Question

Determine the mass of $N{a}^{22}$ which has an activity of $5mCl$. Half life of $N{a}^{22}$ is 2.6 years. Avagadro number $=6.023\times {10}^{23}$ atoms.

Answer 1

We know $1Ci=3.7\times {10}^{10}$ dps

Activity of $N{a}^{22}$ $A=5\times {10}^{-3}\times (3.7\times {10}^{10})$ dps

We get $A=1.85\times {10}^8$ dps

Half life $T_{1/2}=2.6\times 365\times 86400=8.2\times {10}^7$ s

From radioactive decay equation, activity of radioactive substance $A=\frac{0.693}{T_{1/2}}N$

$\therefore$ $1.85\times {10}^8=\frac{0.693}{8.2\times {10}^7}N$

$⟹$ $N=2.2\times {10}^{16}$ $N{a}^{22}$ atoms

Mass of $N{a}^{22}$ $m=\frac{N}{N_A}M$ where $M=23$ g (atomic weight of $N{a}^{22}$ atom)

$\therefore$ $m=\frac{2.2\times {10}^{16}}{6.023\times {10}^{23}}\times 23$

$⟹m=8.4\times {10}^{-6}g$ $=8.4\mu g$