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 (v) 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 Frequencies**:
- The frequency of the revolution of an electron in the nth orbit is denoted as \( f_n \).
- When the electron transitions from the nth orbit to the first orbit, it emits a photon with frequency \( v \).

2. **Energy of the Electron**:
- The energy of the electron in the nth orbit can be expressed as:
\[ E_n = -\frac{k e^4 m_e}{2 n^2} \]
- Here, \( k \) is a constant, \( e \) is the charge of the electron, and \( m_e \) is the mass of the electron.

3. **Photon Emission**:
- The energy of the emitted photon when the electron transitions from the nth orbit to the first orbit is given by:
\[ E_{photon} = E_n - E_1 \]
- Substituting the energy expressions:
\[ E_{photon} = \left(-\frac{k e^4 m_e}{2 n^2}\right) - \left(-\frac{k e^4 m_e}{2}\right) \]
- Simplifying this gives:
\[ E_{photon} = \frac{k e^4 m_e}{2} \left(1 - \frac{1}{n^2}\right) \]

4. **Relating Energy to Frequency**:
- The energy of the emitted photon can also be expressed in terms of its frequency:
\[ E_{photon} = h v \]
- Where \( h \) is Planck's constant. Setting the two expressions for energy equal gives:
\[ h v = \frac{k e^4 m_e}{2} \left(1 - \frac{1}{n^2}\right) \]

5. **Finding the Frequency of the Photon**:
- Rearranging for \( v \):
\[ v = \frac{k e^4 m_e}{2h} \left(1 - \frac{1}{n^2}\right) \]

6. **Relating to \( f_n \)**:
- The frequency of the electron in the nth orbit can be expressed as:
\[ f_n = \frac{k e^4 m_e}{h n^3} \]
- Now we can express \( v \) in terms of \( f_n \):
\[ v = f_n \cdot \frac{1}{2} \left(1 - \frac{1}{n^2}\right) n^3 \]
- As \( n \) approaches infinity, the term \( \left(1 - \frac{1}{n^2}\right) \) approaches 1, leading to:
\[ v \approx f_n \]

7. **Final Relation**:
- Therefore, the relationship between the frequency of the emitted photon \( v \) and the frequency of revolution of the electron \( f_n \) is:
\[ v = f_n \cdot \text{(some constant)} \]
- For large \( n \), this simplifies to:
\[ v \approx f_n \]

### Conclusion:
Thus, the final relation between the frequency \( v \) of the emitted photon and \( f_n \) is:
\[ v \approx f_n \]