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# Question and Answer

## QuestionPhysicsClass 12

The half-life of $${_{92}^{238}}{{}}{{}}{U}{}$$  undergoing α-decay is $$4.5\times 10^9$$  years. What is the activity of $$1g$$  sample of $${_{92}^{238}}{{}}{{}}{U}{}$$  ?

$$R=1.23\times 10^4\mathit{Bq}$$
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## Solution

Solution -
The half-life of $${_{92}^{238}}{{}}{{}}{U}{}$$  undergoing α-decay is
$${T}_{1/2}=4.5\times {10}^{9}\;years$$
or$$T_{1/2}=1.42\times 10^{17}s$$(By converting into seconds)
Let$$R =$$ activity of given sample.
One mole of any isotope contains Avogadro’s number of atoms. Now we find number of atoms in $$1g$$ of $${_{92}^{238}}{{}}{{}}{U}{}$$ which is given by $$N=\frac{1}{238\times {10}^{-3}}\times 6.023\times {10}^{24}\;=\;25.3\times {10}^{20}\;\mathit{a}\mathit{t}\mathit{o}\mathit{m}\mathit{s}$$
the decay rate $$R$$ is given by
$$R=\mathit{\lambda N}$$
Now $$decay\;constant(\lambda )\;=\;\dfrac{0.693}{T{}_{\frac{1}{2}}}$$
So, $$R=\left(\frac{0.693}{T_{1/2}}\right)N$$
or  $$R=\left(\frac{0.693}{T_{1/2}}\right)N$$
On putting the values we have,
$$R=\left(\frac{0.693}{1.42\times 10^{17}}\right)\times 25.3\times 10^{20}$$
or $$R=1.23\times 10^4\mathit{Bq}$$.
Hence activity of given sample of uranium is$$R=1.23\times 10^4\mathit{Bq}$$.