When a particle as an electron exist in a box, three quantum numbers are necessary to describe the spatial distributioni fo electrons in atoms. The describe an electron in an atom completely, a fourth quantum number, `m_(s)` called the spin quantum number must be specified. This is because every electron has magnetic moment associated with it which is quantized in one of two possible orientations. Electrons having the same spin strongly repel each other and tend to occupy different region of space. This result of a fundamental law of nature is known as Pauli exclusion principle.
The radius of which of the following orbit is same as that of the first Bohr's orbit of hydrogen atom?

Options

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

Answer

To determine which orbit has the same radius as the first Bohr's orbit of the hydrogen atom, we will use the formula for the radius of the nth orbit in a hydrogen-like atom:

\[ r_n = r_0 \times \frac{n^2}{Z} \]

where:
- \( r_n \) is the radius of the nth orbit,
- \( r_0 \) is the radius of the first Bohr's orbit of hydrogen (approximately \( 0.529 \) angstroms),
- \( n \) is the principal quantum number,
- \( Z \) is the atomic number of the atom.

### Step 1: Identify the radius of the first Bohr's orbit
The radius of the first Bohr's orbit of hydrogen is given as:

\[ r_0 = 0.529 \text{ angstroms} \]

### Step 2: Analyze each option
We will calculate the radius for each option using the formula provided.

1. **Option 1: \( \text{He}^+ \) (Z = 2, n = 2)**
\[ r = r_0 \times \frac{2^2}{2} = r_0 \times \frac{4}{2} = 2r_0 \]

2. **Option 2: \( \text{Li}^{2+} \) (Z = 3, n = 2)**
\[ r = r_0 \times \frac{2^2}{3} = r_0 \times \frac{4}{3} \]

3. **Option 3: \( \text{Li}^{2+} \) (Z = 3, n = 3)**
\[ r = r_0 \times \frac{3^2}{3} = r_0 \times \frac{9}{3} = 3r_0 \]

4. **Option 4: \( \text{Be}^{3+} \) (Z = 4, n = 2)**
\[ r = r_0 \times \frac{2^2}{4} = r_0 \times \frac{4}{4} = r_0 \]

### Step 3: Compare the calculated radii
Now we compare the calculated radii with the radius of the first Bohr's orbit of hydrogen:

- Option 1: \( 2r_0 \) (not equal)
- Option 2: \( \frac{4}{3}r_0 \) (not equal)
- Option 3: \( 3r_0 \) (not equal)
- Option 4: \( r_0 \) (equal)

### Conclusion
The radius of the orbit for **Option 4: \( \text{Be}^{3+} \)** is the same as that of the first Bohr's orbit of hydrogen.

**Final Answer: Option 4**