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
\(R=1.23\times 10^4\mathit{Bq}\)
To Keep Reading This Answer, Download the App
4.6
Review from Google Play
To Keep Reading This Answer, Download the App
4.6
Review from Google Play
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\(R=1.23\times 10^4\mathit{Bq}\).
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\(R=1.23\times 10^4\mathit{Bq}\).
To Keep Reading This Solution, Download the APP
4.6
Review from Google Play
To Keep Reading This Solution, Download the APP
4.6
Review from Google Play
Correct9
Incorrect0
Still Have Question?
Watch More Related Solutions
Load More
More Solution Recommended For You