Question

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

• A. \(\frac{e^2}{2r^2}\)
• B. \(\frac{e^2}{r}\)
• C. \(\frac{e^2}{2r}\)
• D. \(\frac{e^2}{r^2}\)

Hint:

Set the Coulomb force equal to the centripetal force.

Answer 1

The Correct Option is C

Step 1: Force balance
\(\frac{mv^2}{r}=\frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2}\).

Step 2: Kinetic energy
\(\frac12mv^2=\frac{1}{4\pi\varepsilon_0}\frac{e^2}{2r}\).
So the kinetic energy is proportional to \(\frac{e^2}{2r}\). Option (C).

Final Answer:
The kinetic energy is proportional to \(\frac{e^2}{2r}\), option (C). \[ \boxed{\text{(C)}} \]