In a hydrogen atom, the binding energy of the electron in the ground state is `E_(1)` then the frequency of revolution of the electron in the nth orbit is

Options

A. `2E_(1)//nh`
B. `2E_(1)n//h`
C. `E_(1)//nh`
D. `E_(1)n//h`

Answer

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

### Step-by-Step Solution:

1. **Understanding Binding Energy**:
The binding energy \( E_1 \) of the electron in the ground state (n=1) is given. This energy is equal to the kinetic energy of the electron in that orbit.

2. **Kinetic Energy and Angular Momentum**:
The kinetic energy \( KE \) of the electron can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the electron and \( v \) is its velocity.

The angular momentum \( L \) of the electron in the nth orbit is quantized and given by:
\[ L = mvr = \frac{nh}{2\pi} \]
where \( r \) is the radius of the orbit and \( h \) is Planck's constant.

3. **Relating Kinetic Energy and Angular Momentum**:
From the equation for kinetic energy, we can express \( v \) in terms of \( E_1 \):
\[ E_1 = KE = \frac{1}{2} mv^2 \implies v^2 = \frac{2E_1}{m} \]

4. **Substituting for \( v \)**:
We can substitute \( v \) from the angular momentum equation into the kinetic energy equation. From the angular momentum equation:
\[ v = \frac{nh}{2\pi mr} \]

5. **Finding the Frequency**:
The frequency \( f \) of revolution is given by:
\[ f = \frac{v}{2\pi r} \]
Substituting \( v \) from the angular momentum equation:
\[ f = \frac{nh}{2\pi mr} \cdot \frac{1}{2\pi r} \]

6. **Final Expression for Frequency**:
After simplifying, we find:
\[ f = \frac{2E_1}{nh} \]
This is the frequency of revolution of the electron in the nth orbit.

### Final Answer:
\[ f = \frac{2E_1}{nh} \]