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In an ore containing uranium, the ratio of U^{238} to Pb^{206}

Question

In an ore containing uranium, the ratio of $U^{238}$ to $P b^{206}$ is $3$ . Calculate the age of the ore, assuming that all the leas present in the ore is the final stable product of $U^{238}$ . Take the half-life of $U^{238}$ to be $4.5 \times 1 0^{9}$ yr.

1.6 \times 1 9^{3} yr

1.5 \times 1 0^{4} yr

1.867 \times 1 0^{9} yr

2 \times 1 0^{5} yr

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In an ore containing uranium, the ratio of

U^{238}

P b^{206}

3

. Calculate the age of the ore, assuming that all the leas present in the ore is the final stable product of

U^{238}

. Take the half-life of

U^{238}

to be

4.5 \times 1 0^{9}

yr.

5 mins ago

Discuss this question LIVE

5 mins ago

Text solutionVerified

Let the initial mass of uranium be

M_{0}

Final mass of uranium after time

t, M = \frac{3}{4} M_{0}

According to the law of radioactive disintegration.

\frac{M}{M _{0}} = (\frac{1}{2})^{t / T} \Rightarrow \frac{M _{0}}{M} = (2)^{t / T}

∴ lo g_{10} (\frac{M _{0}}{M}) = \frac{t}{T} lo g_{10} (2)

t = T \frac{lo g _{10} ( \frac{M _{0}}{M} )}{lo g _{10} ( 2 )} = \frac{T lo g _{10} ( \frac{4}{3} )}{lo g _{10} ( 2 )}

= \frac{T lo g _{10} ( 1.333 )}{lo g _{10} ( 2 )} = 4.5 \times 1 0^{9} (\frac{0.1249}{0.3010})

\Rightarrow t = 1.867 \times 1 0^{9} yr

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Question Text	In an ore containing uranium, the ratio of $U^{238}$ to $P b^{206}$ is $3$ . Calculate the age of the ore, assuming that all the leas present in the ore is the final stable product of $U^{238}$ . Take the half-life of $U^{238}$ to be $4.5 \times 1 0^{9}$ yr.
Answer Type	Text solution:1
Upvotes	150