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
$n\log 2=\log 10-\log 6$
$\Rightarrow n\times 0.3=1-0.778=0.22$
$\Rightarrow n=\frac{0.222}{0.3}=0.74$
So, $t=n{T}_{12}=0.74\times 4.47\times 10^8$
or $t=3.3\times 10^8$ yr