If the I.P. of `Li ^(+2)is 122.4 ` eV. Find out `6 ^(th)I.P.` of carbon -

Options

A. `122. 4 xx 4 eV`
B. `122.4 xx2 eV`
C. `122.4xx3eV`
D. `122.4 xx5eV`

Answer

To find the 6th ionization potential (I.P.) of carbon given the I.P. of `Li^(+2)` is 122.4 eV, we can follow these steps:

### Step 1: Understand the Ionization Potential (I.P.)
The ionization potential (I.P.) is the energy required to remove an electron from an atom or ion in its gaseous state.

### Step 2: Determine the Energy Levels
Using Bohr's model, the energy of the nth orbit for a hydrogen-like atom is given by:
\[ E_n = -\frac{13.6 \, Z^2}{n^2} \, \text{eV} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number.

### Step 3: Calculate the Energy for `Li^(+2)`
For `Li^(+2)`, the atomic number \( Z = 3 \). The first ionization potential corresponds to the transition from the first energy level (n=1) to infinity (n=∞):
- \( E_{\infty} = 0 \) eV (as n approaches infinity)
- \( E_1 = -\frac{13.6 \cdot 3^2}{1^2} = -\frac{13.6 \cdot 9}{1} = -122.4 \, \text{eV} \)

Thus, the first ionization energy (I.P.) is:
\[ I.P. = E_{\infty} - E_1 = 0 - (-122.4) = 122.4 \, \text{eV} \]

### Step 4: Determine the Electronic Configuration of Carbon
The electronic configuration of carbon (C) is \( 1s^2 \, 2s^2 \, 2p^2 \). When 5 electrons are removed, the configuration of \( C^{+5} \) will be \( 1s^1 \).

### Step 5: Calculate the Energy for Carbon
For \( C^{+5} \) (where \( Z = 6 \)):
- The energy for the first level (n=1) is:
\[ E_1 = -\frac{13.6 \cdot 6^2}{1^2} = -\frac{13.6 \cdot 36}{1} = -489.6 \, \text{eV} \]

### Step 6: Calculate the 6th Ionization Potential
The 6th ionization potential corresponds to the energy required to remove the last electron from \( C^{+5} \):
\[ I.P. = E_{\infty} - E_1 = 0 - (-489.6) = 489.6 \, \text{eV} \]

### Final Answer
The 6th ionization potential of carbon is **489.6 eV**.

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