At a given instant there are `25 %` undecayed radioactive nuclei in a same. After `10 sec` the number of undecayed nuclei reduces to `6.25 %`, the mean life of the nuclei is.

Options

A. 14.43 sec
B. 7.21 sec
C. 5 sec
D. 10 sec

Explanation

To solve the problem, we will use the concept of radioactive decay and the relationship between the number of undecayed nuclei, the decay constant, and the mean life.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Initially, we have 25% of undecayed nuclei.
- After 10 seconds, this reduces to 6.25% of undecayed nuclei.

2. **Expressing the Decay**:
- Let the initial number of undecayed nuclei be \( N_0 \).
- After 10 seconds, the number of undecayed nuclei is \( N(t) \).
- We can express this as:
\[ N(t) = N_0 e^{-\lambda t} \]
- Where \( \lambda \) is the decay constant and \( t \) is time.

3. **Setting Up the Equations**:
- Initially, \( N_0 = 0.25 N \) (where \( N \) is the total number of nuclei).
- After 10 seconds, \( N(10) = 0.0625 N \).
- We can set up the equation:
\[ 0.0625 N = 0.25 N e^{-\lambda \cdot 10} \]

4. **Simplifying the Equation**:
- Dividing both sides by \( N \) (assuming \( N \neq 0 \)):
\[ 0.0625 = 0.25 e^{-\lambda \cdot 10} \]
- Dividing both sides by 0.25:
\[ \frac{0.0625}{0.25} = e^{-\lambda \cdot 10} \]
- This simplifies to:
\[ 0.25 = e^{-\lambda \cdot 10} \]

5. **Taking the Natural Logarithm**:
- Taking the natural logarithm of both sides:
\[ \ln(0.25) = -\lambda \cdot 10 \]
- We know that \( \ln(0.25) = \ln\left(\frac{1}{4}\right) = -\ln(4) \):
\[ -\ln(4) = -\lambda \cdot 10 \]
- Therefore:
\[ \lambda = \frac{\ln(4)}{10} \]

6. **Finding the Mean Life**:
- The mean life \( \tau \) is related to the decay constant by:
\[ \tau = \frac{1}{\lambda} \]
- Substituting for \( \lambda \):
\[ \tau = \frac{10}{\ln(4)} \]

7. **Calculating the Value**:
- We know \( \ln(4) \approx 1.386 \):
\[ \tau \approx \frac{10}{1.386} \approx 7.22 \text{ seconds} \]

### Final Answer:
The mean life of the nuclei is approximately **7.22 seconds**.