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. 1
B. 2
C. 3
D. 4

Answer

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 hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Relevant Parameters:**
- For the hydrogen atom (H), the atomic number \( Z_2 = 1 \) and the principal quantum number for the ground state \( n_2 = 1 \).
- For the triply ionized Beryllium ion (Be^(3+)), the atomic number \( Z_1 = 4 \) (since Beryllium has an atomic number of 4).

2. **Use the Formula for the Radius of Electron Orbits:**
- The formula for the radius of the electron orbit in a hydrogen-like atom is given by:
\[ R_n = a_0 \frac{n^2}{Z} \]
where \( R_n \) is the radius of the orbit, \( a_0 \) is the Bohr radius, \( n \) is the principal quantum number, and \( Z \) is the atomic number.

3. **Set Up the Equation for Equal Radii:**
- We need to find \( n_1 \) for the Beryllium ion such that the radius of its orbit is equal to that of the ground state of hydrogen. Therefore, we set:
\[ R_1 = R_2 \]
This gives us:
\[ a_0 \frac{n_1^2}{Z_1} = a_0 \frac{n_2^2}{Z_2} \]

4. **Substitute Known Values:**
- Substituting the known values into the equation:
\[ a_0 \frac{n_1^2}{4} = a_0 \frac{1^2}{1} \]
- The \( a_0 \) cancels out:
\[ \frac{n_1^2}{4} = 1 \]

5. **Solve for \( n_1 \):**
- Rearranging the equation gives:
\[ n_1^2 = 4 \]
- Taking the square root of both sides:
\[ n_1 = 2 \]

### Final Answer:
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 \( n_1 = 2 \).