The energies of three consecutive 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 three consecutive energy levels \(L_3\), \(L_2\), and \(L_1\).

### Step-by-Step Solution:

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

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

3. **Relate Energy Change to Wavelength**:
- According to Planck's law, the energy of a photon is given by:
\[ E = \frac{hc}{\lambda} \]
- Setting the energy change equal to the energy of the emitted photon:
\[ \frac{3E_0}{4} = \frac{hc}{\lambda} \]
- Rearranging gives us:
\[ \lambda = \frac{hc}{\frac{3E_0}{4}} = \frac{4hc}{3E_0} \]

4. **Calculate the Energy Change for Transition \(L_2 \to L_1\)**:
- The energy change (\(\Delta E_{21}\)) 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:
\[ \Delta E_{21} = \left(\frac{4E_0}{9} \cdot \frac{4}{4}\right) - \left(\frac{E_0}{4} \cdot \frac{9}{9}\right) = \frac{16E_0}{36} - \frac{9E_0}{36} = \frac{7E_0}{36} \]

5. **Relate Energy Change to Wavelength for Transition \(L_2 \to L_1\)**:
- Using Planck's law again:
\[ \Delta E_{21} = \frac{hc}{\lambda_1} \]
- Setting the energy change equal to the energy of the emitted photon gives:
\[ \frac{7E_0}{36} = \frac{hc}{\lambda_1} \]
- Rearranging gives us:
\[ \lambda_1 = \frac{hc}{\frac{7E_0}{36}} = \frac{36hc}{7E_0} \]

6. **Relate the Two Wavelengths**:
- From the previous steps, we have:
\[ \lambda = \frac{4hc}{3E_0} \]
- Now, we can express \(\lambda_1\) in terms of \(\lambda\):
\[ \lambda_1 = \frac{36hc}{7E_0} = \frac{36}{7} \cdot \frac{hc}{E_0} = \frac{36}{7} \cdot \frac{4}{3} \lambda = \frac{48}{7} \lambda \]

### Final Answer:
The wavelength of emission for the transition \(L_2 \to L_1\) is:
\[ \lambda_1 = \frac{48}{7} \lambda \]