n number of `alpha`- particles are being emitted by N atoms of a radioactive elemtn. The half life of element will be

Options

A. `(N/n)s`
B. `(n/N)s`
C. `(0.693N)/ns`
D. `(0.693n)/Ns`

Answer

To find the half-life of a radioactive element that emits `n` alpha particles from `N` atoms, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Decay Law**:
The rate of decay of a radioactive sample is given by the equation:
\[ \frac{dN}{dt} = -\lambda N \]
where \(N\) is the number of atoms, \(\lambda\) is the decay constant, and \(t\) is time.

2. **Relating Alpha Particle Emission to Decay**:
According to the problem, `n` alpha particles are emitted by `N` atoms. This means that the rate of decay (number of decays per unit time) is equal to `n`:
\[ \frac{dN}{dt} = n \]

3. **Setting Up the Equation**:
By equating the two expressions for the rate of decay, we have:
\[ -\lambda N = n \]
Rearranging this gives:
\[ \lambda = -\frac{n}{N} \]
Since \(\lambda\) is a positive quantity, we can ignore the negative sign.

4. **Calculating Half-Life**:
The half-life \(T_{1/2}\) of a radioactive substance is given by the formula:
\[ T_{1/2} = \frac{\ln(2)}{\lambda} \]
Substituting our expression for \(\lambda\):
\[ T_{1/2} = \frac{\ln(2)}{n/N} = \frac{N \ln(2)}{n} \]

5. **Substituting the Value of \(\ln(2)\)**:
The value of \(\ln(2)\) is approximately \(0.693\). Thus:
\[ T_{1/2} = \frac{0.693 N}{n} \]

6. **Final Result**:
Therefore, the half-life of the radioactive element is:
\[ T_{1/2} = 0.693 \frac{N}{n} \text{ seconds} \]

### Conclusion:
The half-life of the element is \(0.693 \frac{N}{n}\) seconds.