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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 \({_{92}^{238}}{{}}{{}}{U}{}\)  ?

Answer

\(R=1.23\times 10^4\mathit{Bq}\)
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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\(R=1.23\times 10^4\mathit{Bq}\).
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