Question 42.41: A 1.00 g sample of samarium emits alpha particles at a rate ......
A 1.00 g sample of samarium emits alpha particles at a rate of 120 particles/s. The responsible isotope is ^{147}Sm, whose natural abundance in bulk samarium is 15.0%. Calculate the half-life.
The number N of undecayed nuclei present at any time and the rate of decay R at that
time are related by R = λN, where λ is the disintegration constant. The disintegration
constant is related to the half-life T^{1/2} by \lambda=(\ln 2) / T_{1 / 2}, \text { so } R=(N \ln 2) / T_{1 / 2} and
T_{1 / 2}=(N \ln 2) / R .
Since 15.0% by mass of the sample is ^{147}Sm, the number of ^{147}Sm nuclei present in the sample is
N=\frac{(0.150)(1.00 \mathrm{~g})}{(147 \mathrm{u})\left(1.661 \times 10^{-24} \mathrm{~g} / \mathrm{u}\right)}=6.143 \times 10^{20}
Thus,
T_{1 / 2}=\frac{\left(6.143 \times 10^{20}\right) \ln 2}{120 \mathrm{~s}^{-1}}=3.55 \times 10^{18} \mathrm{~s}=1.12 \times 10^{11} \mathrm{y}.