Find the frequency of revolution of the electron in the first orbit of H-atom

Options

A. `6xx10^(14) Hz`
B. `6.6xx10^(10) Hz`
C. `6.6xx10^(-10)Hz`
D. `6.6xx10^(15)Hz`

Explanation

To find the frequency of revolution of the electron in the first orbit of a hydrogen atom, we can follow these steps:

### Step 1: Understand the Motion
The electron revolves in a circular path around the nucleus of the hydrogen atom. The distance it covers in one complete revolution is the circumference of the circle, which is given by:
\[ d = 2 \pi r \]
where \( r \) is the radius of the orbit.

### Step 2: Define Frequency
The frequency \( f \) of revolution is defined as the number of revolutions per unit time. It can be expressed in terms of the time period \( T \):
\[ f = \frac{1}{T} \]
The time period \( T \) can be calculated as:
\[ T = \frac{\text{Distance}}{\text{Velocity}} = \frac{2 \pi r}{v} \]
where \( v \) is the velocity of the electron.

### Step 3: Substitute for Frequency
Substituting the expression for \( T \) into the equation for \( f \):
\[ f = \frac{v}{2 \pi r} \]

### Step 4: Find the Velocity of the Electron
For the first orbit of the hydrogen atom, the velocity \( v \) of the electron can be calculated using the formula:
\[ v = \frac{c}{137} \]
where \( c \) is the speed of light, approximately \( 3 \times 10^8 \, \text{m/s} \). Thus:
\[ v = \frac{3 \times 10^8}{137} \approx 2.189 \times 10^6 \, \text{m/s} \]

### Step 5: Determine the Radius
The radius of the first orbit of the hydrogen atom is given as:
\[ r = 0.53 \, \text{Å} = 0.53 \times 10^{-10} \, \text{m} \]

### Step 6: Calculate the Frequency
Now, we can substitute the values of \( v \) and \( r \) into the frequency formula:
\[ f = \frac{2.189 \times 10^6}{2 \pi (0.53 \times 10^{-10})} \]
Calculating the denominator:
\[ 2 \pi (0.53 \times 10^{-10}) \approx 3.34 \times 10^{-10} \]
Now calculating \( f \):
\[ f \approx \frac{2.189 \times 10^6}{3.34 \times 10^{-10}} \approx 0.6579 \times 10^{16} \, \text{Hz} \]
To express this in standard form:
\[ f \approx 6.579 \times 10^{15} \, \text{Hz} \approx 6.6 \times 10^{15} \, \text{Hz} \]

### Conclusion
Thus, the frequency of revolution of the electron in the first orbit of the hydrogen atom is approximately:
\[ f \approx 6.6 \times 10^{15} \, \text{Hz} \]