What is the potential energy of an electron present in `N-` shell of the `Be^(3+)` ion ?

Options

A. `-3.4 eV`
B. `-6.8 eV`
C. `-13.6 eV`
D. `-27.2 eV`

Answer

To find the potential energy of an electron in the N-shell of the \( \text{Be}^{3+} \) ion, we can follow these steps:

### Step 1: Identify the principal quantum number (n)
The N-shell corresponds to the principal quantum number \( n = 4 \).

### Step 2: Determine the atomic number (Z) of Beryllium
Beryllium (Be) has an atomic number \( Z = 4 \).

### Step 3: Use the formula for total energy
The total energy \( E \) of an electron in a hydrogen-like ion is given by the formula:
\[ E = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
Substituting the values for \( Z \) and \( n \):
\[ E = -\frac{13.6 \, \text{eV} \cdot 4^2}{4^2} \]

### Step 4: Calculate the total energy
Calculating the above expression:
\[ E = -\frac{13.6 \, \text{eV} \cdot 16}{16} = -13.6 \, \text{eV} \]

### Step 5: Calculate the potential energy
The potential energy \( U \) of the electron is related to the total energy by the equation:
\[ U = 2E \]
Substituting the total energy we calculated:
\[ U = 2 \cdot (-13.6 \, \text{eV}) = -27.2 \, \text{eV} \]

### Final Answer
The potential energy of an electron present in the N-shell of the \( \text{Be}^{3+} \) ion is:
\[ \boxed{-27.2 \, \text{eV}} \]