The frequency of radiation absorbed or emitted when transition occurs between two stationary states with energies `E_(1)` (lower) and `E_(2)` (higher) is given by

Options

A. `v=(E_(1)+E_(2))/h`
B. `v=(E_(1)-E_(2))/h`
C. `v=(E_(1)xxE_(2))/h`
D. `v=(E_(2)-E_(1))/h`

Answer

To solve the question regarding the frequency of radiation absorbed or emitted during a transition between two stationary states with energies \( E_1 \) (lower) and \( E_2 \) (higher), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Energy Levels**:
- Let \( E_2 \) be the energy of the higher stationary state (n2).
- Let \( E_1 \) be the energy of the lower stationary state (n1).

2. **Understand the Transition**:
- When an electron transitions from a higher energy level (n2) to a lower energy level (n1), it emits energy in the form of radiation (a photon).

3. **Apply the Energy Difference Formula**:
- The energy of the emitted or absorbed photon is equal to the difference in energy between the two states. This can be expressed as:
\[ \Delta E = E_2 - E_1 \]

4. **Relate Energy to Frequency**:
- The energy of the photon is also related to its frequency (\( \nu \)) by the equation:
\[ E = h \nu \]
where \( h \) is Planck's constant.

5. **Combine the Equations**:
- Setting the two expressions for energy equal gives us:
\[ h \nu = E_2 - E_1 \]

6. **Solve for Frequency**:
- Rearranging the equation to solve for frequency (\( \nu \)) yields:
\[ \nu = \frac{E_2 - E_1}{h} \]

### Final Answer:
The frequency of radiation absorbed or emitted during the transition is given by:
\[ \nu = \frac{E_2 - E_1}{h} \]