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(B) Radioactive decay follows first-order kinetics. The decay constant $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{10} \ h^{-1}$.Using the first-order i

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$4.56 	imes 10^{23} \ 	ext{atoms}$

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(B) Radioactive decay follows first-order kinetics. The decay constant $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{10} \ h^{-1}$.Using the first-order integrated rate equation: $k = \frac{2.303}{t} \log \frac{N_0}{N_t}$.Given $N_0 = 1 \ g$-atom, $t = 4 \ h$, and $k = 0.0693 \ h^{-1}$.$0.0693 = \frac{2.303}{4} \log \frac{1}{N_t} \implies \log \frac{1}{N_t} = \frac{0.0693 	imes 4}{2.303} \approx 0.12036$.$\log N_t = -0.12036 = \bar{1}.87964 \implies N_t = 10^{-0.12036} \approx 0.7579 \ g$-atoms.Number of atoms $= N_t 	imes N_A = 0.7579 	imes 6.022 	imes 10^{23} \approx 4.56 	imes 10^{23} \ 	ext{atoms}$.

The half-life of a radioactive element is $10 \ hours$. How much will be left after $4 \ hours$ in a $1 \ g$-atom sample?

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