80 kg of a radioactive material reduces to 10 kg in 1 h. The decay constant of the material is

Options

A. `5.80 xx 10^(-4) s^(-1)`
B. `1.16 xx 10^(-3) s^(-1)`
C. `2.32 xx 10^(-3) s^(-1)`
D. `4.64 xx 10^(-3) s^(-1)`

Answer

To find the decay constant (λ) of a radioactive material that reduces from 80 kg to 10 kg in 1 hour, we can use the radioactive decay formula:

### Step-by-Step Solution:

1. **Identify the Initial and Final Mass**:
- Initial mass (N₀) = 80 kg
- Final mass (N) = 10 kg

2. **Convert Time to Seconds**:
- Time (t) = 1 hour = 3600 seconds (since 1 hour = 60 minutes and 1 minute = 60 seconds).

3. **Use the Radioactive Decay Formula**:
The formula for radioactive decay is:
\[ N = N₀ e^{-\lambda t} \]
Rearranging it gives:
\[ \frac{N}{N₀} = e^{-\lambda t} \]

4. **Substitute the Values**:
\[ \frac{10}{80} = e^{-\lambda \cdot 3600} \]
Simplifying the left side:
\[ \frac{1}{8} = e^{-\lambda \cdot 3600} \]

5. **Take Natural Logarithm on Both Sides**:
\[ \ln\left(\frac{1}{8}\right) = -\lambda \cdot 3600 \]
This can be rewritten as:
\[ -\ln(8) = -\lambda \cdot 3600 \]
Thus:
\[ \lambda \cdot 3600 = \ln(8) \]

6. **Solve for λ**:
\[ \lambda = \frac{\ln(8)}{3600} \]

7. **Calculate ln(8)**:
Since \(8 = 2^3\), we can use the property of logarithms:
\[ \ln(8) = \ln(2^3) = 3 \ln(2) \]
Therefore:
\[ \lambda = \frac{3 \ln(2)}{3600} \]

8. **Substitute the Value of ln(2)**:
Using \( \ln(2) \approx 0.693\):
\[ \lambda = \frac{3 \times 0.693}{3600} \]
\[ \lambda \approx \frac{2.079}{3600} \approx 5.775 \times 10^{-4} \text{ s}^{-1} \]

### Final Answer:
The decay constant (λ) is approximately \(5.775 \times 10^{-4} \text{ s}^{-1}\).

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