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  • A radioactive material decays by simultaneous emission of two particles with half lives 1620 and 810 years. The time after which one fourth of the material remains, is<s

A radioactive material decays by simultaneous emission of two particles with half lives 1620 and 810 years. The time after which one fourth of the material remains, is

Option: 1

1080 years


Option: 2

2430 years


Option: 3

3240 years 


Option: 4

4840 years


Answers (1)

best_answer

Decay of radioactive material follows first order kinetics.

\lambda=\lambda_1+\lambda_2

\frac{.693}{t_{1 / 2}}=\frac{.693}{\left(t_{1 / 2}\right)_1}+\frac{.693}{\left(t_{1 / 2}\right)_2}

\frac{1}{t_{1 / 2}}=\frac{1}{\left(t_{1 / 2}\right)_1}+\frac{1}{\left(t_{1 / 2}\right)_2}

=\frac{1}{1620}+\frac{1}{810}=\frac{1+2}{1620}

\mathrm{t}_{1 / 2}=540 \, \mathrm{yrs} .

\text { So } 1 / 4^{\text {th }} \text { of the material remains after } 540 \times 2=1080 \text { years. }

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