The radius of which of the following orbit is same as that of the first Bohr's orbit of Hydrogen atom?
(a). `He^(+) (n=2)`
(b) `Li^(2+)(n=2)`
(c). `Li^(2+)(n=3)`
(d). `Be^(3+)(n=2)`

Options

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

Answer

To determine which of the given options has the same radius as the first Bohr's orbit of the hydrogen atom, we can use the formula for the radius of an electron orbit in a hydrogen-like atom:

\[ r = r_H \cdot \frac{n^2}{Z} \]

Where:
- \( r_H \) is the radius of the first Bohr orbit of hydrogen (0.529 Å).
- \( n \) is the principal quantum number (the orbit number).
- \( Z \) is the atomic number of the atom.

### Step-by-Step Solution:

1. **Identify the radius of the first Bohr orbit of Hydrogen:**
- For hydrogen (\( Z = 1 \)) and \( n = 1 \):
\[ r_H = 0.529 \, \text{Å} \]

2. **Calculate the radius for each option:**

**(a) Helium ion \( He^+ \) with \( n = 2 \):**
- \( Z = 2 \), \( n = 2 \):
\[ r = r_H \cdot \frac{n^2}{Z} = 0.529 \cdot \frac{2^2}{2} = 0.529 \cdot \frac{4}{2} = 0.529 \cdot 2 = 1.058 \, \text{Å} \]

**(b) Lithium ion \( Li^{2+} \) with \( n = 2 \):**
- \( Z = 3 \), \( n = 2 \):
\[ r = r_H \cdot \frac{n^2}{Z} = 0.529 \cdot \frac{2^2}{3} = 0.529 \cdot \frac{4}{3} \approx 0.7067 \, \text{Å} \]

**(c) Lithium ion \( Li^{2+} \) with \( n = 3 \):**
- \( Z = 3 \), \( n = 3 \):
\[ r = r_H \cdot \frac{n^2}{Z} = 0.529 \cdot \frac{3^2}{3} = 0.529 \cdot \frac{9}{3} = 0.529 \cdot 3 = 1.587 \, \text{Å} \]

**(d) Beryllium ion \( Be^{3+} \) with \( n = 2 \):**
- \( Z = 4 \), \( n = 2 \):
\[ r = r_H \cdot \frac{n^2}{Z} = 0.529 \cdot \frac{2^2}{4} = 0.529 \cdot \frac{4}{4} = 0.529 \, \text{Å} \]

3. **Compare the calculated radii:**
- \( r_H = 0.529 \, \text{Å} \) (Hydrogen)
- \( r(He^+, n=2) = 1.058 \, \text{Å} \)
- \( r(Li^{2+}, n=2) \approx 0.7067 \, \text{Å} \)
- \( r(Li^{2+}, n=3) = 1.587 \, \text{Å} \)
- \( r(Be^{3+}, n=2) = 0.529 \, \text{Å} \)

4. **Conclusion:**
- The radius of the first Bohr orbit of the hydrogen atom is the same as that of the beryllium ion \( Be^{3+} \) when \( n = 2 \).

### Final Answer:
**(d) \( Be^{3+} (n=2) \)**