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  • The present day abundances of the isotopes and <img alt="\mathrm{U}^{235}" src="htt

The present day abundances of the isotopes \mathrm{U}^{238} and \mathrm{U}^{235} are in the ratio of 128: 1 . They have half lives of 4.5 \times 10^9 years and 7 \times 10^8 years respectively. If age of earth in \frac{49 \mathrm{X}}{76} \times 10^9years, then \mathrm{X} is (Assume equal amount of each isotope existed at the time of formation of the earth)

Option: 1

3


Option: 2

6


Option: 3

9


Option: 4

12


Answers (1)

best_answer

128 \mathrm{~N}=\mathrm{N}_0 \mathrm{e}^{-\lambda_1t}                      ...(i)

\mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda_2 {\mathrm{t}}}                               ...(ii)

128=\mathrm{e}^{-\lambda_1 \mathrm{t}+\lambda_2 \mathrm{t}}

\Rightarrow \quad \ln 128=\left(-\frac{\ln 2}{4.5 \times 10^9}+\frac{\ln 2}{7 \times 10^8}\right) \mathrm{t}

\Rightarrow \quad 7 \ln 2=\ln 2\left[\frac{1}{7 \times 10^8}-\frac{1}{4.5 \times 10^9}\right] \mathrm{t}

\Rightarrow \quad 7=\left[\frac{(45-7)}{7 \times 45 \times 10^8}\right] \mathrm{t} \quad \Rightarrow \quad \frac{49 \times 9}{38 \times 2} \times 10^9=\mathrm{t}

\text { Hence } \mathrm{X}=9

Posted by

avinash.dongre

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