Question

In an agricultural experiment, a solution containing 1 mole of a radioactive material (t1/2 = 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

Explanation:

It is given:

No. of atoms initially, $N_{0}=6 \times 10^{23} \text { atoms }$ = 1 mole

Radioactive material has the half-life, $T_{1 / 2}=14.3$ days

To settle down, the plant takes the time, t = 70 h

Constant of disintegration,

$\lambda=0.693 \mathrm{t} 1 / 2=0.69314 .3 \times 24 \mathrm{h}-1$

$N=N_{0} e^{-\lambda} t$

$=6 \times \mathrm{e}-0.693 \times 1023 \times 7014.3 \times 24=5.209 \times 1023$

Activity, $\mathrm{R}=\mathrm{d} \mathrm{Ndt}=5.209 \times 0.69314 .3 \times 1023 \times 24=2.9 \times 1017 \mathrm{dis} / \mathrm{sec}$

Transmitted fraction of activity = $1 \mu \mathrm{Ci} 2.9 \times 1017 \times 100 \%$

$=1.275 \times 10-11 \%$