The ionization energy of `He^(+)` is x times that of H. The ionization energy of `Li^(2+)` is y times that of H. find `|y-x|`

Explanation

To solve the problem, we need to find the values of \( x \) and \( y \) based on the ionization energies of \( He^+ \) and \( Li^{2+} \) in relation to that of hydrogen (\( H \)).

### Step-by-Step Solution:

1. **Understand Ionization Energy Formula**:
The ionization energy (IE) for a hydrogen-like atom can be calculated using the formula:
\[ \text{IE} = 13.6 \, \text{eV} \times Z^2 \]
where \( Z \) is the atomic number of the element.

2. **Calculate Ionization Energy for \( He^+ \)**:
- For \( He^+ \), the atomic number \( Z = 2 \).
- Therefore, the ionization energy of \( He^+ \) is:
\[ \text{IE}_{He^+} = 13.6 \, \text{eV} \times 2^2 = 13.6 \, \text{eV} \times 4 = 54.4 \, \text{eV} \]
- According to the problem, this is \( x \) times the ionization energy of hydrogen (\( 13.6 \, \text{eV} \)):
\[ 54.4 \, \text{eV} = 13.6 \, \text{eV} \times x \]
- Dividing both sides by \( 13.6 \, \text{eV} \):
\[ x = \frac{54.4}{13.6} = 4 \]

3. **Calculate Ionization Energy for \( Li^{2+} \)**:
- For \( Li^{2+} \), the atomic number \( Z = 3 \).
- Therefore, the ionization energy of \( Li^{2+} \) is:
\[ \text{IE}_{Li^{2+}} = 13.6 \, \text{eV} \times 3^2 = 13.6 \, \text{eV} \times 9 = 122.4 \, \text{eV} \]
- According to the problem, this is \( y \) times the ionization energy of hydrogen:
\[ 122.4 \, \text{eV} = 13.6 \, \text{eV} \times y \]
- Dividing both sides by \( 13.6 \, \text{eV} \):
\[ y = \frac{122.4}{13.6} = 9 \]

4. **Calculate \( |y - x| \)**:
- Now that we have \( x = 4 \) and \( y = 9 \):
\[ |y - x| = |9 - 4| = 5 \]

### Final Answer:
\[ |y - x| = 5 \]