If the velocity of the electron in the first Bohr orbit having radius 0.53 Å is 2200 km/s . What is the frequency of the electron in the same orbit ?

Options

A. `5.5xx10^15 Hz`
B. `6.60xx10^15 Hz`
C. `7.32xx10^15 Hz`
D. `4.85xx10^15 Hz`

Explanation

To find the frequency of the electron in the first Bohr orbit, we can follow these steps:

### Step 1: Understand the relationship between velocity, radius, and frequency
The velocity \( v \) of an electron in a circular orbit is related to its angular frequency \( \omega \) and the radius \( r \) of the orbit. The angular frequency can be expressed in terms of frequency \( f \) as:
\[ \omega = 2\pi f \]
The relationship between linear velocity and angular frequency is given by:
\[ v = r \cdot \omega \]
Substituting the expression for \( \omega \):
\[ v = r \cdot (2\pi f) \]

### Step 2: Rearranging the equation to find frequency
We can rearrange the equation to solve for frequency \( f \):
\[ f = \frac{v}{2\pi r} \]

### Step 3: Substitute the known values
Given:
- The velocity \( v = 2200 \, \text{km/s} = 2200 \times 10^3 \, \text{m/s} = 2.2 \times 10^6 \, \text{m/s} \)
- The radius \( r = 0.53 \, \text{Å} = 0.53 \times 10^{-10} \, \text{m} \)

Now, substituting these values into the frequency formula:
\[ f = \frac{2.2 \times 10^6}{2\pi \times (0.53 \times 10^{-10})} \]

### Step 4: Calculate the frequency
Calculating the denominator:
\[ 2\pi \times (0.53 \times 10^{-10}) \approx 3.14 \times 0.53 \times 10^{-10} \approx 1.67 \times 10^{-10} \]
Now substituting this back into the frequency equation:
\[ f = \frac{2.2 \times 10^6}{1.67 \times 10^{-10}} \approx 1.315 \times 10^{16} \, \text{Hz} \]

### Step 5: Final result
Thus, the frequency of the electron in the first Bohr orbit is approximately:
\[ f \approx 1.315 \times 10^{16} \, \text{Hz} \]