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 find the velocity of the electron moving in the 2nd orbit of \( \text{Li}^{2+} \) given the velocity of the electron in the 3rd orbit of \( \text{He}^{+} \) is \( v \), we can use the formula for the velocity of an electron in a given orbit according to Bohr's model:

### Step-by-Step Solution:

1. **Understand the Formula**:
The velocity \( v_n \) of an electron in the \( n \)-th orbit is given by:
\[ v_n = 2.18 \times 10^6 \times \frac{Z}{n} \text{ m/s} \]
where \( Z \) is the atomic number and \( n \) is the orbit number.

2. **Determine for \( \text{He}^{+} \)**:
For \( \text{He}^{+} \):
- Atomic number \( Z = 2 \)
- Orbit number \( n = 3 \)

Plugging these values into the formula:
\[ v_3 = 2.18 \times 10^6 \times \frac{2}{3} \]
We know this velocity is given as \( v \):
\[ v = 2.18 \times 10^6 \times \frac{2}{3} \tag{1} \]

3. **Determine for \( \text{Li}^{2+} \)**:
For \( \text{Li}^{2+} \):
- Atomic number \( Z = 3 \)
- Orbit number \( n = 2 \)

Using the formula again:
\[ v_2 = 2.18 \times 10^6 \times \frac{3}{2} \]
This gives us the velocity of the electron in the 2nd orbit of \( \text{Li}^{2+} \):
\[ v_2 = 2.18 \times 10^6 \times \frac{3}{2} \tag{2} \]

4. **Relate the Two Velocities**:
Now, we can relate the two velocities (from equations (1) and (2)):
\[ \frac{v}{v_2} = \frac{2/3}{3/2} \]
Simplifying this:
\[ \frac{v}{v_2} = \frac{2 \times 2}{3 \times 3} = \frac{4}{9} \]
Rearranging gives:
\[ v_2 = \frac{9}{4} v \]

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