The probability of a radioactive atoms to survive 5 times longer than its half-value period is-

Answer

To solve the problem of finding the probability of a radioactive atom surviving five times longer than its half-life, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Half-Life Concept**:
The half-life (T₁/₂) of a radioactive substance is the time required for half of the radioactive atoms in a sample to decay.

2. **Define the Time in the Problem**:
We need to find the probability of an atom surviving for a time \( t \) that is five times the half-life. Thus, we can express this as:
\[ t = 5 \times T_{1/2} \]

3. **Use the Decay Formula**:
The number of radioactive atoms remaining after time \( t \) is given by the formula:
\[ N(t) = N_0 e^{-\lambda t} \]
where \( N_0 \) is the initial number of atoms, \( \lambda \) is the decay constant, and \( t \) is the time elapsed.

4. **Express the Probability**:
The probability \( P \) that an atom survives after time \( t \) is given by the ratio of the remaining atoms to the initial number of atoms:
\[ P = \frac{N(t)}{N_0} = e^{-\lambda t} \]

5. **Substituting for \( t \)**:
Substitute \( t = 5 \times T_{1/2} \) into the probability expression:
\[ P = e^{-\lambda (5 \times T_{1/2})} \]

6. **Relate Half-Life to Decay Constant**:
The decay constant \( \lambda \) is related to the half-life by the equation:
\[ T_{1/2} = \frac{\ln(2)}{\lambda} \quad \Rightarrow \quad \lambda = \frac{\ln(2)}{T_{1/2}} \]

7. **Substituting \( \lambda \)**:
Now substitute \( \lambda \) back into the probability equation:
\[ P = e^{-5 \times \left(\frac{\ln(2)}{T_{1/2}}\right) \times T_{1/2}} = e^{-5 \ln(2)} \]

8. **Simplifying the Expression**:
Using the property of exponents:
\[ P = e^{\ln(2^{-5})} = 2^{-5} \]

9. **Final Probability**:
Therefore, the probability that a radioactive atom survives five times longer than its half-life is:
\[ P = \frac{1}{32} \]

### Final Answer:
The probability of a radioactive atom surviving five times longer than its half-life is \( \frac{1}{32} \).