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 ionization energies of the ions \( He^{+} \) and \( Li^{2+} \) in terms of the ionization energy of hydrogen (\( H \)), and then calculate \( |y - x| \).

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

### Step 2: Calculate Ionization Energy of \( He^{+} \)
For \( He^{+} \):
- The atomic number \( Z \) is 2.
- Thus, the ionization energy of \( He^{+} \) is:
\[ IE(He^{+}) = 13.6 \, \text{eV} \times 2^2 = 13.6 \, \text{eV} \times 4 = 54.4 \, \text{eV} \]
- Since the ionization energy of hydrogen (\( H \)) is 13.6 eV, we can express \( x \) as:
\[ x = \frac{IE(He^{+})}{IE(H)} = \frac{54.4 \, \text{eV}}{13.6 \, \text{eV}} = 4 \]

### Step 3: Calculate Ionization Energy of \( Li^{2+} \)
For \( Li^{2+} \):
- The atomic number \( Z \) is 3.
- Thus, the ionization energy of \( Li^{2+} \) is:
\[ IE(Li^{2+}) = 13.6 \, \text{eV} \times 3^2 = 13.6 \, \text{eV} \times 9 = 122.4 \, \text{eV} \]
- We can express \( y \) as:
\[ y = \frac{IE(Li^{2+})}{IE(H)} = \frac{122.4 \, \text{eV}}{13.6 \, \text{eV}} = 9 \]

### Step 4: Calculate \( |y - x| \)
Now we have:
- \( x = 4 \)
- \( y = 9 \)

Thus,
\[ |y - x| = |9 - 4| = 5 \]

### Final Answer
The value of \( |y - x| \) is \( 5 \).

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