Which state of the triply ionized Beryllium `(Be^(3+))` has the same orbit radius as that of the ground state of hydrogen atom?

Options

A. `n=1`
B. `n=2`
C. `n=3`
D. `n=4`

Explanation

To solve the problem of finding which state of the triply ionized Beryllium `(Be^(3+))` has the same orbit radius as that of the ground state of the hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the formula for orbit radius**:
The radius of an electron orbit in a hydrogen-like atom is given by the formula:
\[ r = 0.059 \times \frac{n^2}{Z} \text{ angstroms} \]
where \( n \) is the principal quantum number and \( Z \) is the atomic number.

2. **Identify the parameters for hydrogen**:
For the hydrogen atom:
- \( Z_H = 1 \) (atomic number of hydrogen)
- The ground state corresponds to \( n_H = 1 \).

3. **Calculate the radius for the ground state of hydrogen**:
Plugging the values into the formula:
\[ r_H = 0.059 \times \frac{1^2}{1} = 0.059 \text{ angstroms} \]

4. **Identify the parameters for triply ionized Beryllium**:
For triply ionized Beryllium `(Be^(3+))`:
- \( Z_{Be} = 4 \) (atomic number of beryllium)
- We need to find \( n \) such that the radius of the orbit matches that of hydrogen.

5. **Set up the equation for Beryllium**:
We want the radius of Beryllium to equal the radius of hydrogen:
\[ 0.059 \times \frac{n^2}{4} = 0.059 \]

6. **Simplify the equation**:
Dividing both sides by \( 0.059 \):
\[ \frac{n^2}{4} = 1 \]

7. **Solve for \( n^2 \)**:
Multiplying both sides by 4:
\[ n^2 = 4 \]

8. **Find \( n \)**:
Taking the square root of both sides:
\[ n = 2 \]

### Conclusion:
The state of the triply ionized Beryllium `(Be^(3+))` that has the same orbit radius as that of the ground state of hydrogen atom is \( n = 2 \).