You visited us 1 times! Enjoying our articles? Unlock Full Access!

Magnetic Susceptibility

In an ore con...

Question

In an ore containing uranium the ratio of $U^{238}$ to $P b^{206}$ nuclei in $3$ . Calculate the age of ore, assuming that all the lead present in the ore is the final stable product of $U^{238} . t_{1 / 2}$ for $U^{238} = 4.5 \times 10^{9} y e a r$ .

Open in App

Solution

Given: $\frac{N_{U}}{N_{P b}} = 3$
Let the initial number of uranium be $M_{0}$ .
Final number of uranium after time $t, M = \frac{3}{4} M_{0}$
According to law of disintegration:
$\frac{M}{M_{0}} = {(\frac{1}{2})}^{t / t_{1 / 2}}$

$\Rightarrow \frac{M_{0}}{M} = (2)^{t / t_{1 / 2}}$
$∴ l o g_{10} \frac{M_{0}}{M} = \frac{t}{t_{1 / 2}} l o g_{10} 2$
$t = t_{1 / 2} \frac{l o g_{10} \frac{M_{0}}{M}}{l o g_{10} 2} = t_{1 / 2} \frac{4 l o g_{10} \frac{4}{3}}{l o g_{10} 2}$

$= t_{1 / 2} \frac{l o g_{10} 1.3333}{l o g_{10} 2}$
$= 4.5 \times 10^{9} (\frac{0.1249}{0.3010})$
$= 1.867 \times 10^{9} y e a r s$

Suggest Corrections

Similar questions

U^{238}

P b^{206}

3

U^{238}

U^{238}

4.5 \times 10^{9}

^{238} U

^{206} P b

3

^{238} U

^{238} U

4.5 \times 10^{9} y e a r s

_{92} U^{238}

_{82} {Pb}^{206}

U^{238}

P b^{206}

P b^{206}

U^{238}

U^{238} = 4.5 \times 10^{9} y e a r s

U^{238}

P b^{206}

P b^{206}

4.5 \times 10^{9}

U^{238}

= 4.5 \times 10^{9}

23.8 g

U^{238}

20.6 g

P b^{206}

X \times 10_{9}

2 X

Join BYJU'S Learning Program