What will be half life period of a nucleus if at the end of 4.2 days, N `=0.798 N_(0)` ?

Options

A. 10 days
B. 12.83 days
C. 15 days
D. 20 days

Answer

To solve the problem of finding the half-life period of a nucleus when at the end of 4.2 days, \( N = 0.798 N_0 \), we can follow these steps:

### Step 1: Understand the relationship between N, N0, and time
We know that:
\[ N = N_0 e^{-\lambda t} \]
Where:
- \( N \) = amount of substance remaining after time \( t \)
- \( N_0 \) = initial amount of substance
- \( \lambda \) = decay constant
- \( t \) = time elapsed

### Step 2: Substitute the given values
From the problem, we have:
\[ N = 0.798 N_0 \quad \text{and} \quad t = 4.2 \text{ days} \]
Substituting these values into the equation gives:
\[ 0.798 N_0 = N_0 e^{-\lambda \cdot 4.2} \]

### Step 3: Simplify the equation
Dividing both sides by \( N_0 \) (assuming \( N_0 \neq 0 \)):
\[ 0.798 = e^{-\lambda \cdot 4.2} \]

### Step 4: Take the natural logarithm of both sides
Taking the natural logarithm (ln) of both sides:
\[ \ln(0.798) = -\lambda \cdot 4.2 \]

### Step 5: Solve for the decay constant \( \lambda \)
Rearranging the equation to solve for \( \lambda \):
\[ \lambda = -\frac{\ln(0.798)}{4.2} \]

### Step 6: Calculate \( \lambda \)
Using a calculator, we find:
\[ \ln(0.798) \approx -0.2231 \]
Thus:
\[ \lambda = -\frac{-0.2231}{4.2} \approx 0.053 \text{ per day} \]

### Step 7: Use the decay constant to find the half-life \( t_{1/2} \)
The half-life \( t_{1/2} \) is given by the formula:
\[ t_{1/2} = \frac{0.693}{\lambda} \]
Substituting the value of \( \lambda \):
\[ t_{1/2} = \frac{0.693}{0.053} \approx 13.09 \text{ days} \]

### Step 8: Conclusion
Thus, the half-life period of the nucleus is approximately 13 days.