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Standard XII

Chemistry

First Order Reaction

92U238 by suc...

Question

$_{92} U^{238}$ by successive radioactive decays changes to $_{82} P b^{206}$ . A sample of uranium ore was analysed and found to contain $1.0 g U^{238}$ and $0.1 g P b^{206}$ has accumulated due to decay of uranium, find out the age of ore. $t_{1 / 2}$ for $U^{238} = 4.5 \times 10^{9} y e a r$ .

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Solution

$U^{238} \to P b^{206}$
Weight at $t = t 1 g 0.1 g$
Mole at $t = t \frac{1}{238} \frac{0.1}{206}$
$∴ N_{0 U^{238}} = \frac{1}{238} + \frac{0.1}{206}$ and $N_{U^{238}} = \frac{1}{238}$
$t = \frac{2.303}{K} log \frac{N_{0}}{N}$
$= \frac{2.303 \times 4.5 \times 10^{9}}{0.693} log \frac{\frac{1}{238} + \frac{0.1}{206}}{\frac{1}{238}}$
$= 7.1 \times 10^{8} y r$ .

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92U238 by successive radioactive decays changes to 82Pb206. A sample of uranium ore was analysed and found to contain 1.0 g U238 and 0.1 g Pb206 has accumulated due to decay of uranium, find out the age of ore. t1/2 for U238=4.5×109year.

U238→Pb206Weight at t=t 1g 0.1 gMole at t=t 1238 0.1206∴N0U238=1238+0.1206 and NU238=1238t=2.303KlogN0N=2.303×4.5×1090.693log1238+0.12061238=7.1×108yr.

$_{92} U^{238}$ by successive radioactive decays changes to $_{82} P b^{206}$ . A sample of uranium ore was analysed and found to contain $1.0 g U^{238}$ and $0.1 g P b^{206}$ has accumulated due to decay of uranium, find out the age of ore. $t_{1 / 2}$ for $U^{238} = 4.5 \times 10^{9} y e a r$ .