Question

The kinetic energy of the electron in an orbit of radius $r$ in hydrogen atom is proportional to ($e$ = electronic charge)

• A. $\dfrac{e^2}{2r^2}$
• B. $\dfrac{e^2}{r^2}$
• C. $\dfrac{e^2}{2r}$
• D. $\dfrac{e^2}{4r}$

Hint:

The Coulomb force supplies the centripetal force, which gives kinetic energy proportional to 1 over r.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
In a Bohr orbit the electrostatic force provides the centripetal force on the electron.

Step 2: Force balance
In Gaussian-style units (the form in which the options are written):
\[ \frac{mv^2}{r}=\frac{e^2}{r^2}\Rightarrow mv^2=\frac{e^2}{r} \]

Step 3: Kinetic energy
\[ KE=\frac12mv^2=\frac{e^2}{2r} \]
In SI units this reads \(\dfrac{e^2}{8\pi\varepsilon_0r}\), which has the same dependence on \(r\).

Step 4: Conclusion
The kinetic energy is proportional to \(\dfrac{e^2}{r}\). Among the options, \(\dfrac{e^2}{2r}\) has exactly this form, so the answer is option (C).

Final Answer:
The kinetic energy equals e squared over 2r, which varies as one over r, option (C). \[ \boxed{\frac{e^2}{2r}} \]