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BITSAT 2015

Exam: BITSAT
Year: 2015
Subject: Physics
Chapter: Nuclear Physics

BITSAT 2015 · Physics · Nuclear Physics

In an ore containing uranium, the ratio of U^ 238 to Pb^ 206 is 3....

In an ore containing uranium, the ratio of

U^{238}

to

\mathrm{Pb}^{206}

is 3. Calculate the àge of the ore, assuming that all the lead present in the ore is the final stable product of

U^{238}

. Take the half-life of

U^{238}

to be

4.5 \times 10^{9} \mathrm{yr}

.

Options

A $1.6 \times 19^{3} \mathrm{yr}$
B $1.5 \times 10^{4} \mathrm{yr}$
C $1.867 \times 10^{9} \mathrm{yr}$
D $2 \times 10^{5} \mathrm{yr}$

Correct answer

C. $1.867 \times 10^{9} \mathrm{yr}$

Step-by-step solution

Let the initial mass of uranium be

\mathrm{M}_{0}

Final mass of uranium after time

\mathrm{t}\mathrm{M}=\frac{3}{4} \mathrm{M}_{0}

According to the law of radioactive disintegration.

\begin{array}{l}\frac{\mathrm{M}}{\mathrm{M}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}} \Rightarrow \frac{\mathrm{M}_{0}}{\mathrm{M}}=(2)^{\mathrm{t}/ \mathrm{T}} \\ \therefore \log _{10}\left(\frac{\mathrm{M}_{0}}{\mathrm{M}}\right)=\frac{\mathrm{t}}{\mathrm{T}} \log _{10}(2) \\ \mathrm{t}=\mathrm{T} \frac{\log _{10}\left(\frac{\mathrm{M}_{0}}{\mathrm{M}}\right)}{\log _{10}(2)}=\frac{\mathrm{T} \log _{10}\left(\frac{4}{3}\right)}{\log _{10}(2)}\\ =\frac{\mathrm{T} \log _{10}(1.333)}{\log _{10}(2)}=4.5 \times10^{9}\left(\frac{0.1249}{0.3010}\right)\\ \Rightarrow \mathrm{t}=1.867 \times10^{9} \mathrm{yr} . \end{array}

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