8 kg of `Cu^(66)` undergones radioactive decay and after 15 minutes only 1 kg remains . The half - life , in minutes , is then

Options

A. 5
B. 15/8
C. `8//15`
D. 15 In (2)

Answer

To solve the problem, we need to determine the half-life of the radioactive substance \( Cu^{66} \) given that 8 kg decays to 1 kg in 15 minutes.

### Step-by-Step Solution:

1. **Identify Initial and Remaining Amounts**:
- Initial amount (\( N_0 \)) = 8 kg
- Remaining amount after 15 minutes (\( N_t \)) = 1 kg

2. **Determine the Time Interval**:
- Time interval (\( t \)) = 15 minutes

3. **Use the Radioactive Decay Formula**:
The relationship between the initial amount, remaining amount, and time can be expressed using the decay constant (\( \lambda \)):
\[ N_t = N_0 e^{-\lambda t} \]
Rearranging gives:
\[ \lambda = -\frac{1}{t} \ln \left(\frac{N_t}{N_0}\right) \]

4. **Substitute the Values**:
Substitute \( N_0 = 8 \) kg, \( N_t = 1 \) kg, and \( t = 15 \) minutes:
\[ \lambda = -\frac{1}{15} \ln \left(\frac{1}{8}\right) \]

5. **Simplify the Logarithm**:
We can simplify \( \ln \left(\frac{1}{8}\right) \) as follows:
\[ \ln \left(\frac{1}{8}\right) = \ln(1) - \ln(8) = 0 - \ln(8) = -\ln(8) \]
Thus:
\[ \lambda = \frac{1}{15} \ln(8) \]

6. **Express \( \ln(8) \)**:
Since \( 8 = 2^3 \):
\[ \ln(8) = \ln(2^3) = 3 \ln(2) \]
Therefore:
\[ \lambda = \frac{3 \ln(2)}{15} = \frac{\ln(2)}{5} \]

7. **Calculate the Half-Life**:
The half-life (\( T_{1/2} \)) is given by the formula:
\[ T_{1/2} = \frac{\ln(2)}{\lambda} \]
Substituting the value of \( \lambda \):
\[ T_{1/2} = \frac{\ln(2)}{\frac{\ln(2)}{5}} = 5 \text{ minutes} \]

### Final Answer:
The half-life of \( Cu^{66} \) is **5 minutes**.