Question

The half-life of a radioisotope is 10 h. Find the total number of disintegration in the tenth hour measured from a time when the activity was 1 Ci.

Answer 1

Given:
Half-life of radioisotope, $T_{1/2}$ = 10 hrs
Initial activity, A₀ = 1 Ci
Disintegration constant, $\lambda =\frac{0.693}{10\times 3600}{\mathrm{s}}^{-1}$
Activity of radioactive sample, $A=A_0{e}^{-\lambda t}$
Here, A₀ = Initial activity
$\lambda$ = Disintegration constant
t = Time taken

After 9 hours,
Activity, $A=A_0{e}^{-\lambda t}=1\times e^{-\frac{0.693}{10\times 3600}\times 9}=0.536\mathrm{Ci}$

∴ Number of atoms left, N = $\frac{A}{\lambda }$ = $\frac{0.536\times 10\times 3.7\times {10}^{10}\times 3600}{0.693}=103.023\times {10}^{13}$
After 10 hrs,

$$\begin{gathered} Activity,A\mathit{"}=A_0{e}^{-\lambda t} \\ =1\times e^{\frac{-0.693}{10}\times 10}=0.5\mathrm{Ci} \end{gathered}$$

Number of atoms left after the 10^th hour $\left(N"\right)$ will be
$$\begin{gathered} A"=\lambda N" \\ N"=\frac{A"}{\lambda } \\ =\frac{0.5\times 3.7\times {10}^{10}\times 3.600}{0.693/10} \\ =26.37\times {10}^{10}\times 3600=96.103\times {10}^{13} \end{gathered}$$

Number of disintegrations = (103.023 − 96.103) × 10¹³
= 6.92 × 10¹³