#### 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?

#### Solution

Given:

Initial no of atoms, N_{0} = 1 mole = 6 × 10^{23} 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 = 0.693/t_"1/2"` = `0.693/(14.3 xx 24) "h"^-1`

`N = N_0e^(-lambdat)`

= `6 xx 10^23 xx e^((-0.693 xx 70)/(14.3 xx 24))`

= `6 xx 10^23 xx 0.868`

= `5.209 xx 10^23`

Activity , `R = "dN"/"dt" = 5.209 xx 10^23 xx 0.693/(14.3 xx 24)`

= `(0.0105 xx 10^23)/3600 "dis/hr"`

= `2.9 xx 10^-6 xx 10^23 "dis/sec"`

= `2.9 xx 10^17 "dis/sec"`

Fraction of activity transmitted = `((1 µCi)/(2.9 xx 10^17)) xx 100%`

= `[((1 xx 3.7 xx 10^4)/(2.9 xx 10^17)) xx 100%]`

= `1.275 xx 10^-11 %`