The ratio of kinetic energy and potential energy of an electron in a Bohr orbit of a hydrogen`-` like species is `:`

Options

A. `1//2`
B. `-1//2`
C. 1
D. -1

Answer

To find the ratio of kinetic energy (KE) to potential energy (PE) of an electron in a Bohr orbit of a hydrogen-like species, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Expressions for KE and PE**:
- The kinetic energy (KE) of an electron is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
- The potential energy (PE) of an electron in the electric field of a nucleus is given by:
\[ PE = -\frac{e^2}{r} \]
where \( e \) is the charge of the electron and \( r \) is the radius of the orbit.

2. **Centripetal and Centrifugal Forces**:
- For an electron moving in a circular orbit, the centripetal force required to keep it in that orbit is provided by the electrostatic force of attraction between the electron and the nucleus.
- The centripetal force can be expressed as:
\[ F_{centripetal} = \frac{mv^2}{r} \]
- The electrostatic force (Coulomb's law) acting on the electron is given by:
\[ F_{electrostatic} = \frac{e^2}{r^2} \]

3. **Equating Forces**:
- Setting the centripetal force equal to the electrostatic force gives us:
\[ \frac{mv^2}{r} = \frac{e^2}{r^2} \]
- Rearranging this equation, we find:
\[ mv^2 = \frac{e^2}{r} \]

4. **Substituting into KE**:
- Now, substituting \( mv^2 \) into the kinetic energy formula:
\[ KE = \frac{1}{2} mv^2 = \frac{1}{2} \left(\frac{e^2}{r}\right) = \frac{e^2}{2r} \]

5. **Finding the Ratio of KE to PE**:
- Now we can find the ratio of kinetic energy to potential energy:
\[ \frac{KE}{PE} = \frac{\frac{e^2}{2r}}{-\frac{e^2}{r}} = \frac{1}{2} \cdot (-1) = -\frac{1}{2} \]

### Final Answer:
The ratio of kinetic energy to potential energy of an electron in a Bohr orbit of a hydrogen-like species is:
\[ \frac{KE}{PE} = -\frac{1}{2} \]