Radius of `1^(st)` orbit of H & some orbit of `Be^(3+)` is same .
Energy of their orbit of `Be^(3+)` is :

Options

A. `-54.4 eV`
B. `-13.6 eV`
C. `-108.8eV`
D. `-27.2eV`

Explanation

To solve the problem, we need to find the energy of the orbit of \( \text{Be}^{3+} \) given that the radius of the first orbit of hydrogen (H) is the same as that of some orbit of \( \text{Be}^{3+} \).

### Step-by-Step Solution:

1. **Understand the relationship between radius and quantum numbers**:
The radius of an electron orbit in a hydrogen-like atom is given by the formula:
\[ R_n = \frac{n^2}{Z} \cdot R_0 \]
where \( R_0 \) is the Bohr radius, \( n \) is the principal quantum number, and \( Z \) is the atomic number.

2. **Set up the equation for hydrogen and beryllium**:
For hydrogen (\( Z = 1 \)), the radius of the first orbit (\( n = 1 \)) is:
\[ R_1 = \frac{1^2}{1} \cdot R_0 = R_0 \]
For \( \text{Be}^{3+} \) (\( Z = 4 \)), let the principal quantum number be \( n \). The radius of the orbit for \( \text{Be}^{3+} \) is:
\[ R_n = \frac{n^2}{4} \cdot R_0 \]

3. **Equate the radii**:
Since the radius of the first orbit of hydrogen is the same as that of some orbit of \( \text{Be}^{3+} \):
\[ R_0 = \frac{n^2}{4} \cdot R_0 \]
Dividing both sides by \( R_0 \) (assuming \( R_0 \neq 0 \)):
\[ 1 = \frac{n^2}{4} \]
Rearranging gives:
\[ n^2 = 4 \quad \Rightarrow \quad n = 2 \]

4. **Calculate the energy of the orbit for \( \text{Be}^{3+} \)**:
The energy of an electron in a hydrogen-like atom is given by:
\[ E_n = -\frac{13.6 Z^2}{n^2} \text{ eV} \]
For \( \text{Be}^{3+} \) with \( Z = 4 \) and \( n = 2 \):
\[ E_2 = -\frac{13.6 \cdot 4^2}{2^2} = -\frac{13.6 \cdot 16}{4} = -\frac{217.6}{4} = -54.4 \text{ eV} \]

5. **Final answer**:
The energy of the orbit of \( \text{Be}^{3+} \) is:
\[ E = -54.4 \text{ eV} \]