Find the excitation energy of n = 3 level of He atom

Options

A. 48.4 eV
B. 2.4 eV
C. 3.8 eV
D. 24.2 eV

Answer

To find the excitation energy of the n = 3 level of the helium atom, we will follow these steps:

### Step 1: Understand the Concept of Excitation Energy
Excitation energy is the energy required to move an electron from a lower energy level (ground state) to a higher energy level. For the helium atom, we will consider the transition from the ground state (n = 1) to the excited state (n = 3).

### Step 2: Write the Formula for Energy Levels
The energy of an electron in a hydrogen-like atom is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
where:
- \( E_n \) is the energy of the level,
- \( Z \) is the atomic number (for helium, \( Z = 2 \)),
- \( n \) is the principal quantum number.

### Step 3: Calculate the Energy for n = 1 and n = 3
1. **For n = 1 (Ground State)**:
\[ E_1 = -\frac{13.6 \, \text{eV} \cdot 2^2}{1^2} = -\frac{13.6 \cdot 4}{1} = -54.4 \, \text{eV} \]

2. **For n = 3 (Excited State)**:
\[ E_3 = -\frac{13.6 \, \text{eV} \cdot 2^2}{3^2} = -\frac{13.6 \cdot 4}{9} = -6.04 \, \text{eV} \]

### Step 4: Calculate the Excitation Energy
The excitation energy (\( E_{EX} \)) required to move from the ground state to the excited state is given by:
\[ E_{EX} = E_3 - E_1 \]
Substituting the values we calculated:
\[ E_{EX} = (-6.04 \, \text{eV}) - (-54.4 \, \text{eV}) = -6.04 + 54.4 = 48.36 \, \text{eV} \]

### Step 5: Final Result
Thus, the excitation energy of the n = 3 level of the helium atom is approximately:
\[ E_{EX} \approx 48.4 \, \text{eV} \]