In an agriculture experiment, a solution containing 1 mole of a radioactive meterial `(t_(1//2)=14.3 days)`was injected into the roots of a plants.the plant was allowed 70 hours to settle down and then activity eas 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

A, B

Explanation

Given `n = 1 mole = 6 xx 10^(23) ` atoms ,
` t _(1/2) = 14.3 ` days ,
` t = 70 ` hrs
` (dN)/(dt) (after time t) = lambda N `
` N = N_(0) e^(-lambda t) `
` = 6 xx 10^(23) xx e^ (-0.693 xx 70)/(143xx24)`
` = 6 xx 10^(23) xx 0.868 `
` = 5.209 xx 10^(23) `
` (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^(-5) xx 10^(23) ` dis// sec `
` = 2.9 xx 10^(17) ` dis//sec `
Fraction of activity transmitted
` = (( 1 mu Ci)/(2.9 xx10 ^(17))) xx 100 % `
` = [((1 xx 3.7 xx 10^(4))/(2.9 xx10^(17)) xx 100)] %`
` = 1.275 xx 10^(-11) % ` .