A radioactive substance disintegrates 1/64 of initial value in 60 s. The half-life of this substance is

Options

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

Explanation

To find the half-life of a radioactive substance that disintegrates to \( \frac{1}{64} \) of its initial value in 60 seconds, we can follow these steps:

### Step 1: Understand the relationship between disintegration and half-life
The half-life (\( t_{1/2} \)) is the time required for a substance to reduce to half of its initial amount. If a substance disintegrates to \( \frac{1}{64} \) of its initial value, we can express this in terms of half-lives.

### Step 2: Express \( \frac{1}{64} \) in terms of powers of 2
We know that:
\[ \frac{1}{64} = \frac{1}{2^6} \]
This means that the substance has gone through 6 half-lives to reach \( \frac{1}{64} \) of its initial value.

### Step 3: Set up the equation using the given time
Given that the total time taken to reach \( \frac{1}{64} \) of the initial value is 60 seconds, we can relate this to the number of half-lives:
\[ \text{Total time} = \text{Number of half-lives} \times t_{1/2} \]
Substituting the known values:
\[ 60 \, \text{s} = 6 \times t_{1/2} \]

### Step 4: Solve for the half-life
Now, we can solve for \( t_{1/2} \):
\[ t_{1/2} = \frac{60 \, \text{s}}{6} = 10 \, \text{s} \]

### Conclusion
The half-life of the radioactive substance is \( 10 \, \text{seconds} \).

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