Question

The half-life of ²²⁶Ra is 1602 y. Calculate the activity of 0.1 g of RaCl₂ in which all the radium is in the form of ²²⁶Ra. Taken atomic weight of Ra to be 226 g mol⁻¹ and that of Cl to be 35.5 g mol⁻¹.

Answer 1

Given:
Half-life of radium, T_1/2 = 1602 years
Atomic weight of radium = 226 g/mole
Atomic weight of chlorine = 35.5 g/mole
Now,
1 mole of RaCl₂ = 226 + 71 = 297 g
297 g = 1 mole of RaCl₂
0.1 g = $\frac{1}{297}\times 0.1\mathrm{mole}\mathrm{of}{\mathrm{RaCl}}_2$
Total number of atoms in 0.1 g of RaCl₂, N $=\frac{0.1\times 6.023\times {10}^{23}}{297}=0.02027\times {10}^{22}$
∴ No of atoms, N = 0.02027 $\times$ 10²²

$$\begin{gathered} \mathrm{Disintegration}\mathrm{constant},\lambda =\frac{0.693}{T_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ =\frac{0.693}{1602\times 365\times 24\times 3600} \\ =1.371\times {10}^{-11} \end{gathered}$$

Activity of radioactive sample, A = $\lambda N$
= $1.371\times {10}^{-11}\times 2.027\times {10}^{20}$
= $2.8\times {10}^9\mathrm{disintegrations}/\mathrm{second}$