The ionization potential for the electron in the ground state of hydrogen atom is 13.6eV. what would be the ionization potential for the electron in the first excited state of` Li^(2+)`?

Options

A. `54.4`eV
B. `5.4` eV
C. `30.6`eV
D. `84.4e`V

Answer

To find the ionization potential for the electron in the first excited state of the lithium ion \( \text{Li}^{2+} \), we can use the formula for the ionization energy of hydrogen-like atoms. The ionization energy (or ionization potential) for a hydrogen-like atom can be calculated using the formula:

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

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

### Step 1: Identify the values for \( Z \) and \( n \)

For \( \text{Li}^{2+} \):
- The atomic number \( Z = 3 \) (since lithium has 3 protons).
- The first excited state corresponds to \( n = 2 \).

### Step 2: Substitute the values into the formula

Now we will substitute \( Z = 3 \) and \( n = 2 \) into the formula:

\[ E_2 = -\frac{3^2 \cdot 13.6 \, \text{eV}}{2^2} \]

### Step 3: Calculate \( E_2 \)

Calculating \( E_2 \):

\[ E_2 = -\frac{9 \cdot 13.6 \, \text{eV}}{4} \]

\[ E_2 = -\frac{122.4 \, \text{eV}}{4} \]

\[ E_2 = -30.6 \, \text{eV} \]

### Step 4: Determine the ionization potential

The ionization potential is the energy required to remove the electron from the atom, which is the absolute value of \( E_2 \):

\[ \text{Ionization Potential} = |E_2| = 30.6 \, \text{eV} \]

### Final Answer

The ionization potential for the electron in the first excited state of \( \text{Li}^{2+} \) is **30.6 eV**.

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