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

Options

A. 8 seconds
B. 15 seconds
C. 7.5 seconds
D. 5 seconds

Answer

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 follow these steps:

### Step 1: Understand the decay process
Radioactive decay can be expressed in terms of half-lives. Each half-life is the time it takes for half of the radioactive substance to decay.

### Step 2: Determine how many half-lives correspond to the decay to \( \frac{1}{64} \)
We know that:
- After 1 half-life, the remaining amount is \( \frac{1}{2} \) of the initial amount.
- After 2 half-lives, the remaining amount is \( \frac{1}{4} \) of the initial amount.
- After 3 half-lives, the remaining amount is \( \frac{1}{8} \) of the initial amount.
- After 4 half-lives, the remaining amount is \( \frac{1}{16} \) of the initial amount.
- After 5 half-lives, the remaining amount is \( \frac{1}{32} \) of the initial amount.
- After 6 half-lives, the remaining amount is \( \frac{1}{64} \) of the initial amount.

Thus, it takes 6 half-lives to reduce the activity to \( \frac{1}{64} \).

### Step 3: Calculate the duration of one half-life
We know that the total time for these 6 half-lives is 30 seconds. Therefore, we can calculate the duration of one half-life as follows:

\[ \text{Half-life} = \frac{\text{Total time}}{\text{Number of half-lives}} = \frac{30 \text{ seconds}}{6} = 5 \text{ seconds} \]

### Conclusion
The half-life of the radioactive element is 5 seconds.

### Final Answer
The half-life of the radioactive element is **5 seconds**.

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