The ionization potential for the electron in the ground state of the hydrogen atom is 13.6 eV `"atom"^(-1).` What would be the inization potential for the electron in the first excited state of `Li^(+)` ?

Options

A. `3.4` eV
B. `10.2` eV
C. `30.6` eV
D. `6.8` eV

Answer

To find the ionization potential for the electron in the first excited state of the lithium ion (Li⁺), we can follow these steps:

### Step 1: Understand Ionization Potential
Ionization potential (or ionization energy) is the energy required to remove an electron from an atom or ion. For hydrogen, the ionization potential in the ground state is given as 13.6 eV.

### Step 2: Use the Formula for Ionization Energy
The ionization energy for an electron in a hydrogen-like atom can be calculated using the formula:
\[ E = -13.6 \, \text{eV} \cdot Z^2 \cdot \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \(Z\) is the atomic number,
- \(n_1\) is the principal quantum number of the initial state,
- \(n_2\) is the principal quantum number of the final state (infinity for ionization).

### Step 3: Identify Values for Lithium Ion (Li⁺)
For Li⁺:
- The atomic number \(Z = 3\).
- For the first excited state, the electron is in \(n_1 = 2\).
- The final state for ionization is \(n_2 = \infty\).

### Step 4: Substitute Values into the Formula
Now we can substitute these values into the formula:
\[ E = -13.6 \, \text{eV} \cdot 3^2 \cdot \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) \]
Calculating this gives:
\[ E = -13.6 \, \text{eV} \cdot 9 \cdot \left( \frac{1}{4} - 0 \right) \]
\[ E = -13.6 \, \text{eV} \cdot 9 \cdot \frac{1}{4} \]
\[ E = -30.6 \, \text{eV} \]

### Step 5: Determine the Ionization Potential
Since ionization potential is the energy required to remove the electron, we take the positive value:
\[ \text{Ionization Potential} = 30.6 \, \text{eV} \]

### Final Answer
The ionization potential for the electron in the first excited state of Li⁺ is **30.6 eV**.

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