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Beta Decay

_921U^238 by...

Question

$_{921} U^{238}$ by successive radioactivity decays changes to $_{82} P b^{206}$ .A sample of uranium ore was analyzed and found to contain 1.0g of $U^{238}$ and $0.1$ g of $P b^{206}$ . Assuming that all the $P b^{206}$ had accumulated due to decay of $U^{238}$ ,find out the age of the ore ?(Half life of $U^{238} = 4.5 * 10^{9} y e a r s) .$

t = 7.1 * 10^{5}

years

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t = 7.1 * 10^{8}

years

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t = 7.1 * 10^{6}

years

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t = 7.1 * 10^{3}

years

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Solution

The correct option is B $t = 7.1 * 10^{8}$ years
Number of moles of U- $238$ present $= \frac{1.0}{238} = 0.0042$

Number of moles of Pb- $206$ present $= \frac{0.1}{206} = 0.000485$

$λ = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{4.5 \times 10^{9}}$

$t = \frac{2.303}{λ} l o g \frac{a}{a - x}$

$t = \frac{2.303}{0.693} \times 4.5 \times 10^{9} \times l o g \frac{0.0042 + 0.000485}{0.0042}$

$t = 7.1 \times 10^{8}$ years

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921U238 by successive radioactivity decays changes to 82Pb206.A sample of uranium ore was analyzed and found to contain 1.0g of U238and 0.1g of Pb206. Assuming that all the Pb206had accumulated due to decay of U238,find out the age of the ore ?(Half life of U238=4.5∗109years).

The correct option is B t=7.1∗108 yearsNumber of moles of U-238 present =1.0238=0.0042Number of moles of Pb-206 present =0.1206=0.000485λ=0.693t1/2=0.6934.5×109t=2.303λlogaa−xt=2.3030.693×4.5×109×log0.0042+0.0004850.0042t=7.1×108 years