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(B) The decay constant $\lambda$ is given by $\lambda = \frac{\ln(2)}{T_{1/2}} = \frac{0.693}{6.92 \ s} \approx 0.1 \ s^{-1}$.The number of nuclei rem

Vedclass · Accepted Answer

$0.63$

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(B) The decay constant $\lambda$ is given by $\lambda = \frac{\ln(2)}{T_{1/2}} = \frac{0.693}{6.92 \ s} \approx 0.1 \ s^{-1}$.The number of nuclei remaining after time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.The fractional change in the number of nuclei is given by $\frac{N_0 - N(t)}{N_0} = 1 - e^{-\lambda t}$.Substituting the values $t = 10 \ s$ and $\lambda = 0.1 \ s^{-1}$:Fractional change $= 1 - e^{-(0.1)(10)} = 1 - e^{-1}$.Since $e^{-1} \approx 0.37$,the fractional change is $1 - 0.37 = 0.63$.

$A$ radioactive element is disintegrating with a half-life of $6.92 \ s$. The fractional change in the number of nuclei of the radioactive element during $10 \ s$ is:

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