The velocity of electron moving in 3rd orbit of `He^(+)` is v. The velocity of electron moving in 2nd orbit of `Li^(+2)` is

Options

A. `(9)/(4)V`
B. `(4)/(9)V`
C. v
D. None of these

Answer

To solve the problem of finding the velocity of the electron moving in the 2nd orbit of `Li^(2+)`, given that the velocity of the electron in the 3rd orbit of `He^(+)` is `v`, we can use the formula for the velocity of an electron in a hydrogen-like atom:

### Step-by-Step Solution:

1. **Understand the Formula**:
The velocity of an electron in the nth orbit of a hydrogen-like atom is given by the formula:
\[ v_n = \frac{Z e^2}{2 \varepsilon_0 h n} \]
where:
- \( Z \) = atomic number
- \( e \) = elementary charge
- \( \varepsilon_0 \) = permittivity of free space
- \( h \) = Planck's constant
- \( n \) = principal quantum number (orbit number)

2. **Velocity in the 3rd Orbit of `He^(+)`**:
For `He^(+)`, the atomic number \( Z = 2 \) and the electron is in the 3rd orbit (\( n = 3 \)):
\[ v_3 = \frac{2 e^2}{2 \varepsilon_0 h \cdot 3} = \frac{e^2}{3 \varepsilon_0 h} \]
Given that this velocity is equal to \( v \), we can express it as:
\[ v = \frac{e^2}{3 \varepsilon_0 h} \]

3. **Velocity in the 2nd Orbit of `Li^(2+)`**:
For `Li^(2+)`, the atomic number \( Z = 3 \) and the electron is in the 2nd orbit (\( n = 2 \)):
\[ v_2 = \frac{3 e^2}{2 \varepsilon_0 h \cdot 2} = \frac{3 e^2}{4 \varepsilon_0 h} \]

4. **Relate \( v_2 \) to \( v \)**:
Now, we can relate \( v_2 \) to \( v \):
\[ v_2 = \frac{3 e^2}{4 \varepsilon_0 h} = \frac{3}{4} \cdot \frac{e^2}{\varepsilon_0 h} = \frac{3}{4} \cdot \frac{3v}{1} = \frac{9}{4}v \]

5. **Final Result**:
Therefore, the velocity of the electron moving in the 2nd orbit of `Li^(2+)` is:
\[ v_2 = \frac{9}{4}v \]

### Conclusion:
The velocity of the electron in the 2nd orbit of `Li^(2+)` is \( \frac{9}{4}v \).