The energies of three conservative energy levels `L_3,L_2` and `L_1` of hydrogen atom are `E_0`. `(4E_0)/9` and `E_0/4` respectively. A photon of wavelength `lambda` is emitted for a transition `L_3` to `L_1` .What will be the wavelength of emission for transition `L_2` to `L_1` ?

Options

A. `(16lambda)/31`
B. `(27lambda)/7`
C. `19/20 lambda`
D. `lambda`

Answer

To solve the problem, we need to find the wavelength of emission for the transition from energy level \( L_2 \) to \( L_1 \) in a hydrogen atom, given the energies of the levels and the wavelength for the transition from \( L_3 \) to \( L_1 \).

### Step-by-Step Solution:

1. **Identify the Energy Levels:**
- Given:
- Energy of \( L_3 = E_0 \)
- Energy of \( L_2 = \frac{4E_0}{9} \)
- Energy of \( L_1 = \frac{E_0}{4} \)

2. **Calculate the Energy Difference for Transition \( L_3 \) to \( L_1 \):**
- The energy difference \( \Delta E_{31} \) for the transition from \( L_3 \) to \( L_1 \) is:
\[ \Delta E_{31} = E_{L_3} - E_{L_1} = E_0 - \frac{E_0}{4} = \frac{4E_0}{4} - \frac{E_0}{4} = \frac{3E_0}{4} \]

3. **Relate Energy Difference to Wavelength:**
- The energy of the emitted photon is related to its wavelength \( \lambda \) by the equation:
\[ E = \frac{hc}{\lambda} \]
- For the transition \( L_3 \) to \( L_1 \):
\[ \frac{hc}{\lambda} = \frac{3E_0}{4} \]
- Rearranging gives:
\[ hc = \frac{3E_0 \lambda}{4} \]

4. **Calculate the Energy Difference for Transition \( L_2 \) to \( L_1 \):**
- The energy difference \( \Delta E_{21} \) for the transition from \( L_2 \) to \( L_1 \) is:
\[ \Delta E_{21} = E_{L_2} - E_{L_1} = \frac{4E_0}{9} - \frac{E_0}{4} \]
- To subtract these fractions, we need a common denominator. The least common multiple of 9 and 4 is 36:
\[ \Delta E_{21} = \left(\frac{4E_0 \cdot 4}{36}\right) - \left(\frac{E_0 \cdot 9}{36}\right) = \frac{16E_0}{36} - \frac{9E_0}{36} = \frac{7E_0}{36} \]

5. **Relate Energy Difference to Wavelength for Transition \( L_2 \) to \( L_1 \):**
- Using the energy-wavelength relationship for this transition:
\[ \frac{hc}{\lambda'} = \frac{7E_0}{36} \]
- Rearranging gives:
\[ hc = \frac{7E_0 \lambda'}{36} \]

6. **Equate the Two Expressions for \( hc \):**
- From the two expressions for \( hc \):
\[ \frac{3E_0 \lambda}{4} = \frac{7E_0 \lambda'}{36} \]
- Cancel \( E_0 \) from both sides:
\[ \frac{3\lambda}{4} = \frac{7\lambda'}{36} \]

7. **Solve for \( \lambda' \):**
- Cross-multiplying gives:
\[ 3 \cdot 36 \lambda' = 7 \cdot 4 \lambda \]
\[ 108 \lambda' = 28 \lambda \]
\[ \lambda' = \frac{28 \lambda}{108} = \frac{7 \lambda}{27} \]

### Final Answer:
The wavelength of emission for the transition from \( L_2 \) to \( L_1 \) is:
\[ \lambda' = \frac{7}{27} \lambda \]