The ratio activity of an element becomes `1//64 th` of its original value in `60 sec`. Then the half-life period is

Options

A. 5 sec s
B. 10 sec s
C. 20 sec s
D. 30 sec s

Explanation

To solve the problem, we need to find the half-life period of an element given that its activity becomes \( \frac{1}{64} \) of its original value in 60 seconds.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The activity ratio becomes \( \frac{1}{64} \) in \( 60 \) seconds.
- We need to find the half-life period \( T_0 \).

2. **Use the Activity Formula**:
The relationship between the activity at any time \( t \) and the initial activity \( A_0 \) is given by the formula:
\[ A = A_0 \left( \frac{1}{2} \right)^{\frac{t}{T_0}} \]
where \( A \) is the activity at time \( t \), \( A_0 \) is the initial activity, and \( T_0 \) is the half-life period.

3. **Set Up the Equation**:
Since we know that \( A = \frac{1}{64} A_0 \) at \( t = 60 \) seconds, we can substitute this into the formula:
\[ \frac{1}{64} A_0 = A_0 \left( \frac{1}{2} \right)^{\frac{60}{T_0}} \]

4. **Cancel \( A_0 \)**:
Since \( A_0 \) is common on both sides, we can cancel it out (assuming \( A_0 \neq 0 \)):
\[ \frac{1}{64} = \left( \frac{1}{2} \right)^{\frac{60}{T_0}} \]

5. **Express \( \frac{1}{64} \) as a Power of 2**:
We know that \( \frac{1}{64} = \frac{1}{2^6} \):
\[ \frac{1}{2^6} = \left( \frac{1}{2} \right)^{\frac{60}{T_0}} \]

6. **Equate the Exponents**:
Since the bases are the same, we can equate the exponents:
\[ 6 = \frac{60}{T_0} \]

7. **Solve for \( T_0 \)**:
Rearranging the equation gives:
\[ T_0 = \frac{60}{6} = 10 \text{ seconds} \]

### Final Answer:
The half-life period \( T_0 \) is \( 10 \) seconds.