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The half life for $\alpha$-decay of uranium ${ }_{92} \mathrm{U}^{228}$ is $4.47 \times 10^{8} \mathrm{yr}$. If a rock contains $60 \%$ of original ${ }_{92} \mathrm{U}^{228}$ atoms, then its age is [take $\log 6=0.778, \log 2=0.3$ ]
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The correct answer is:
$3.3 \times 10^{8} \mathrm{yr}$
Given: $\mathrm{T}_{1 / 2}=4.47 \times 10^{8} \mathrm{yr}$
$\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{60}{100}=\left(\frac{1}{2}\right)^{\mathrm{n}} \Rightarrow 2^{\mathrm{n}}=\frac{10}{6}$
Apply logarithm on both sides
$\begin{aligned}
& \mathrm{n} \log 2=\log 10-\log 6 \\
\Rightarrow & \mathrm{n} \times 0.3=1-0.778=0.22 \\
\Rightarrow & \mathrm{n}=\frac{0.222}{0.3}=0.74
\end{aligned}$
So, $\mathrm{t}=\mathrm{nT}_{1 / 2}=0.74 \times 4.47 \times 10^{8}$ or, $\mathrm{t}=3.3 \times 10^{8} \mathrm{yr}$
$\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{60}{100}=\left(\frac{1}{2}\right)^{\mathrm{n}} \Rightarrow 2^{\mathrm{n}}=\frac{10}{6}$
Apply logarithm on both sides
$\begin{aligned}
& \mathrm{n} \log 2=\log 10-\log 6 \\
\Rightarrow & \mathrm{n} \times 0.3=1-0.778=0.22 \\
\Rightarrow & \mathrm{n}=\frac{0.222}{0.3}=0.74
\end{aligned}$
So, $\mathrm{t}=\mathrm{nT}_{1 / 2}=0.74 \times 4.47 \times 10^{8}$ or, $\mathrm{t}=3.3 \times 10^{8} \mathrm{yr}$
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