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 find the half-life period of the element, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We know that the activity of the element becomes \( \frac{1}{64} \) of its original value in 60 seconds. We need to determine the half-life period (\( \tau \)) of the element.

2. **Using the Radioactive Decay Formula**: The relationship between the remaining quantity of a radioactive substance and its half-life can be expressed as:
\[ n = n_0 \left(\frac{1}{2}\right)^{\frac{T}{\tau}} \]
where:
- \( n \) is the remaining quantity after time \( T \),
- \( n_0 \) is the initial quantity,
- \( T \) is the time elapsed,
- \( \tau \) is the half-life.

3. **Substituting the Known Values**: From the problem, we know:
- \( n = \frac{n_0}{64} \)
- \( T = 60 \) seconds

Substituting these values into the formula gives:
\[ \frac{n_0}{64} = n_0 \left(\frac{1}{2}\right)^{\frac{60}{\tau}} \]

4. **Canceling \( n_0 \)**: Since \( n_0 \) is common on both sides, we can cancel it out:
\[ \frac{1}{64} = \left(\frac{1}{2}\right)^{\frac{60}{\tau}} \]

5. **Expressing \( \frac{1}{64} \) as a Power of 2**: We know that:
\[ \frac{1}{64} = \frac{1}{2^6} \]
Therefore, we can rewrite the equation as:
\[ \frac{1}{2^6} = \left(\frac{1}{2}\right)^{\frac{60}{\tau}} \]

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

7. **Solving for \( \tau \)**: Rearranging the equation to solve for \( \tau \):
\[ \tau = \frac{60}{6} = 10 \text{ seconds} \]

8. **Conclusion**: The half-life period of the element is \( 10 \) seconds.