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Question

$$_{92}U^{238}$$ decays to $$_{90}Th^{234}$$ with half-life $$4.5\times 10^9$$ year. The resulting $$_{90}Th^{234}$$ is in excited state and hence, emits a further a gamma ray to come to the ground state, with half-life $$10^{-8}\ s$$. A sample of $$_{92}U^{238}$$ emits $$20$$ gamma rays per second. In what time, the emission rate will drop to $$5$$ gamma ray per second ?

A
$$9\times 10^9$$ year
B
$$2\times 10^{-8}\ s$$
C
$$0.25\times 10^{-8}\ s$$
D
$$1.125\times 10^9$$ year
Solution
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Correct option is C. $$9\times 10^9$$ year
$$\dfrac{dN}{dt}=-\lambda N$$ [$$\dfrac{dN}{dt}$$ rate of decay $$N$$= no. of nucleus]

$$\Rightarrow \dfrac{dN}{N}=-\lambda dt$$

$$\Rightarrow \log N/N_0=-\lambda t$$............(1)

Now,
$$\lambda N_0=20$$ and $$\lambda N=5$$

$$N/N_0=1/4$$

Now from (1) we get

$$\Rightarrow \log 1/4=-\dfrac{\log 2}{t_{1/2}}t$$

$$\Rightarrow t=-t_{1/2}\dfrac{\log 2^{-2}}{\log 2}$$

$$\Rightarrow t=2t_{1/2}=2\times 4.5\times 10^9$$

$$=9\times 10^9$$ Years.

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