A radioactive substance decays to` ( 1/10)^th`of its original value in 56 days. Calculate its decay constant.

Answer

To calculate the decay constant (λ) of a radioactive substance that decays to \( \frac{1}{10} \) of its original value in 56 days, we can use the formula for radioactive decay:

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

Where:
- \( N \) is the remaining quantity of the substance,
- \( N_0 \) is the initial quantity of the substance,
- \( \lambda \) is the decay constant,
- \( t \) is the time elapsed.

### Step 1: Set up the equation
Given that the substance decays to \( \frac{1}{10} \) of its original value, we can express this as:

\[ N = \frac{N_0}{10} \]

Substituting this into the decay formula gives:

\[ \frac{N_0}{10} = N_0 e^{-\lambda t} \]

### Step 2: Simplify the equation
We can divide both sides by \( N_0 \) (assuming \( N_0 \neq 0 \)):

\[ \frac{1}{10} = e^{-\lambda t} \]

### Step 3: Take the natural logarithm
To solve for \( \lambda \), we take the natural logarithm of both sides:

\[ \ln\left(\frac{1}{10}\right) = -\lambda t \]

### Step 4: Substitute the time value
We know that \( t = 56 \) days, so we can substitute this into the equation:

\[ \ln\left(\frac{1}{10}\right) = -\lambda \cdot 56 \]

### Step 5: Solve for \( \lambda \)
Rearranging the equation gives:

\[ \lambda = -\frac{\ln\left(\frac{1}{10}\right)}{56} \]

Using the property of logarithms, we can rewrite \( \ln\left(\frac{1}{10}\right) \):

\[ \ln\left(\frac{1}{10}\right) = \ln(1) - \ln(10) = 0 - \ln(10) = -\ln(10) \]

Thus, we have:

\[ \lambda = \frac{\ln(10)}{56} \]

### Step 6: Calculate \( \lambda \)
Using the approximate value \( \ln(10) \approx 2.303 \):

\[ \lambda = \frac{2.303}{56} \]

Calculating this gives:

\[ \lambda \approx 0.0411 \, \text{days}^{-1} \]

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

\[ \lambda \approx 0.0411 \, \text{days}^{-1} \]

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