The radioactivity of a certain radioactive element drops to `1//64` of its initial value in `30` seconds. Its half-life is.

Options

A. 2 seconds
B. 4 seconds
C. 5 seconds
D. 6 seconds

Explanation

To find the half-life of a radioactive element whose radioactivity drops to \( \frac{1}{64} \) of its initial value in 30 seconds, we can use the formula for radioactive decay:

\[ N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{\tau}} \]

Where:
- \( N \) is the remaining quantity of the substance.
- \( N_0 \) is the initial quantity of the substance.
- \( t \) is the time elapsed.
- \( \tau \) is the half-life of the substance.

### Step 1: Identify the values
From the problem, we know:
- The radioactivity drops to \( \frac{1}{64} \) of its initial value, so:
\[ N = \frac{N_0}{64} \]
- The time elapsed \( t = 30 \) seconds.

### Step 2: Substitute values into the decay formula
Substituting the known values into the decay formula gives:
\[ \frac{N_0}{64} = N_0 \left( \frac{1}{2} \right)^{\frac{30}{\tau}} \]

### Step 3: Cancel \( N_0 \)
Since \( N_0 \) is present on both sides of the equation, we can cancel it out:
\[ \frac{1}{64} = \left( \frac{1}{2} \right)^{\frac{30}{\tau}} \]

### Step 4: Express \( \frac{1}{64} \) as a power of \( \frac{1}{2} \)
We know that:
\[ \frac{1}{64} = \left( \frac{1}{2} \right)^6 \]
So we can rewrite the equation as:
\[ \left( \frac{1}{2} \right)^6 = \left( \frac{1}{2} \right)^{\frac{30}{\tau}} \]

### Step 5: Set the exponents equal to each other
Since the bases are the same, we can set the exponents equal to each other:
\[ 6 = \frac{30}{\tau} \]

### Step 6: Solve for \( \tau \)
To find \( \tau \), we rearrange the equation:
\[ \tau = \frac{30}{6} = 5 \text{ seconds} \]

### Conclusion
The half-life \( \tau \) of the radioactive element is \( 5 \) seconds.

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