In hydrogen atom electron jumps from `(n+1)th` state to `nth` state the frequency of emitted photon is directly proportional to `(n>>1)`

Options

A. `n`
B. `1/n`
C. `1/n^2`
D. `1/n^3`

Answer

To solve the problem of the frequency of the emitted photon when an electron in a hydrogen atom jumps from the \((n+1)\)th state to the \(n\)th state, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy Levels**:
The energy of an electron in the \(n\)th state of a hydrogen atom is given by the formula:
\[ E_n = -\frac{R_H}{n^2} \]
where \(R_H\) is the Rydberg constant.

2. **Calculate Energy for \(n\) and \(n+1\)**:
- For the \(n\)th state:
\[ E_n = -\frac{R_H}{n^2} \]
- For the \((n+1)\)th state:
\[ E_{n+1} = -\frac{R_H}{(n+1)^2} \]

3. **Determine the Change in Energy (\(\Delta E\))**:
The change in energy when the electron jumps from the \((n+1)\)th state to the \(n\)th state is:
\[ \Delta E = E_n - E_{n+1} = \left(-\frac{R_H}{n^2}\right) - \left(-\frac{R_H}{(n+1)^2}\right) \]
Simplifying this gives:
\[ \Delta E = \frac{R_H}{(n+1)^2} - \frac{R_H}{n^2} \]

4. **Combine the Terms**:
To combine the fractions, we find a common denominator:
\[ \Delta E = R_H \left( \frac{n^2 - (n+1)^2}{n^2(n+1)^2} \right) \]
Expanding \((n+1)^2\) gives:
\[ (n+1)^2 = n^2 + 2n + 1 \]
Thus:
\[ n^2 - (n^2 + 2n + 1) = -2n - 1 \]
Therefore:
\[ \Delta E = R_H \cdot \frac{-2n - 1}{n^2(n+1)^2} \]

5. **Approximate for Large \(n\)**:
Since \(n \gg 1\), we can approximate:
\[ \Delta E \approx -\frac{2R_H}{n^2(n^2)} = -\frac{2R_H}{n^4} \]

6. **Relate Energy Change to Frequency**:
The frequency (\(f\)) of the emitted photon is related to the energy change by:
\[ \Delta E = h f \]
Therefore:
\[ f = \frac{\Delta E}{h} = \frac{-2R_H}{h n^4} \]

7. **Conclusion**:
Since we are interested in the magnitude and the proportionality, we can express:
\[ f \propto \frac{1}{n^3} \]
Thus, the frequency of the emitted photon is inversely proportional to \(n^3\) when \(n \gg 1\).

### Final Answer:
The frequency of the emitted photon is directly proportional to \(\frac{1}{n^3}\).