Question

An uranium mineral containing ${}^{238}\text{U}$ and ${}^{206}\text{Pb}$ in the ratio 4 : 1 by mass.$t_{1/2}{}^{238}\text{U}=4.5\times {10}^{9}year$. Assume all the lead present in mineral is formed from disintegration of ${}^{238}\text{U}$. Age of mineral is:

• A. $[1.648\times {10}^{9}year]$
• B. $[3.656\times {10}^{9}year]$
• C. $[4.243\times {10}^{9}year]$
• D. $[1.449\times {10}^{9}year]$

Answer 1

The correct option is D $[1.449\times {10}^{9}year]$

$\lambda =\frac{0.693}{t_{1/2}}=\frac{0.693}{4.5\times {10}^9}$
$N_0=U-238+pb-206=4+1=5$
$N=U-238=4$
$t=\frac{2.303}{\lambda }log\frac{N_0}{N}=\frac{2.303}{\frac{0.693}{4.5\times {10}^9}}log\frac{5}{4}=1.449\times {10}^{9}year$