The energy of a I,II and III energy levels of a certain atom are `E,(4E)/(3)` and 2E respectively. A photon of wavelength `lambda` is emitted during a transition from III to I. what will be the wavelength of emission for II to I?

Options

A. `(lambda)/(2)`
B. `lambda`
C. `2lambda`
D. `3lambda`

Answer

To solve the problem, we need to find the wavelength of the photon emitted during the transition from the second energy level (E2) to the first energy level (E1) of a certain atom, given the energy levels and the wavelength of a photon emitted during the transition from the third energy level (E3) to the first energy level (E1).

### Step-by-Step Solution:

1. **Identify the Energy Levels**:
- E1 = E (Energy of the first level)
- E2 = (4/3)E (Energy of the second level)
- E3 = 2E (Energy of the third level)

2. **Calculate the Energy Difference for the Transition from E3 to E1**:
- The energy difference (ΔE) when a transition occurs from E3 to E1 is:
\[ \Delta E_{3 \to 1} = E3 - E1 = 2E - E = E \]

3. **Relate Energy to Wavelength**:
- The energy of the emitted photon is related to its wavelength (λ) by the equation:
\[ E = \frac{hc}{\lambda} \]
- From the transition E3 to E1, we have:
\[ E = \frac{hc}{\lambda} \]
- This gives us our first equation:
\[ E = \frac{hc}{\lambda} \quad \text{(Equation 1)} \]

4. **Calculate the Energy Difference for the Transition from E2 to E1**:
- Now, we calculate the energy difference for the transition from E2 to E1:
\[ \Delta E_{2 \to 1} = E2 - E1 = \frac{4E}{3} - E = \frac{4E}{3} - \frac{3E}{3} = \frac{E}{3} \]

5. **Relate Energy Difference to Wavelength for the Transition from E2 to E1**:
- Using the same energy-wavelength relationship:
\[ \Delta E_{2 \to 1} = \frac{hc}{\lambda'} \]
- Substituting the energy difference we found:
\[ \frac{E}{3} = \frac{hc}{\lambda'} \]

6. **Substituting Equation 1 into the New Equation**:
- From Equation 1, we know that \( E = \frac{hc}{\lambda} \). Substituting this into the equation gives:
\[ \frac{\frac{hc}{\lambda}}{3} = \frac{hc}{\lambda'} \]

7. **Canceling hc and Rearranging**:
- Cancel \( hc \) from both sides:
\[ \frac{1}{3\lambda} = \frac{1}{\lambda'} \]
- Rearranging gives:
\[ \lambda' = 3\lambda \]

### Final Answer:
The wavelength of emission for the transition from the second energy level to the first energy level is:
\[ \lambda' = 3\lambda \]