The energy, radius and velocity of the electron in the hydrogen atom in the ground state are `-13.6 eV, 0.53 Å` and `2.188 xx 10^(8) m s^(-1)` respectively.
The velocity of the electron in the third orbit of hydrogen atom will be

Options

A. `0.729 xx 10^(8) cm s^(-1)`
B. `6.564 xx 10^(8) cm s^(-1)`
C. `19.692 xx 10^(8) cms^(-1)`
D. `0.243 xx 10^(8) cm s^(-1)`

Explanation

To find the velocity of the electron in the third orbit of a hydrogen atom, we can use the relationship between the velocities in different orbits.

### Step-by-Step Solution:

1. **Identify the given values**:
- The velocity of the electron in the ground state (first orbit, n=1) is given as:
\[ V_0 = 2.188 \times 10^8 \text{ m/s} \]

2. **Recall the formula for velocity in different orbits**:
- The velocity of the electron in the nth orbit can be expressed as:
\[ V_n = \frac{V_0}{n} \]
- Where \( V_n \) is the velocity in the nth orbit and \( n \) is the principal quantum number.

3. **Substitute the values for the third orbit**:
- For the third orbit, \( n = 3 \):
\[ V_3 = \frac{V_0}{3} \]

4. **Calculate the velocity**:
- Substitute \( V_0 \) into the equation:
\[ V_3 = \frac{2.188 \times 10^8 \text{ m/s}}{3} \]
- Performing the division:
\[ V_3 = 0.72933 \times 10^8 \text{ m/s} \]
- Rounding to three significant figures, we get:
\[ V_3 \approx 0.729 \times 10^8 \text{ m/s} \]

5. **Final Answer**:
- The velocity of the electron in the third orbit of the hydrogen atom is approximately:
\[ V_3 \approx 0.729 \times 10^8 \text{ m/s} \]