There are n number of radioactive nuclei in a sample that undergoes beta decay. If from the sample, `n'` number of `beta`-particels are emitted every `2s`, then half-life of nuclei is .

Options

A. `n'//2`
B. `0.693 xx(2n//n')`
C. `0.693 1n(2n//n')`
D. `0.693 xx(n//n')`

Explanation

To find the half-life of the radioactive nuclei in the given problem, we can follow these steps:

### Step 1: Understand the relationship between activity and decay constant
The activity (A) of a radioactive sample is defined as the number of decays (or emissions) per unit time. In this case, the number of beta particles emitted every 2 seconds is given as \( n' \). Therefore, the activity in terms of beta particles emitted per second can be expressed as:
\[ A = \frac{n'}{2} \quad \text{(since it is emitted every 2 seconds)} \]

### Step 2: Relate activity to decay constant
The activity can also be expressed in terms of the number of radioactive nuclei (n) and the decay constant (\( \lambda \)):
\[ A = n \lambda \]
Equating the two expressions for activity, we have:
\[ \frac{n'}{2} = n \lambda \]

### Step 3: Solve for the decay constant (\( \lambda \))
From the equation above, we can solve for the decay constant (\( \lambda \)):
\[ \lambda = \frac{n'}{2n} \]

### Step 4: Relate decay constant to half-life
The decay constant (\( \lambda \)) is also related to the half-life (\( t_{1/2} \)) of the radioactive substance by the formula:
\[ \lambda = \frac{0.693}{t_{1/2}} \]
We can set the two expressions for \( \lambda \) equal to each other:
\[ \frac{0.693}{t_{1/2}} = \frac{n'}{2n} \]

### Step 5: Solve for half-life (\( t_{1/2} \))
Now, we can solve for the half-life (\( t_{1/2} \)):
\[ t_{1/2} = \frac{0.693 \cdot 2n}{n'} \]
This simplifies to:
\[ t_{1/2} = \frac{1.386n}{n'} \]

### Final Result
Thus, the half-life of the nuclei is:
\[ t_{1/2} = \frac{1.386n}{n'} \]

---