The wavelength of the first line of Lyman series for hydrogen is identical to that of the second line of Balmer series for some hydrogen like ion x. Calculate energies of the first four levels of x.

Explanation

To solve the problem, we need to find the energies of the first four levels of a hydrogen-like ion \( x \) whose atomic number \( Z \) we will determine based on the given condition regarding the wavelengths of the spectral lines.

### Step-by-Step Solution:

1. **Identify the transitions:**
- The first line of the Lyman series corresponds to the transition from \( n = 2 \) to \( n = 1 \) in hydrogen.
- The second line of the Balmer series corresponds to the transition from \( n = 4 \) to \( n = 2 \) in hydrogen-like ion \( x \).

2. **Use the Rydberg formula:**
The Rydberg formula for the wavelength \( \lambda \) of a spectral line is given by:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
where \( R \) is the Rydberg constant, \( Z \) is the atomic number, \( n_f \) is the final energy level, and \( n_i \) is the initial energy level.

3. **Calculate the wavelength for the Lyman series (hydrogen):**
For the Lyman series (first line):
- \( n_f = 1 \), \( n_i = 2 \)
\[ \frac{1}{\lambda_{Lyman}} = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \cdot \frac{3}{4} \]

4. **Calculate the wavelength for the Balmer series (ion \( x \)):**
For the Balmer series (second line):
- \( n_f = 2 \), \( n_i = 4 \)
\[ \frac{1}{\lambda_{Balmer}} = RZ^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = RZ^2 \left( \frac{1}{4} - \frac{1}{16} \right) = RZ^2 \cdot \frac{3}{16} \]

5. **Set the wavelengths equal:**
Since the wavelengths are identical:
\[ R \cdot \frac{3}{4} = RZ^2 \cdot \frac{3}{16} \]
Cancel \( R \) and \( 3 \) from both sides:
\[ \frac{1}{4} = \frac{Z^2}{16} \]
Cross-multiplying gives:
\[ Z^2 = 4 \implies Z = 2 \]

6. **Calculate the energies of the first four levels of ion \( x \):**
The energy levels for a hydrogen-like atom are given by:
\[ E_n = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]
Substituting \( Z = 2 \):
\[ E_n = -\frac{2^2 \cdot 13.6}{n^2} = -\frac{54.4}{n^2} \, \text{eV} \]

7. **Calculate energies for \( n = 1, 2, 3, 4 \):**
- For \( n = 1 \):
\[ E_1 = -\frac{54.4}{1^2} = -54.4 \, \text{eV} \]
- For \( n = 2 \):
\[ E_2 = -\frac{54.4}{2^2} = -13.6 \, \text{eV} \]
- For \( n = 3 \):
\[ E_3 = -\frac{54.4}{3^2} = -6.04 \, \text{eV} \]
- For \( n = 4 \):
\[ E_4 = -\frac{54.4}{4^2} = -3.4 \, \text{eV} \]

### Final Energies:
- \( E_1 = -54.4 \, \text{eV} \)
- \( E_2 = -13.6 \, \text{eV} \)
- \( E_3 = -6.04 \, \text{eV} \)
- \( E_4 = -3.4 \, \text{eV} \)