The radius of which of the following orbit is same as that of the first Bohr's orbit of hydrogen atom.

Options

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

Explanation

To solve the problem of determining which orbit has the same radius as the first Bohr orbit of the hydrogen atom, we can follow these steps:

### Step 1: Understand the Formula for Bohr's Radius
The radius of the nth orbit in the Bohr model is given by the formula:
\[ R_n = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2} \]
Where:
- \( R_n \) is the radius of the nth orbit,
- \( n \) is the principal quantum number,
- \( h \) is Planck's constant,
- \( \epsilon_0 \) is the permittivity of free space,
- \( m \) is the mass of the electron,
- \( Z \) is the atomic number,
- \( e \) is the charge of the electron.

### Step 2: Calculate the Radius for the First Bohr Orbit of Hydrogen
For the hydrogen atom:
- The principal quantum number \( n = 1 \),
- The atomic number \( Z = 1 \).

Substituting these values into the formula, we can simplify it to:
\[ R_1 = \frac{1^2 h^2 \epsilon_0}{\pi m \cdot 1 \cdot e^2} \]

This simplifies to a known constant value:
\[ R_1 = 0.53 \text{ Å} \]

### Step 3: Set Up the Condition for Other Atoms
We need to find the condition where the radius of another atom's orbit equals \( R_1 \). From the formula, we can see that:
\[ R_n = R_1 \Rightarrow \frac{n^2}{Z} = 1 \]
This implies:
\[ n^2 = Z \]

### Step 4: Check Each Option
Now, we will check the given options to see which one satisfies \( n^2 = Z \).

1. **Option A: Lithium (Li)**
- \( Z = 3 \)
- Check \( n^2 = 2^2 = 4 \) (for n=2)
- \( 4 \neq 3 \)

2. **Option B: Beryllium (Be)**
- \( Z = 4 \)
- Check \( n^2 = 3^2 = 9 \) (for n=3)
- \( 9 \neq 4 \)

3. **Option C: Beryllium Ion (Be\(^{3+}\))**
- \( Z = 4 \)
- Check \( n^2 = 2^2 = 4 \) (for n=2)
- \( 4 = 4 \) (This is equal)

4. **Option D: Helium (He)**
- \( Z = 2 \)
- Check \( n^2 = 2^2 = 4 \) (for n=2)
- \( 4 \neq 2 \)

### Conclusion
The only option that satisfies the condition \( n^2 = Z \) is **Option C: Beryllium Ion (Be\(^{3+}\))**.

### Final Answer
The radius of the orbit that is the same as that of the first Bohr's orbit of the hydrogen atom is that of **Beryllium Ion (Be\(^{3+}\))**.

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