If `v_(0)` is the velocity of electron in the first orbit of hydrogen atom, then the velocity of electron in the first orbit of `Li^(2+)` ion will be.....

Options

A. `v_(0)`
B. `2v_(0)`
C. `3v_(0)`
D. `5v_(0)`

Explanation

To solve the problem of finding the velocity of an electron in the first orbit of the \( \text{Li}^{2+} \) ion, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Velocity**:
The velocity of an electron in the nth orbit of a hydrogen-like atom can be expressed as:
\[ v_n = v_0 \cdot \frac{Z}{n} \]
where:
- \( v_n \) is the velocity of the electron in the nth orbit,
- \( v_0 \) is the velocity of the electron in the first orbit of hydrogen (\( Z = 1 \)),
- \( Z \) is the atomic number of the ion,
- \( n \) is the principal quantum number (the orbit number).

2. **Identify the Values for \( \text{Li}^{2+} \)**:
For the \( \text{Li}^{2+} \) ion:
- The atomic number \( Z \) of lithium is 3.
- The electron is in the first orbit, so \( n = 1 \).

3. **Substitute the Values into the Formula**:
Now, substituting the values into the formula:
\[ v_1 = v_0 \cdot \frac{Z}{n} = v_0 \cdot \frac{3}{1} \]

4. **Calculate the Velocity**:
This simplifies to:
\[ v_1 = 3v_0 \]

5. **Conclusion**:
Therefore, the velocity of the electron in the first orbit of the \( \text{Li}^{2+} \) ion is \( 3v_0 \).

### Final Answer:
The velocity of the electron in the first orbit of \( \text{Li}^{2+} \) ion is \( 3v_0 \).

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