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  • A radioactive isotope A with a half life of years decays into which is st

A radioactive isotope A with a half life of 1.25 \times 10^{10} years decays into B which is stable. A sample of rock from a planet is found to contain both \mathrm{A} and \mathrm{B} present in the ratio 1: 15. The age of the rock is:

Option: 1

9.6 \times 10^{10}\, years


Option: 2

4.2 \times 10^{10}\, years


Option: 3

5 \times 10^{10}\, years


Option: 4

1.95 \times 10^{10}\, years


Answers (1)

best_answer

According to Rutherford and Soddy law for radioactive decay,

Number of atoms remained undecayed after time t is

\mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}}

\mathrm{e}^{\lambda \mathrm{t}}=\frac{\mathrm{N}_0}{\mathrm{~N}} \quad \text { or } \quad \lambda t=\ln \frac{\mathrm{N}_0}{\mathrm{~N}} \quad \text { or } \quad \mathrm{t}=\frac{1}{\lambda} \ln \frac{\mathrm{N}_0}{\mathrm{~N}} \quad \text { or, } \quad \mathrm{t}=\frac{\mathrm{T}_{1 / 2}}{\ln 2} \ln \frac{\mathrm{N}_0}{\mathrm{~N}}

=\frac{1.25 \times 10^{10}}{\ln 2} \ln \frac{16}{1}=\frac{1.25 \times 10^{10} \times 4 \times \ln 2}{\ln 2}=5 \times 10^{10} \text { years }

Posted by

Kuldeep Maurya

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