During an agricultural experiment, a solution containing 1 mole of...

The initial number of radioactive atoms in $1\text{ mole}$ is:

$N_0 = 6.023 \times 10^{23}$

The decay constant $\lambda$ is given by:

$\lambda = \dfrac{\ln 2}{t_{1/2}} = \dfrac{0.693}{14.3 \times 24 \times 3600} = 5.61 \times 10^{-7}\text{ s}^{-1}$

The initial activity $A_0$ of the injected material is:

$A_0 = \lambda N_0 = (5.61 \times 10^{-7}) \times (6.023 \times 10^{23}) = 3.38 \times 10^{17}\text{ Bq}$

The plant is left to settle for $t = 70\text{ hours} = \dfrac{70}{24}\text{ days} = 2.92\text{ days}$. The total activity of the radioactive material remaining in the plant after this time is:

$A(t) = A_0 e^{-\lambda t} = A_0 e^{-\dfrac{0.693 \times 2.92}{14.3}}$

$A(t) = 3.38 \times 10^{17} \times e^{-0.141} \approx 3.38 \times 10^{17} \times 0.868 = 2.93 \times 10^{17}\text{ Bq}$

The measured activity in the fruit is:

$A_f = 1\mu\text{Ci} = 10^{-6} \times 3.7 \times 10^{10}\text{ Bq} = 3.7 \times 10^4\text{ Bq}$

In steady state, the percentage of activity transmitted from the root to the fruit is the ratio of the activity in the fruit to the total remaining activity in the plant:

$\text{Percentage transmitted} = \dfrac{A_f}{A(t)} \times 100$

$\text{Percentage transmitted} = \dfrac{3.7 \times 10^4}{2.93 \times 10^{17}} \times 100 = 1.26 \times 10^{-13} \times 100 = 1.26 \times 10^{-11} \%$

Answer: $1.26 \times 10^{-11} \%$