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Question

(C) The rate of decay of a radioactive substance is given by the law of radioactive decay: $\frac{dN}{dt} = -\lambda N$.Here,the magnitude of the rate

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$\frac{0.693 N}{n} \, s$

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(C) The rate of decay of a radioactive substance is given by the law of radioactive decay: $\frac{dN}{dt} = -\lambda N$.Here,the magnitude of the rate of emission of alpha particles is given as $n = |\frac{dN}{dt}|$.Therefore,$n = \lambda N$.From this,the decay constant $\lambda$ is calculated as $\lambda = \frac{n}{N}$.The half-life $T_{1/2}$ of a radioactive element is related to the decay constant by the formula $T_{1/2} = \frac{\ln(2)}{\lambda} \approx \frac{0.693}{\lambda}$.Substituting the value of $\lambda$ into the half-life formula,we get $T_{1/2} = \frac{0.693}{n/N} = \frac{0.693 N}{n} \, s$.

$N$ atoms of a radioactive element emit $n$ alpha particles per second. The half-life of the element is:

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