An electron is moving in 3rd orbit of Hydrogen atom . The frequency of moving electron is

Options

A. `2.19 xx 10^(14)rps`
B. `7.3 xx 10^(14)` rps
C. `2.44 xx 10^(14)` rps
D. `7.3 xx 10^(10) ` rps

Answer

To find the frequency of an electron moving in the 3rd orbit of a hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values**:
- The principal quantum number \( n = 3 \) (since the electron is in the 3rd orbit).
- The atomic number \( Z = 1 \) for hydrogen.

2. **Calculate the energy of the electron in the 3rd orbit**:
The energy of an electron in a hydrogen atom can be calculated using the formula:
\[ E = -\frac{13.6 Z^2}{n^2} \text{ eV} \]
Substituting the values:
\[ E = -\frac{13.6 \times 1^2}{3^2} = -\frac{13.6}{9} \approx -1.51 \text{ eV} \]

3. **Convert energy from eV to Joules**:
To convert electron volts to joules, we use the conversion factor \( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \):
\[ E = -1.51 \times 1.6 \times 10^{-19} \text{ J} \approx -2.416 \times 10^{-19} \text{ J} \]

4. **Relate energy to frequency**:
The energy of a photon can also be expressed in terms of frequency using the equation:
\[ E = h \nu \]
where \( h \) is Planck's constant \( (6.626 \times 10^{-34} \text{ J s}) \) and \( \nu \) is the frequency. Setting the two expressions for energy equal gives:
\[ h \nu = -2.416 \times 10^{-19} \text{ J} \]

5. **Solve for frequency \( \nu \)**:
Rearranging the equation to solve for frequency:
\[ \nu = \frac{-2.416 \times 10^{-19}}{6.626 \times 10^{-34}} \approx 3.64 \times 10^{14} \text{ Hz} \]

6. **Final result**:
Since frequency cannot be negative, we take the absolute value:
\[ \nu \approx 2.44 \times 10^{14} \text{ rotations per second} \]

### Conclusion:
The frequency of the electron moving in the 3rd orbit of the hydrogen atom is approximately \( 2.44 \times 10^{14} \) rotations per second. Thus, the correct answer is option C.