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The isotopes { }^{238} \mathrm{U} and { }^{236} \mathrm{U} occur in nature in the mass ratio 150: 1.Initially if they were found in equal mass and half lives of { }^{238} \mathrm{U} and { }^{236} \mathrm{U} are 9 \times 10^{9} and 1 \times 10^{9} year respectively, then the age of earth is approx.?

Option: 1

8.13 \times 10^{9} \mathrm{years}


Option: 2

4.23 \times 10^{9} \mathrm{years}


Option: 3

3.26 \times 10^{8} \mathrm{years}


Option: 4

2.54 \times 10^{8} \mathrm{years}


Answers (1)

best_answer

\mathrm{For \, { }^{238} \mathrm{U}, \lambda_{1} t=\ln \left(\frac{n_{0}}{n_{1}}\right)}
\mathrm{\text { For }{ }^{236} U, \quad \lambda_{2} t =\ln \left(\frac{n_{0}^{\prime}}{n_{2}}\right)}
\mathrm{\left(\lambda_{1}-\lambda_{2}\right) t =\ln \left(\frac{n_{0}}{n_{1}} \times \frac{n_{2}}{n_{0}^{\prime}}\right) \quad\left\{\because n_{0}=n_{0}^{\prime}\right\}}
\mathrm{\ln 2\left(\frac{1}{9 \times 10^{9}}-\frac{1}{1 \times 10^{9}}\right) t=\ln \left(\frac{1}{150}\right) }

                                                      \mathrm{t =\frac{9}{8} \frac{\ln (150)}{\ln (2)} \times 10^{9} }

                                                      \mathrm{t =8.13 \times 10^{9} \text { years } }

Posted by

Ritika Jonwal

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