The relationship between kinetic energy (K) and potential energy (U) of electron moving in a orbit around the nucleus is

Options

A. `U=-K`
B. `U=-2K`
C. `U=-3K`
D. `U=-(1)/(2)K`

Explanation

To find the relationship between the kinetic energy (K) and potential energy (U) of an electron moving in an orbit around the nucleus, we can follow these steps:

### Step-by-Step Solution:

1. **Define Kinetic Energy (K)**:
The kinetic energy of an electron in orbit is given by the formula:
\[ K = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the electron and \( v \) is its velocity.

2. **Define Potential Energy (U)**:
The potential energy of an electron due to its interaction with the nucleus (assuming a Coulombic attraction) is given by:
\[ U = -\frac{e^2}{4 \pi \epsilon_0 r} \]
where \( e \) is the charge of the electron, \( \epsilon_0 \) is the permittivity of free space, and \( r \) is the distance from the nucleus.

3. **Relate Kinetic Energy and Potential Energy**:
For an electron in a stable orbit, the centripetal force required for circular motion is provided by the electrostatic force of attraction between the electron and the nucleus. Therefore, we can write:
\[ \frac{mv^2}{r} = \frac{e^2}{4 \pi \epsilon_0 r^2} \]
Rearranging this gives:
\[ mv^2 = \frac{e^2}{4 \pi \epsilon_0 r} \]

4. **Substitute into Kinetic Energy**:
Now, substituting \( mv^2 \) into the kinetic energy formula:
\[ K = \frac{1}{2} mv^2 = \frac{1}{2} \left(\frac{e^2}{4 \pi \epsilon_0 r}\right) \]
This simplifies to:
\[ K = \frac{e^2}{8 \pi \epsilon_0 r} \]

5. **Express Potential Energy in Terms of Kinetic Energy**:
From the expression for potential energy:
\[ U = -\frac{e^2}{4 \pi \epsilon_0 r} \]
We can see that:
\[ U = -2K \]

6. **Conclusion**:
Therefore, the relationship between the kinetic energy and potential energy of the electron is:
\[ U = -2K \]

### Final Answer:
The relationship between kinetic energy (K) and potential energy (U) of an electron moving in an orbit around the nucleus is:
\[ U = -2K \]