Question

A 2.71 g sample of KCl from the chemistry stockroom is found to be radioactive, and it is decaying at a constant rate of 4490 Bq. The decays are traced to the element potassium and in particular to the isotope ${}^{40}K$, which constitutes 1.17% of normal potassium. Calculate the half-life of this nuclide.
[Given $M_K=39.102$ g/mol, $M_{Cl}=35.453$ g/mol]

• A. $1.25\times {10}^9$ year
• B. $1.25\times {10}^{12}$ year
• C. $4\times {10}^{15}$ year
• D. $4\times {10}^9$ year

Answer 1

The correct option is A

$1.25\times {10}^9$ year

$T_{\frac{1}{2}}=\frac{Nln2}{R}$

We know that N in this equation is 1.17% of the total number $N_K$ of potassium atoms in the sample. We also know that $N_K$ must equal the number $N_{KCl}$ of molecules in the sample. We can obtain $N_{KCl}$ from the molar mass $M_{KCl}$ of KCl (the mass of one mole of KCl) and the given mass Msam of the sample by combining$n=N/N_A$ and $n=M_{sam}/M$ to write.
$N_{KCl}$ =(Number of moles in sample) $N_A=\frac{M_{sam}}{M_{KCl}}{N}_A$, (41 – 22)

Where $N_A$ is Abogadro’s number $(6.02\times {10}^{23}mo{l}^{-1})$. We see that the molar mass of potassium is 39.102 g/mol and the molar mass of chlorine is 35.453 g/mol; thus, the molar mass of KCl is 74.555 g/mol.
$N_{KCl}=\frac{\left(2.71g\right)(6.02\times {10}^{23}mo{l}^{-1})}{74.555g/mol}=2.188\times {10}^{22}$

as the number of KCl molecules in the sample. Thus, the total number $N_K$ of potassium atoms is also $2.188\times {10}^{22}$, and the number of ${}^{40}K$ in the sample must be
$N=0.0117N_K=\left(0.0117\right)(2.188\times {10}^{22})$

Now,

$T_{\frac{1}{2}}=\frac{(2.560\times {10}^{20})ln2}{4490s^{-1}}$

$=3.95\times {10}^{16}s=1.25\times {10}^{9}y$.