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

Options

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

Answer

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

### Step 1: Identify the parameters
The hydrogen atom has an atomic number \( z = 1 \) and we are considering the first stationary orbit, which corresponds to \( n = 1 \).

### Step 2: Calculate the radius of the first orbit
The radius \( r \) of the \( n \)-th orbit in a hydrogen atom is given by the formula:
\[ r = r_0 \frac{z}{n^2} \]
where \( r_0 \) is a constant approximately equal to \( 0.53 \) Å (or \( 0.53 \times 10^{-10} \) m). For \( z = 1 \) and \( n = 1 \):
\[ r = 0.53 \times 10^{-10} \text{ m} \]

### Step 3: Calculate the velocity of the electron
The velocity \( v \) of the electron in the \( n \)-th orbit is given by:
\[ v = v_0 \frac{n^2}{z} \]
where \( v_0 \) is a constant approximately equal to \( 2.2 \times 10^6 \) m/s. For \( z = 1 \) and \( n = 1 \):
\[ v = 2.2 \times 10^6 \text{ m/s} \]

### Step 4: Calculate the time period \( T \)
The time period \( T \) for one complete revolution is given by:
\[ T = \frac{2 \pi r}{v} \]
Substituting the values of \( r \) and \( v \):
\[ T = \frac{2 \pi (0.53 \times 10^{-10})}{2.2 \times 10^6} \]

### Step 5: Calculate the frequency \( f \)
The frequency \( f \) is the reciprocal of the time period:
\[ f = \frac{1}{T} \]
Substituting the expression for \( T \):
\[ f = \frac{v}{2 \pi r} \]

### Step 6: Substitute the values to find frequency
Now substituting \( v = 2.2 \times 10^6 \) m/s and \( r = 0.53 \times 10^{-10} \) m into the frequency equation:
\[ f = \frac{2.2 \times 10^6}{2 \pi (0.53 \times 10^{-10})} \]

### Step 7: Calculate the numerical value
Calculating the above expression:
1. Calculate \( 2 \pi \approx 6.28 \).
2. Calculate \( 2 \pi (0.53 \times 10^{-10}) \approx 3.34 \times 10^{-10} \).
3. Now calculate \( f \):
\[ f \approx \frac{2.2 \times 10^6}{3.34 \times 10^{-10}} \approx 6.6 \times 10^{15} \text{ Hz} \]

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