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 (\( \lambda \)) of the radioactive material, we can use the radioactive decay formula:

\[ N = N_0 e^{-\lambda t} \]

Where:
- \( N \) is the remaining quantity of the substance (10 kg)
- \( N_0 \) is the initial quantity of the substance (80 kg)
- \( \lambda \) is the decay constant
- \( t \) is the time (1 hour)

### Step 1: Identify the given values
- Initial quantity, \( N_0 = 80 \, \text{kg} \)
- Remaining quantity, \( N = 10 \, \text{kg} \)
- Time, \( t = 1 \, \text{hour} = 3600 \, \text{seconds} \)

### Step 2: Substitute the values into the decay formula
We can rearrange the decay formula to isolate \( \lambda \):

\[ 10 = 80 e^{-\lambda \cdot 3600} \]

### Step 3: Divide both sides by 80
\[ \frac{10}{80} = e^{-\lambda \cdot 3600} \]

This simplifies to:

\[ \frac{1}{8} = e^{-\lambda \cdot 3600} \]

### Step 4: Take the natural logarithm of both sides
Taking the natural logarithm (ln) gives us:

\[ \ln\left(\frac{1}{8}\right) = -\lambda \cdot 3600 \]

### Step 5: Simplify the left side using properties of logarithms
Using the property of logarithms, we can rewrite \( \ln\left(\frac{1}{8}\right) \):

\[ \ln(1) - \ln(8) = 0 - \ln(8) = -\ln(8) \]

Thus, we have:

\[ -\ln(8) = -\lambda \cdot 3600 \]

### Step 6: Remove the negative sign
Removing the negative sign from both sides gives:

\[ \ln(8) = \lambda \cdot 3600 \]

### Step 7: Solve for \( \lambda \)
Now, we can solve for \( \lambda \):

\[ \lambda = \frac{\ln(8)}{3600} \]

### Step 8: Calculate \( \ln(8) \)
Using the property of logarithms, we know that:

\[ \ln(8) = \ln(2^3) = 3 \ln(2) \]

Given that \( \ln(2) \approx 0.693 \):

\[ \ln(8) \approx 3 \times 0.693 = 2.079 \]

### Step 9: Substitute back to find \( \lambda \)
Now substituting back into the equation for \( \lambda \):

\[ \lambda = \frac{2.079}{3600} \]

### Step 10: Calculate \( \lambda \)
Calculating the value:

\[ \lambda \approx 5.775 \times 10^{-4} \, \text{s}^{-1} \]

### Final Answer
The decay constant \( \lambda \) is approximately:

\[ \lambda \approx 5.775 \times 10^{-4} \, \text{s}^{-1} \]