In a hydrogen atom the electron makes a transition from `(n+1)^(th)` level to the`n^(th)` level .If ` n gt gt 1 ` the frequency of radiation emitted is proportional to :

Options

A. `1/(n^(2))`
B. `1/n`
C. `1/(n^(3))`
D. `1/(n^(4))`

Explanation

To solve the problem, we need to determine the frequency of radiation emitted when an electron in a hydrogen atom transitions from the \((n+1)^{th}\) energy level to the \(n^{th}\) energy level.

### Step-by-Step Solution:

1. **Understanding Energy Levels**:
The energy of an electron in the \(n^{th}\) level of a hydrogen atom is given by the formula:
\[ E_n = -\frac{Rhc}{n^2} \]
where \(R\) is the Rydberg constant, \(h\) is Planck's constant, and \(c\) is the speed of light.

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

3. **Energy Difference**:
The energy difference \(\Delta E\) between these two levels when the electron transitions from \((n+1)\) to \(n\) is:
\[ \Delta E = E_n - E_{n+1} \]
Substituting the expressions for \(E_n\) and \(E_{n+1}\):
\[ \Delta E = -\frac{Rhc}{n^2} - \left(-\frac{Rhc}{(n+1)^2}\right) \]
This simplifies to:
\[ \Delta E = \frac{Rhc}{(n+1)^2} - \frac{Rhc}{n^2} \]

4. **Finding a Common Denominator**:
To combine these fractions, we find a common denominator:
\[ \Delta E = Rhc \left( \frac{n^2 - (n+1)^2}{n^2(n+1)^2} \right) \]
Simplifying \(n^2 - (n+1)^2\) gives:
\[ n^2 - (n^2 + 2n + 1) = -2n - 1 \]
Thus:
\[ \Delta E = Rhc \left( \frac{-2n - 1}{n^2(n+1)^2} \right) \]

5. **Frequency of the Emitted Radiation**:
The frequency \(f\) of the emitted radiation is related to the energy difference by Planck's equation:
\[ \Delta E = hf \]
Therefore:
\[ f = \frac{\Delta E}{h} = \frac{R c}{n^2(n+1)^2} \cdot \frac{-2n - 1}{h} \]

6. **Asymptotic Behavior for Large \(n\)**:
If \(n \gg 1\), we can approximate \((n+1)^2 \approx n^2\). Thus:
\[ f \approx \frac{R c (-2n - 1)}{h n^4} \]
The dominant term for large \(n\) is:
\[ f \propto \frac{1}{n^3} \]

### Final Result:
The frequency of radiation emitted is proportional to:
\[ f \propto \frac{1}{n^3} \]