A small quantity of solution containing `Na^24` radio nuclide `(half-life=15 h)` of activity `1.0 microcurie` is injected into the blood of a person. A sample of the blood of volume `1cm^3` taken after `5h` shows an activity of `296` disintegrations per minute. Determine the total volume of the blood in the body of the person. Assume that the radioactive solution mixes uniformly in the blood of person.
(1 curie `=3.7 xx 10^10` disintegrations per second)

Options

A. `5.94 L`
B. 2 L
C. 317 L
D. 1 L

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Convert the initial activity from microcuries to disintegrations per second
Given:
- Initial activity \( A_0 = 1.0 \, \text{microcurie} \)

Using the conversion:
\[ 1 \, \text{curie} = 3.7 \times 10^{10} \, \text{disintegrations/second} \]
So,
\[ A_0 = 1.0 \times 10^{-6} \, \text{curie} = 1.0 \times 10^{-6} \times 3.7 \times 10^{10} \, \text{disintegrations/second} \]
Calculating this gives:
\[ A_0 = 3.7 \times 10^4 \, \text{disintegrations/second} \]

### Step 2: Calculate the decay constant \( \lambda \)
The half-life \( t_{1/2} \) is given as 15 hours. The decay constant \( \lambda \) is calculated using the formula:
\[ \lambda = \frac{0.693}{t_{1/2}} \]
Substituting the value:
\[ \lambda = \frac{0.693}{15 \, \text{hours}} \]
Converting hours to seconds (1 hour = 3600 seconds):
\[ \lambda = \frac{0.693}{15 \times 3600} = \frac{0.693}{54000} \approx 0.00001283 \, \text{seconds}^{-1} \]

### Step 3: Calculate the activity after 5 hours
The activity \( A \) after time \( t \) can be calculated using:
\[ A = A_0 e^{-\lambda t} \]
Where \( t = 5 \, \text{hours} = 5 \times 3600 \, \text{seconds} = 18000 \, \text{seconds} \).

Substituting the values:
\[ A = 3.7 \times 10^4 \times e^{-0.00001283 \times 18000} \]
Calculating the exponent:
\[ e^{-0.00001283 \times 18000} \approx e^{-0.231} \approx 0.793 \]
So,
\[ A \approx 3.7 \times 10^4 \times 0.793 \approx 29300 \, \text{disintegrations/second} \]

### Step 4: Convert the activity of the blood sample
The activity of the blood sample taken is given as 296 disintegrations per minute. Converting this to disintegrations per second:
\[ \text{Activity of blood sample} = \frac{296}{60} \approx 4.93 \, \text{disintegrations/second} \]

### Step 5: Calculate the total volume of blood in the body
Let \( V \) be the total volume of blood in the body. The relationship between the activities is given by:
\[ A = \frac{A_{\text{sample}}}{V_{\text{sample}}} \times V \]
Where:
- \( A_{\text{sample}} = 4.93 \, \text{disintegrations/second} \)
- \( V_{\text{sample}} = 1 \, \text{cm}^3 \)

Rearranging gives:
\[ V = \frac{A}{A_{\text{sample}}} \]

Substituting the values:
\[ V = \frac{29300 \, \text{disintegrations/second}}{4.93 \, \text{disintegrations/second}} \approx 5940 \, \text{cm}^3 \]

### Step 6: Convert the volume to liters
Since \( 1 \, \text{liter} = 1000 \, \text{cm}^3 \):
\[ V \approx \frac{5940}{1000} \approx 5.94 \, \text{liters} \]

### Final Answer
The total volume of blood in the body of the person is approximately **5.94 liters**.

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