Determine the Bohr orbit of `Li^(2+)` ion in which electron is moving at speed equal to the speed of electron in the first Bohr orbit of H-atom

Explanation

To determine the Bohr orbit of the `Li^(2+)` ion where the electron is moving at the same speed as the electron in the first Bohr orbit of the hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the speed of the electron in the first Bohr orbit of Hydrogen (H)**:
The speed of the electron in the first orbit (n=1) of the hydrogen atom is given by the formula:
\[ v = 2.18 \times 10^6 \frac{Z}{n} \text{ m/s} \]
For hydrogen, \(Z = 1\) and \(n = 1\):
\[ v_H = 2.18 \times 10^6 \frac{1}{1} = 2.18 \times 10^6 \text{ m/s} \]

2. **Determine the speed of the electron in the `Li^(2+)` ion**:
For the `Li^(2+)` ion, which has \(Z = 3\), we will use the same formula:
\[ v_{Li^{2+}} = 2.18 \times 10^6 \frac{Z}{n} = 2.18 \times 10^6 \frac{3}{n} \text{ m/s} \]

3. **Set the speeds equal**:
Since we want the speed of the electron in the `Li^(2+)` ion to be equal to that of the hydrogen atom:
\[ v_H = v_{Li^{2+}} \]
Therefore:
\[ 2.18 \times 10^6 = 2.18 \times 10^6 \frac{3}{n} \]

4. **Cancel the common terms**:
We can cancel \(2.18 \times 10^6\) from both sides:
\[ 1 = \frac{3}{n} \]

5. **Solve for n**:
Rearranging gives:
\[ n = 3 \]

6. **Conclusion**:
The Bohr orbit of the `Li^(2+)` ion where the electron is moving at the same speed as the electron in the first Bohr orbit of the hydrogen atom is \(n = 3\).

### Final Answer:
The Bohr orbit of `Li^(2+)` is \(n = 3\).