Question

${}_{92}^{238}\text{U}$ by successive radioactive decays changes to ${}_{82}^{206}\text{Pb}$. A sample of uranium ore was analysed and found to contain 1:0 g ${}^{238}\text{U}$ and 0.1 g ${}^{206}\text{Pb}$. $t_{1/2}$ for $4.5\times {10}^{9}year$.
Assuming the ${}^{206}\text{Pb}$ has accumulated due to its decay of uranium, the age of ore is:

• A. $[7.1\times {10}^{8}year]$
• B. $[2.4\times {10}^{8}year]$
• C. $[3.3\times {10}^{8}year]$
• D. None of these

Answer 1

Answer: (A)

$[7.1\times {10}^{8}year]$

$\lambda =\frac{0.693}{t_{1/2}}=\frac{0.693}{4.51\times {10}^9}$

$N=\frac{1}{238}=0.0042$

$N_0=\frac{1}{238}+\frac{0.1}{206}=0.0042+0.00048=0.004685$

$t=\frac{2.303}{\lambda }\times log\frac{N_0}{N}$

$t=\frac{2.303}{0.693}\times 4.51\times {10}^9\times log\frac{0.004685}{0.0042}=7.097\times {10}^{8}years$