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

Options

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

Answer

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 radius of a hydrogen-like atom:

\[ R_n = \frac{0.529 \, n^2}{Z} \, \text{angstroms} \]

where:
- \( R_n \) is the radius of the nth orbit,
- \( n \) is the principal quantum number (energy level),
- \( Z \) is the atomic number of the species.

The Bohr radius for hydrogen (where \( Z = 1 \) and \( n = 1 \)) is approximately \( 0.529 \, \text{angstroms} \).

### Step 1: Calculate the radius for each option

1. **Option 1: Beryllium (Z = 4, n = 2)**
\[ R = \frac{0.529 \times n^2}{Z} = \frac{0.529 \times 2^2}{4} = \frac{0.529 \times 4}{4} = 0.529 \, \text{angstroms} \]

2. **Option 2: Lithium (Z = 3, n = 2)**
\[ R = \frac{0.529 \times n^2}{Z} = \frac{0.529 \times 2^2}{3} = \frac{0.529 \times 4}{3} \approx 0.7067 \, \text{angstroms} \]

3. **Option 3: Helium (Z = 2, n = 2)**
\[ R = \frac{0.529 \times n^2}{Z} = \frac{0.529 \times 2^2}{2} = \frac{0.529 \times 4}{2} = 1.058 \, \text{angstroms} \]

4. **Option 4: Lithium (Z = 3, n = 3)**
\[ R = \frac{0.529 \times n^2}{Z} = \frac{0.529 \times 3^2}{3} = \frac{0.529 \times 9}{3} = 1.587 \, \text{angstroms} \]

### Step 2: Compare the calculated radii

- For **Option 1** (Beryllium): \( R = 0.529 \, \text{angstroms} \)
- For **Option 2** (Lithium): \( R \approx 0.7067 \, \text{angstroms} \)
- For **Option 3** (Helium): \( R = 1.058 \, \text{angstroms} \)
- For **Option 4** (Lithium): \( R = 1.587 \, \text{angstroms} \)

### Conclusion

The only species that has the same radius as that of the Bohr orbit of the hydrogen atom is **Option 1: Beryllium**.