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?

Explanation

To solve the problem step by step, we will follow the outlined procedure to calculate the percentage of activity transmitted from the roots to the fruit of the plant.

### Step 1: Identify Given Data
- Number of moles of radioactive material, \( n_0 = 1 \, \text{mole} \)
- Half-life, \( t_{1/2} = 14.3 \, \text{days} \)
- Time allowed for the plant to settle, \( t = 70 \, \text{hours} \)
- Activity measured in the fruit, \( A = 1 \, \mu \text{Ci} = 1 \times 10^{-6} \, \text{Ci} \)

### Step 2: Convert Half-life to Hours
Convert the half-life from days to hours:
\[ t_{1/2} = 14.3 \, \text{days} \times 24 \, \text{hours/day} = 343.2 \, \text{hours} \]

### Step 3: Calculate Decay Constant (\( \lambda \))
The decay constant \( \lambda \) is given by:
\[ \lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{343.2 \, \text{hours}} \approx 0.00202 \, \text{hours}^{-1} \]

### Step 4: Calculate Initial Number of Atoms
Using Avogadro's number, the initial number of atoms \( N_0 \) in 1 mole is:
\[ N_0 = 6.022 \times 10^{23} \, \text{atoms} \]

### Step 5: Calculate Remaining Number of Atoms after 70 Hours
Using the formula for radioactive decay:
\[ N = N_0 e^{-\lambda t} \]
Substituting the values:
\[ N = 6.022 \times 10^{23} \times e^{-0.00202 \times 70} \]
Calculating \( e^{-0.1414} \):
\[ N \approx 6.022 \times 10^{23} \times 0.868 \approx 5.22 \times 10^{23} \, \text{atoms} \]

### Step 6: Calculate Activity (\( R \))
The activity \( R \) is given by:
\[ R = \lambda N \]
Substituting the values:
\[ R = 0.00202 \times 5.22 \times 10^{23} \approx 1.05 \times 10^{21} \, \text{disintegrations per hour} \]
Convert to disintegrations per second:
\[ R \approx \frac{1.05 \times 10^{21}}{3600} \approx 2.92 \times 10^{17} \, \text{disintegrations per second} \]

### Step 7: Convert Activity Measured in the Fruit to Disintegrations per Second
Given that \( 1 \, \mu \text{Ci} = 3.7 \times 10^4 \, \text{disintegrations per second} \):
\[ A = 3.7 \times 10^4 \, \text{disintegrations per second} \]

### Step 8: Calculate the Percentage of Activity Transmitted
The percentage of activity transmitted from the root to the fruit is given by:
\[ \text{Percentage} = \left( \frac{A}{R} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage} = \left( \frac{3.7 \times 10^4}{2.92 \times 10^{17}} \right) \times 100 \approx 1.27 \times 10^{-11} \% \]

### Final Answer
The percentage of activity transmitted from the root to the fruit is approximately \( 1.27 \times 10^{-11} \% \).

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