The activity of a radioactive sample is measured as `9750` counts per minute at `t = 0` and as `975` counts per minute at `t = 5` minutes. The decay constant is approximately

Options

A. `0.922//min`
B. `0.691//min`
C. `0.461//min`
D. `0.230//min`

Answer

To find the decay constant (λ) of a radioactive sample, we can use the formula for radioactive decay:

\[ A = A_0 e^{-\lambda t} \]

Where:
- \( A \) is the activity at time \( t \)
- \( A_0 \) is the initial activity
- \( \lambda \) is the decay constant
- \( t \) is the time elapsed

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial activity \( A_0 = 9750 \) counts per minute (at \( t = 0 \))
- Activity after 5 minutes \( A = 975 \) counts per minute (at \( t = 5 \) minutes)
- Time \( t = 5 \) minutes

2. **Set Up the Decay Equation:**
Using the decay formula:
\[ A = A_0 e^{-\lambda t} \]
Substitute the known values:
\[ 975 = 9750 e^{-\lambda \cdot 5} \]

3. **Rearrange the Equation:**
Divide both sides by \( 9750 \):
\[ \frac{975}{9750} = e^{-5\lambda} \]
Simplifying the left side gives:
\[ \frac{1}{10} = e^{-5\lambda} \]

4. **Take the Natural Logarithm:**
Taking the natural logarithm (ln) of both sides:
\[ \ln\left(\frac{1}{10}\right) = -5\lambda \]

5. **Calculate the Natural Logarithm:**
We know that:
\[ \ln\left(\frac{1}{10}\right) = -\ln(10) \]
Therefore, we can write:
\[ -\ln(10) = -5\lambda \]
Simplifying gives:
\[ \lambda = \frac{\ln(10)}{5} \]

6. **Substitute the Value of \( \ln(10) \):**
Using \( \ln(10) \approx 2.303 \):
\[ \lambda = \frac{2.303}{5} \]

7. **Calculate the Decay Constant:**
Performing the division:
\[ \lambda \approx 0.4606 \text{ per minute} \]

### Final Result:
The decay constant \( \lambda \) is approximately \( 0.461 \) per minute.

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