Four thousand active nuclei of a radioactive material are present at `t=`0. After 60 minutes 500 active nuclei are left in the sample. The decay constant of the sample is

Options

A. `("In"(20))/60` per minute
B. `("In"(2))/20` per minute
C. `20"In"(2)` perminute
D. `60"In"(2)` per minute

Answer

To find the decay constant of the radioactive material, we can use the formula for the decay constant (\( \lambda \)) given by:

\[ \lambda = \frac{1}{T} \ln \left( \frac{A_0}{A} \right) \]

Where:
- \( A_0 \) = initial number of nuclei
- \( A \) = number of nuclei remaining after time \( T \)
- \( T \) = time elapsed

### Step-by-Step Solution:

1. **Identify the values**:
- Initial number of nuclei, \( A_0 = 4000 \)
- Number of nuclei remaining after 60 minutes, \( A = 500 \)
- Time elapsed, \( T = 60 \) minutes

2. **Substitute the values into the formula**:
\[ \lambda = \frac{1}{60} \ln \left( \frac{4000}{500} \right) \]

3. **Calculate the fraction**:
\[ \frac{4000}{500} = 8 \]

4. **Substitute back into the equation**:
\[ \lambda = \frac{1}{60} \ln(8) \]

5. **Use the property of logarithms**:
\[ \ln(8) = \ln(2^3) = 3 \ln(2) \]

6. **Substitute this back into the equation**:
\[ \lambda = \frac{1}{60} \cdot 3 \ln(2) = \frac{3 \ln(2)}{60} \]

7. **Simplify the expression**:
\[ \lambda = \frac{\ln(2)}{20} \]

### Final Answer:
The decay constant \( \lambda \) is:
\[ \lambda = \frac{\ln(2)}{20} \text{ min}^{-1} \]