Question

The half life for $\alpha$-decay of uranium $_{92}U^{228}$ is $4.47 \times 10^8 \, yr$. If a rock contains 60% of original $_{92}U^{228}$ atoms, then its age is
[take $\log \, 6 = 0.778, \log \, 2 = 0.3$]

• A. $1.2 \times 10^7 yr $
• B. $3.3 \times 10^8 yr $
• C. $4.2 \times 10^9 yr $
• D. $6.5 \times 10^9 yr $

Answer 1

Answer: (B)

Given : $T_{1/2} = 4.47 \times 10^8$ yr
$\frac{N}{N_{0} } = \frac{60}{100} = \left(\frac{1}{2}\right)^{n} $
$\Rightarrow 2^{n} = \frac{10}{6} $
Apply logarithm on both sides
$ n \log2 = \log10 - \log6 $
$ \Rightarrow n \times0.3 = 1 - 0.778 = 0.22$
$ \Rightarrow n = \frac{0.222}{0.3} = 0.74 $
So, $t = nT_{12} = 0.74 \times4.47 \times10^{8}$
or $ t = 3.3 \times10^{8} $ yr