The half life of \( {}^{238}\mathrm{U} \) when 1 gram of it emits \( 1.23\times10^{4} \) alpha particle per second is: (taking 1 year \( =3.155\times10^{7} \) second)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 4.519\times10^{9} \) years
The activity of a sample is \( A = \lambda N \), and the half-life is \( T_{1/2} = \frac{0.693}{\lambda} \). One alpha particle is emitted per decay, so the rate of emission is the activity.
Number of nuclei in 1 g (molar mass 238 g)
\( \displaystyle N = \frac{1}{238}\times6.022\times10^{23} = 2.53\times10^{21} \)
Decay constant
\( \displaystyle \lambda = \frac{A}{N} = \frac{1.23\times10^{4}\,\mathrm{s^{-1}}}{2.53\times10^{21}} = 4.86\times10^{-18}\,\mathrm{s^{-1}} \)
Half-life
\( \displaystyle T_{1/2} = \frac{0.693}{4.86\times10^{-18}\,\mathrm{s^{-1}}} = 1.43\times10^{17}\,\mathrm{s} \)
\( \displaystyle T_{1/2} = \frac{1.43\times10^{17}}{3.155\times10^{7}}\,\mathrm{years} = 4.52\times10^{9}\,\mathrm{years} \)