Calculate the decay constant and time taken to decay by `7/8` of initial value if its half-life is 10s.

Answer

To solve the problem, we need to calculate the decay constant (λ) and the time taken to decay by \( \frac{7}{8} \) of the initial value of a radioactive substance, given that its half-life (t₁/₂) is 10 seconds.

### Step 1: Calculate the Decay Constant (λ)

The decay constant (λ) can be calculated using the formula:

\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]

Where:
- \( \ln(2) \approx 0.693 \)
- \( t_{1/2} = 10 \, \text{s} \)

Substituting the values:

\[ \lambda = \frac{0.693}{10} = 0.0693 \, \text{s}^{-1} \]

### Step 2: Determine the Initial and Final Amounts

Let the initial amount of the substance be \( N_0 \). If the substance decays by \( \frac{7}{8} \) of its initial value, the remaining amount \( N \) will be:

\[ N = N_0 - \frac{7}{8} N_0 = \frac{1}{8} N_0 \]

### Step 3: Use the Decay Formula

The decay of a radioactive substance can be described by the equation:

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

Substituting the values we have:

\[ \frac{1}{8} N_0 = N_0 e^{-\lambda t} \]

Dividing both sides by \( N_0 \):

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

### Step 4: Take the Natural Logarithm of Both Sides

Taking the natural logarithm of both sides gives:

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

### Step 5: Solve for Time (t)

We can express \( \frac{1}{8} \) as \( 2^{-3} \):

\[ \ln\left(2^{-3}\right) = -3 \ln(2) \]

Thus, we have:

\[ -3 \ln(2) = -\lambda t \]

Substituting \( \lambda = 0.0693 \):

\[ -3 \cdot 0.693 = -0.0693 t \]

Now, solving for \( t \):

\[ t = \frac{3 \cdot 0.693}{0.0693} \approx 30 \, \text{s} \]

### Final Results

1. The decay constant \( \lambda \) is approximately \( 0.0693 \, \text{s}^{-1} \).
2. The time taken to decay by \( \frac{7}{8} \) of the initial value is approximately \( 30 \, \text{s} \).

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