Which hydrogen -like species will have the same r adius as that of Bohr orbit of hydrogen atom ?

Options

A. `n = 2,Li^(2+)`
B. `n = 2,Be^(3+)`
C. `n = 2,He^(Theta)`
D. `n = 3,Li^(2+)`

Explanation

To determine which hydrogen-like species has the same radius as that of the Bohr orbit of a hydrogen atom, we will use the formula for the Bohr radius:

\[ r = \frac{n^2 h^2}{Z e^2 4 \pi m} \]

Where:
- \( r \) is the radius of the orbit,
- \( n \) is the principal quantum number,
- \( Z \) is the atomic number,
- \( h \) is Planck's constant,
- \( e \) is the charge of the electron,
- \( m \) is the mass of the electron.

### Step 1: Identify the values for Hydrogen
For the hydrogen atom:
- \( n = 1 \)
- \( Z = 1 \)

Substituting these values into the formula gives us the Bohr radius for hydrogen.

### Step 2: Analyze the options
We need to analyze the given options to find their respective \( n \) and \( Z \) values.

1. **Helium negative ion (He⁻)**: This is not a hydrogen-like species because it has more than one electron. Therefore, we can eliminate this option.

2. **Lithium ion (Li²⁺)**:
- For Li²⁺, \( n = 2 \) and \( Z = 3 \).
- The radius \( r \) will be proportional to \( \frac{n^2}{Z} = \frac{2^2}{3} = \frac{4}{3} \).

3. **Beryllium ion (Be³⁺)**:
- For Be³⁺, \( n = 1 \) and \( Z = 4 \).
- The radius \( r \) will be proportional to \( \frac{n^2}{Z} = \frac{1^2}{4} = \frac{1}{4} \).

4. **Hydrogen ion (H⁺)**:
- For H⁺, \( n = 1 \) and \( Z = 1 \).
- The radius \( r \) will be proportional to \( \frac{n^2}{Z} = \frac{1^2}{1} = 1 \).

### Step 3: Compare the results
Now we compare the proportional values obtained:
- For Li²⁺: \( \frac{4}{3} \)
- For Be³⁺: \( \frac{1}{4} \)
- For H⁺: \( 1 \)

The Bohr radius for hydrogen is \( 1 \).

### Conclusion
The only species that has the same radius as that of the Bohr orbit of the hydrogen atom is the hydrogen ion (H⁺).

Thus, the correct answer is **H⁺**.