Question

The activity for a sample of radioactive element found in an experiment is $$6400$$ decays per minute. Repeating this experiment after $$6$$ days, activity was found to be $$400$$ decays per minute. Calculate half life of the given element.

Solution

We know that Activity $$R=\lambda N$$where,$$N=N {_0e^{-\lambda t}}$$$$R=\lambda N {_0e^{-\lambda t}}$$$$R=R {_0e^{-\lambda t}}$$$$\dfrac{R}{R_0}=e^{-\lambda t}$$....................$$(1)$$From Rutherford soddy law, $$N=N {_0e^{-\lambda t}}$$$$\dfrac{N}{N_0}= e^{-\lambda t}$$......$$(2)$$From equation $$(1)$$ and $$(2)$$,$$\dfrac{R}{R_0}=\dfrac{N}{N_0}$$where, $$N=\dfrac{N_0}{2^n}$$        $$\Rightarrow \dfrac{R}{R_0}=\dfrac{1}{2^{t/T_{1/2}}}$$$$N=\dfrac{N_0}{2^{t/T_{1/2}}}$$$$\Rightarrow \dfrac{1}{16}=\dfrac{1}{2^{6/T_{1/2}}}$$$$\dfrac{N}{N_0}=\dfrac{1}{2^{t/T_{1/2}}}$$$$\Rightarrow \dfrac{1}{2^4}=\dfrac{1}{2^{6/T_{1/2}}}$$Given,$$\dfrac{R}{R_0}=\dfrac{400}{6400}$$     $$\Rightarrow \dfrac{R}{R_0}=\dfrac{1}{16}$$$$t=6days$$Comparing powers$$\dfrac{6}{T_{1/2}}=4$$$$T_{1/2}=\dfrac{6}{4}$$$$T_{1/2}=1.5days$$PhysicsNCERTStandard XII

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