The ratio of the third ionization energy of lithium and ionization energy of H atom is of H-atom and that of `Li^(+2)` ion is:

Options

A. `1:1`
B. `3:1`
C. `3:4`
D. `9:1`

Explanation

To solve the question regarding the ratio of the third ionization energy of lithium (Li) and the ionization energy of hydrogen (H), we will follow these steps:

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

### Step 2: Identify the Ionization Energies
1. **For Hydrogen (H)**:
- Atomic number \( Z_H = 1 \)
- For the first ionization (removing the only electron), \( n = 1 \):
\[ \text{IE}_H = \frac{13.6 \cdot 1^2}{1^2} = 13.6 \, \text{eV} \]

2. **For Lithium Ion (Li²⁺)**:
- Atomic number \( Z_{Li} = 3 \)
- After losing 2 electrons, Li²⁺ has 1 electron left, so \( n = 1 \):
\[ \text{IE}_{Li^{2+}} = \frac{13.6 \cdot 3^2}{1^2} = \frac{13.6 \cdot 9}{1} = 122.4 \, \text{eV} \]

### Step 3: Calculate the Ratio of Ionization Energies
Now we can find the ratio of the third ionization energy of lithium to the ionization energy of hydrogen:
\[ \text{Ratio} = \frac{\text{IE}_{Li^{2+}}}{\text{IE}_H} = \frac{122.4 \, \text{eV}}{13.6 \, \text{eV}} \]

### Step 4: Simplify the Ratio
Calculating the ratio:
\[ \text{Ratio} = \frac{122.4}{13.6} = 9 \]
Thus, the ratio of the third ionization energy of lithium to the ionization energy of hydrogen is:
\[ \text{Ratio} = 9:1 \]

### Final Answer
The ratio of the third ionization energy of lithium to the ionization energy of hydrogen is \( 9:1 \).

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