Question

4 × 10²³ tritium atoms are contained in a vessel. The half-life of decay tritium nuclei is 12.3 y. Find (a) the activity of the sample, (b) the number of decay in the next 10 hours (c) the number of decays in the next 6.15 y.

Answer 1

Given:
Number of tritium atoms, N₀ = 4 × 10²³
Half-life of tritium nuclei, $T_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$= 12.3 years

Disintegration constant, $\lambda =\frac{0.693}{T_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}=\frac{0.693}{12.3}$ years^$-1$

Activity of the sample $\left(A\right)$ is given by
A₀ = $\frac{dN}{dt}$ = $\lambda N_0$
$$\begin{gathered} \Rightarrow A_0=\frac{0.693}{t_{1/2}}{N}_0 \\ =\frac{0.693}{12.3}\times 4\times {10}^{23}\mathrm{disintegration}/\mathrm{year} \\ =\frac{0.693\times 4\times {10}^{23}}{12.3\times 3600\times 24\times 365}\mathrm{disintegration}/\sec \\ =7.146\times {10}^{14}\mathrm{disintegration}/\sec \end{gathered}$$

(b) Activity of the sample, A = 7.146$\times$10¹⁴ disintegration/sec

$$\begin{gathered} \mathrm{Number}\mathrm{of}\mathrm{decays}\mathrm{in}\mathrm{the}\mathrm{next}10\mathrm{h}\mathrm{ours}=7.146\times {10}^{14}\times 10\times 3600 \\ =257.256\times {10}^{17} \\ =2.57\times {10}^{19} \end{gathered}$$

(c) Number of atoms left undecayed, $N=N_0{e}^{-\lambda t}$
Here, N₀ = Initial number of atoms
$\therefore N=4\times {10}^{23}\times e^{\frac{-0.693}{12.3}\times 6.15}=2.83\times {10}^{23}$

Number of atoms disintegrated = (N_{0 $-$} N) = (4$-$2.83) × 10²³ = 1.17 × 10²³