Question

(a) On analysis a sample of uranium ore was found to contain 0.277g of ${}_{82}P{b}^{206}$ and 1.667g of ${}_{92}{U}^{238}$.The half life period of $U^{238}$ is $4.51\times {10}^9$ year. If all the lead were assumed to have come from decay of ${}_{92}{U}^{238}$. What is the age of the earth?
(b) An ore of ${}_{92}{U}^{238}$ is found to contain ${}_{92}{U}^{238}$ and ${}_{92}P{b}^{206}$ in the weight ratio of 1:0.1.The half life period of ${}_{92}{U}^{238}$ is $4.5\times {10}^9$ year. Calculate the age of ore?

• A. $\left(a\right)1.147\times {10}^{9}year,\left(b\right)7.044\times {10}^8$ year
• B. $\left(a\right)11.43\times {10}^{11}year,\left(b\right)7.097\times {10}^8$ year
• C. $\left(a\right)1.143\times {10}^{10}year,\left(b\right)7.097\times {10}^9$ year
• D. None of these

Answer 1

Answer: (A)

$\left(a\right)1.147\times {10}^{9}year,\left(b\right)7.044\times {10}^8$ year

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

$N=\frac{1.667}{238}=0.007$

$N_0=\frac{1.667}{238}+\frac{0.277}{206}=0.007+0.00135=0.00835$

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

$t=\frac{2.303}{0.693}\times 4.51\times {10}^9\times log\frac{0.00835}{0.007}=1.147\times {10}^{9}years$

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

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

$N_0=\frac{1}{238}+\frac{0.1}{206}=0.0042+0.00048=0.00468$

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

$t=\frac{2.303}{0.693}\times 4.51\times {10}^9\times log\frac{0.00468}{0.0042}=7.044\times {10}^{8}years$