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The half-life of $\mathrm{Ra}^{226}$ is 1620 years. Then the number of atoms decay in one second in $1 \mathrm{~g}$ of radium (Avogadro number $=6.023 \times 10^{23}$ )
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The correct answer is:
$3.61 \times 10^{10}$
The number of atoms decay in one second $\frac{d N}{d t}=\lambda N$
$\begin{aligned} & =\frac{0.693}{1620 \times 365 \times 86 \times 400} \times \frac{6.023 \times 10^{23}}{226} \\ & =3.61 \times 10^{10}\end{aligned}$
$\begin{aligned} & =\frac{0.693}{1620 \times 365 \times 86 \times 400} \times \frac{6.023 \times 10^{23}}{226} \\ & =3.61 \times 10^{10}\end{aligned}$
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