The velocity of an electron in the second orbit of sodium atom (atomic number `= 11`) is `v`. The veocity of an electron in its fifth orboit wil be

Options

A. `v`
B. `(22)/(5)v`
C. `(5)/(2)v`
D. `(2)/(5)v`

Explanation

To solve the problem of finding the velocity of an electron in the fifth orbit of a sodium atom (atomic number = 11), given that the velocity in the second orbit is \( v \), we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Relationship of Velocity and Orbit Number**:
The velocity of an electron in a given orbit is inversely proportional to the principal quantum number \( n \). This can be expressed as:
\[ v_n \propto \frac{1}{n} \]
where \( v_n \) is the velocity of the electron in the \( n \)-th orbit.

2. **Setting Up the Proportionality**:
We can express the velocities in terms of the known velocity \( v \) in the second orbit:
\[ v_2 = v \quad \text{(for n = 2)} \]
\[ v_n = k \cdot \frac{1}{n} \quad \text{(where k is a constant)} \]

3. **Finding the Constant k**:
For the second orbit:
\[ v_2 = k \cdot \frac{1}{2} = v \]
Thus, we can solve for \( k \):
\[ k = 2v \]

4. **Calculating the Velocity for the Fifth Orbit**:
Now, we can find the velocity in the fifth orbit (\( n = 5 \)):
\[ v_5 = k \cdot \frac{1}{5} = (2v) \cdot \frac{1}{5} = \frac{2v}{5} \]

5. **Final Result**:
Therefore, the velocity of the electron in the fifth orbit is:
\[ v_5 = \frac{2v}{5} \]

### Summary
The velocity of an electron in the fifth orbit of a sodium atom is \( \frac{2v}{5} \).