imes e^{-3 \ln 2} = 600 	imes e^{\ln(2^{-3})} = 600 	imes 2^{-3} = 600 	imes \frac{1}{8} = 75$.The number of nuclei that have undergone alpha-decay is the difference between the initial number and the remaining number: $\Delta N = N_{0,\alpha} - N(60) = 600 - 75 = 525$.

$A$ heavy nucleus $Q$ of half-life $20 \text{ minutes}$ undergoes alpha-decay with a probability of $60 \%$ and beta-decay with a probability of $40 \%$. Initially,the number of $Q$ nuclei is $1000$. The number of alpha-decays of $Q$ in the first one hour is:

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