According to maxwell's theiory of electrodnamics, an electron going in a circle should emit redastion of frequency equal to the frequency of revolution what should be the wavelength of the radiation emitted by a hydrogen atom in ground state if this rule is follewed?

Explanation

To solve the problem step by step, we will follow the reasoning and calculations based on the information provided in the video transcript.

### Step 1: Understand the relationship between frequency, time period, and velocity
The time period \( T \) of an electron moving in a circular orbit can be expressed in terms of its frequency \( f \):
\[ T = \frac{1}{f} \]
The time period can also be calculated using the formula:
\[ T = \frac{2\pi r}{v} \]
where \( r \) is the radius of the orbit and \( v \) is the velocity of the electron.

### Step 2: Set up the equation for frequency
From the two expressions for time period, we can equate them:
\[ \frac{1}{f} = \frac{2\pi r}{v} \]
Rearranging gives us:
\[ f = \frac{v}{2\pi r} \]

### Step 3: Relate wavelength to frequency
According to electromagnetic wave theory, the wavelength \( \lambda \) is related to frequency \( f \) by the equation:
\[ \lambda = \frac{c}{f} \]
where \( c \) is the speed of light.

### Step 4: Substitute frequency into the wavelength equation
Substituting the expression for frequency into the wavelength equation, we get:
\[ \lambda = \frac{c}{\frac{v}{2\pi r}} = \frac{2\pi r c}{v} \]

### Step 5: Substitute known values
For a hydrogen atom in the ground state:
- The radius \( r_0 \) (Bohr radius) is approximately \( 5.3 \times 10^{-11} \) m.
- The speed of light \( c \) is \( 3 \times 10^8 \) m/s.
- The velocity of the electron in the ground state \( v_0 \) is approximately \( 2.18 \times 10^6 \) m/s.

Substituting these values into the equation for wavelength:
\[ \lambda = \frac{2 \pi (5.3 \times 10^{-11}) (3 \times 10^8)}{2.18 \times 10^6} \]

### Step 6: Calculate the wavelength
Calculating the above expression:
1. Calculate \( 2\pi \times 5.3 \times 10^{-11} \):
\[ 2\pi \approx 6.28 \quad \Rightarrow \quad 6.28 \times 5.3 \times 10^{-11} \approx 3.34 \times 10^{-10} \text{ m} \]
2. Now multiply by \( 3 \times 10^8 \):
\[ 3.34 \times 10^{-10} \times 3 \times 10^8 \approx 1.002 \times 10^{-1} \text{ m} \]
3. Finally, divide by \( 2.18 \times 10^6 \):
\[ \lambda \approx \frac{1.002 \times 10^{-1}}{2.18 \times 10^6} \approx 4.59 \times 10^{-8} \text{ m} \approx 45.9 \text{ nm} \]

### Final Answer
The wavelength of the radiation emitted by a hydrogen atom in the ground state, according to Maxwell's theory, is approximately:
\[ \lambda \approx 45.9 \text{ nm} \]