The initial activity of a certain radioactive iostope was measured as `16000` counts `min.^(-1)`. Given that the only activity measured was due to this isotaope and that its activity after `12 h` was `2100` counts `min.^(-1)`, its half-life, in hours, is nearest to [Given `log_(e) (7.2)=2`].

Options

A. `9.0`
B. `6.0`
C. `4.0`
D. `3.0`

Explanation

To find the half-life of the radioactive isotope given its initial and final activity, we can follow these steps:

### Step 1: Understand the relationship between activity and decay constant
The activity \( A \) of a radioactive substance is related to the number of radioactive nuclei \( N \) and the decay constant \( \lambda \) by the equation:
\[ A = \lambda N \]
The activity decreases over time according to the formula:
\[ A(t) = A_0 e^{-\lambda t} \]
where \( A_0 \) is the initial activity and \( A(t) \) is the activity at time \( t \).

### Step 2: Substitute the given values
From the problem, we know:
- Initial activity \( A_0 = 16000 \) counts/min
- Activity after \( t = 12 \) hours \( A(12) = 2100 \) counts/min

We can substitute these values into the activity decay equation:
\[ 2100 = 16000 e^{-\lambda \cdot 12} \]

### Step 3: Solve for \( e^{-\lambda \cdot 12} \)
Rearranging the equation gives:
\[ e^{-\lambda \cdot 12} = \frac{2100}{16000} \]
Calculating the fraction:
\[ \frac{2100}{16000} = \frac{21}{160} = 0.13125 \]
Thus,
\[ e^{-\lambda \cdot 12} = 0.13125 \]

### Step 4: Take the natural logarithm
Taking the natural logarithm of both sides:
\[ -\lambda \cdot 12 = \ln(0.13125) \]
Using the property of logarithms, we can express \( \ln(0.13125) \) as:
\[ \ln(0.13125) = \ln\left(\frac{21}{160}\right) = \ln(21) - \ln(160) \]
However, we can also use the approximation given in the problem. We know \( \log_e(7.2) = 2 \), and since \( 0.13125 \) is approximately \( \frac{1}{7.2} \), we can say:
\[ \ln(0.13125) \approx -2 \]

### Step 5: Solve for \( \lambda \)
Substituting back, we have:
\[ -\lambda \cdot 12 = -2 \]
Thus,
\[ \lambda = \frac{2}{12} = \frac{1}{6} \text{ h}^{-1} \approx 0.1667 \text{ h}^{-1} \]

### Step 6: Calculate the half-life
The half-life \( t_{1/2} \) is given by:
\[ t_{1/2} = \frac{\ln(2)}{\lambda} \]
Using \( \ln(2) \approx 0.693 \):
\[ t_{1/2} = \frac{0.693}{0.1667} \approx 4.15 \text{ hours} \]

### Conclusion
The half-life of the radioactive isotope is approximately \( 4 \) hours.

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