What is the separation energy (in eV ) for `Be^(3+)` in the first excited state ?

Options

A. (a) 13.6 eV
B. (b) 27.2 eV
C. (c) 40.8 eV
D. (d) 54.5 eV

Answer

To find the separation energy for \( Be^{3+} \) in the first excited state, we can follow these steps:

### Step 1: Understand the Concept of Separation Energy
Separation energy is the energy required to remove an electron from an atom or ion. In the case of \( Be^{3+} \), we are interested in the energy needed to remove the electron from its first excited state.

### Step 2: Identify the Quantum State
For \( Be^{3+} \), the first excited state corresponds to the principal quantum number \( n = 2 \).

### Step 3: Use the Formula for Energy Levels
According to Bohr's model, the energy of an electron in a hydrogen-like atom is given by the formula:
\[ E_n = -\frac{13.6 \, Z^2}{n^2} \, \text{eV} \]
where:
- \( E_n \) is the energy of the electron at level \( n \),
- \( Z \) is the atomic number,
- \( n \) is the principal quantum number.

### Step 4: Substitute the Values
For \( Be^{3+} \):
- The atomic number \( Z = 4 \) (since beryllium has 4 protons).
- The principal quantum number for the first excited state \( n = 2 \).

Now, substituting these values into the formula:
\[ E_2 = -\frac{13.6 \times 4^2}{2^2} \]

### Step 5: Calculate the Energy
Calculating the values:
\[ E_2 = -\frac{13.6 \times 16}{4} = -\frac{217.6}{4} = -54.4 \, \text{eV} \]

### Step 6: Determine the Separation Energy
The separation energy is the energy required to remove the electron from the excited state, which is the positive value of the energy calculated:
\[ \text{Separation Energy} = +54.4 \, \text{eV} \]

### Final Answer
The separation energy for \( Be^{3+} \) in the first excited state is \( +54.4 \, \text{eV} \).

---