Question

A radioactive sample contains 2.3 mg of pure ${}_6^{11}C$ which has half-life period of 1224 second. Calculate
i) the number of atoms present initially
ii) the activity when $5\mu g$ of the sample will be left.

Answer 1

molar mass of ${}_6^{11}C$ is 11g/mol.

So, no of molar of $c-11$ present in 2.3mg is = $\frac{2.3\times {10}^{-3}}{11}$ mol

$\therefore$ No.of atom present initially, No = moles $\times$ Avogadro'n No(NA)

$=\frac{2.3\times {10}^{-3}}{10}\times 6.022\times {10}^{23}$

No $=1.26\times {10}^{20}\to Ans\left(i\right)$

Given half life, $T=1224s.$ no Rate constant, of decay :-

$\lambda =\frac{m_2}{2}=\frac{0.693}{1224}=5.67\times {10}^{-4}{5}^{-1}$

So, when $5\mu g$ of sample is present, the activity in given by :-

$A=\lambda \times N$ (N: No of atoms left for decay)

$=\lambda \times \frac{5\times {10}^{-6}}{11}\times NA$ (obtain no.of atom is $5\mu g$ c-11 sample)

$\Rightarrow A=5.67\times {10}^{-4}\times \frac{5\times {10}^{-6}}{11}\times 6.022\times {10}^{33}$

$\Rightarrow A=1.55\times {10}^{14}dps\to$ (Ans)