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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?
Solution
Given:
Initial no of atoms, N0 = 1 mole = 6 × 1023 atoms
Half-life of the radioactive material, T1/2 = 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 %`
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