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In an agricultural experiment a solution containing $1$ mole of a radioactive material $(T_{1 / 2} = 14.3 d a y s)$ 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 $1 μ C i$ , what percentage of activity is transmitted from the root to the fruit in steady state?

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$P = 1.26 \times 10^{11} %$

$P = 2.26 \times 10^{- 11} %$

$P = 2.26 \times 10^{11} %$

$P = 1.26 \times 10^{- 11} %$

answer is B.

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Detailed Solution

$R_{0} = λ N \Rightarrow R_{0} = \frac{0.693}{14.3 \times 3600 \times 24} \times 6.023 \times 10^{23} p e r s e c$

$\Rightarrow R_{0} = 3.37 \times 10^{17} p e r s e c$

After $70$ hours activity

$R = R_{0} e^{- λ t} \Rightarrow R = (3.37 \times 10^{17}) e^{- (\frac{0.693}{14.3 \times 24}) (70)} = 2.92 \times 10^{17} p e r s e c$

In fruits, the activity was observed $1 μ C i$ or $3.7 \times 10^{4}$ per sec

Therefore percentage $P$ of activity transmitted from root to the fruit is

$P = \frac{3.7 \times 10^{4}}{2.92 \times 10^{17}} \times 100 \Rightarrow P = 1.26 \times 10^{- 11} %$

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