Which is the correct relationship-

Options

A. `E_(1)` of `H=1//2E_(2)` of `He^(+)=1//3E_(3)`of `Li^(2+)=1//4E_(4)`of `Be^(3+)`
B. `E_(1)(H)=E_(2)(He^(+))=E_(3)(Li^(2+))=E_(4)(Be^(3+))`
C. `E_(1)(H)=2E_(2)(He^(+))=3E_(3)(Li^(2+))=4E_(4)(Be^(3+))`
D. No relation

Explanation

To solve the question regarding the correct relationship for the energy of hydrogen, helium, lithium, and beryllium ions, we will use the formula for the energy of an electron in a hydrogen-like atom:

\[ E_n = -\frac{13.6 Z^2}{n^2} \text{ eV} \]

where:
- \( E_n \) is the energy of the electron,
- \( Z \) is the atomic number,
- \( n \) is the principal quantum number.

### Step-by-Step Solution:

1. **Calculate Energy for Hydrogen (H)**:
- For hydrogen, \( Z = 1 \) and \( n = 1 \).
\[ E_1 = -\frac{13.6 \times 1^2}{1^2} = -13.6 \text{ eV} \]

2. **Calculate Energy for Helium Ion (He\(^+\))**:
- For helium ion, \( Z = 2 \) and \( n = 1 \).
\[ E_2 = -\frac{13.6 \times 2^2}{1^2} = -\frac{13.6 \times 4}{1} = -54.4 \text{ eV} \]

3. **Calculate Energy for Lithium Ion (Li\(^{2+}\))**:
- For lithium ion, \( Z = 3 \) and \( n = 1 \).
\[ E_3 = -\frac{13.6 \times 3^2}{1^2} = -\frac{13.6 \times 9}{1} = -122.4 \text{ eV} \]

4. **Calculate Energy for Beryllium Ion (Be\(^{3+}\))**:
- For beryllium ion, \( Z = 4 \) and \( n = 1 \).
\[ E_4 = -\frac{13.6 \times 4^2}{1^2} = -\frac{13.6 \times 16}{1} = -217.6 \text{ eV} \]

### Summary of Energies:
- \( E_1 = -13.6 \text{ eV} \) (Hydrogen)
- \( E_2 = -54.4 \text{ eV} \) (Helium ion)
- \( E_3 = -122.4 \text{ eV} \) (Lithium ion)
- \( E_4 = -217.6 \text{ eV} \) (Beryllium ion)

### Conclusion:
The energies of the hydrogen, helium, lithium, and beryllium ions are not equal. Therefore, the correct relationship is that the energy increases in magnitude (becomes less negative) as we move from hydrogen to beryllium.