The frequency of revolution of an electron in nth orbit is `f_(n)`. If the electron makes a transition from nth orbit to `(n = 1)` th orbit , then the relation between the frequency `(v)` of emitted photon and `f_(n)` will be

Options

A. `v = f_(n)^(2)`
B. `v = sqrt f_(n)`
C. `v = (1)/(f_(n)`
D. `v = f_(n)`

Explanation

To solve the problem, we need to establish the relationship between the frequency of the emitted photon (ν) when an electron transitions from the nth orbit to the first orbit, and the frequency of revolution of the electron in the nth orbit (f_n).

### Step-by-Step Solution:

1. **Understanding the Frequency of Revolution**:
The frequency of revolution of an electron in the nth orbit is given by the formula:
\[ f_n = \frac{4 \pi^2 k^2 m e^4}{n^3 h^3} \]
where:
- \( k \) is Coulomb's constant,
- \( m \) is the mass of the electron,
- \( e \) is the charge of the electron,
- \( h \) is Planck's constant.

2. **Energy of the Electron in nth Orbit**:
The energy of an electron in the nth orbit can be expressed as:
\[ E_n = -\frac{k e^2}{2n^2} \]
The negative sign indicates that the electron is bound to the nucleus.

3. **Energy Transition**:
When the electron transitions from the nth orbit to the first orbit (n=1), the change in energy (ΔE) is given by:
\[ \Delta E = E_n - E_1 = \left(-\frac{k e^2}{2n^2}\right) - \left(-\frac{k e^2}{2}\right) \]
Simplifying this, we have:
\[ \Delta E = \frac{k e^2}{2} - \frac{k e^2}{2n^2} = \frac{k e^2}{2} \left(1 - \frac{1}{n^2}\right) \]

4. **Photon Emission**:
The energy of the emitted photon (E) during this transition is equal to the change in energy:
\[ E = h \nu \]
Therefore, we can write:
\[ h \nu = \frac{k e^2}{2} \left(1 - \frac{1}{n^2}\right) \]

5. **Finding the Frequency of the Emitted Photon**:
Rearranging the equation for ν gives:
\[ \nu = \frac{k e^2}{2h} \left(1 - \frac{1}{n^2}\right) \]

6. **Relating ν to f_n**:
From the expression for \( f_n \):
\[ f_n = \frac{4 \pi^2 k^2 m e^4}{n^3 h^3} \]
We can see that the term \( \frac{k e^2}{2h} \) can be related to \( f_n \) by considering the factors involved and recognizing that:
\[ \nu = f_n \left(1 - \frac{1}{n^2}\right) \]

7. **Final Relation**:
Thus, the relation between the frequency of the emitted photon (ν) and the frequency of revolution in the nth orbit (f_n) is:
\[ \nu = f_n \left(1 - \frac{1}{n^2}\right) \]