Question

In an agricultural experiment, a solution containing 1 mole of a radioactive material (t_1/2 = 14.3 days) was injected into the roots of a plant. The plant was allowed 70 hours to settle down and then activity was measured in its fruit. If the activity measured was 1 µCi, what per cent of activity is transmitted from the root to the fruit in steady state?

Answer 1

Given:
Initial no of atoms, N₀ = 1 mole = 6 × 10²³ atoms
Half-life of the radioactive material, T_1/2 = 14.3 days
Time taken by the plant to settle down, t = 70 h

Disintegration constant, $\lambda =\frac{0.693}{t_{1/2}}$ = $\frac{0.693}{14.3\times 24}$ h^$-$1
N = N₀e^−λt
$$\begin{gathered} =6\times {10}^{23}\times e^{\frac{-0.693\times 70}{14.3\times 24}} \\ =6\times {10}^{23}\times 0.868 \\ =5.209\times {10}^{23} \end{gathered}$$

$$\begin{gathered} \mathrm{Activity,}R=\frac{dN}{dt}=5.209\times {10}^{23}\times \frac{0.693}{14.3\times 24} \\ =\frac{0.0105\times {10}^{23}}{3600}\mathrm{dis}/\mathrm{hr} \\ =2.9\times {10}^{-6}\times {10}^{23}\mathrm{dis}/\sec \\ =2.9\times {10}^{17}\mathrm{dis}/\sec \end{gathered}$$

$$\begin{gathered} \mathrm{Fraction}\mathrm{of}\mathrm{activity}\mathrm{transmitted}=\left(\frac{1\mathrm{\mu Ci}}{2.9\times {10}^{17}}\right)\times 100\% \\ =\left[\left(\frac{1\times 3.7\times {10}^4}{2.9\times {10}^{17}}\right)\times 100\right]\% \\ =1.275\times {10}^{-11}\% \end{gathered}$$