Question

In an agricultural experiment, a solution containing 1 mole of a radioactive material $(t_{\frac{1}{2}}$, = 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 $1\mu Ci$, what per cent of activity is transmitted from the root to the fruit in steady state?

Answer 1

Given n = 1 mole = $6\times {10}^{2}3$ atoms,

$t_{\frac{1}{2}}$ = 14.3 days,

t= 70 hrs.

$\frac{dN}{dt}$ (after time t) = $\lambda N$

N = $N_0{e}^{-\lambda t}$

= $6\times {10}^{23}\times e^{\frac{-0.693\times 70}{143\times 24}}$

= $6\times {10}^{23}\times 0.868$

= $5.209\times {10}^{2}3$

$\frac{dN}{dt}=5.209\times {10}^{23}\times \frac{0.693}{14.3\times 24}$

= $\frac{0.0105\times {10}^{23}}{3600}\frac{\text{dis}}{\text{hr}}$

= $2.9\times {10}^{-5}\times {10}^{2}3\frac{\text{dis}}{\text{sec}}$

= $2.9\times {10}^{17}\frac{\text{dis}}{\text{sec}}$.

Fraction of activity transmitted

= $\left(\frac{1\mu Ci}{2.9\times {10}^{17}}\right)\times 100$ %

=$[\left(\frac{1\times 3.7\times {10}^4}{2.9\times {10}^{17}}\right)\times 100]$ %

= $1.275\times {10}^{-11}$ %