Question

In hydrogen atom, the kinetic energy of electron in an orbit of radius \( r \), is given by:

• A. \( \frac{1}{4\pi \epsilon_0} . \frac{e^2}{r} \)
• B. \( -\frac{1}{4\pi \epsilon_0} . \frac{e^2}{r} \)
• C. \( -\frac{1}{4\pi \epsilon_0} . \frac{e^2}{2r} \)
• D. \( \frac{1}{4\pi \epsilon_0} . \frac{e^2}{2r} \)

Hint:

The kinetic energy of an electron in a hydrogen atom is related to the electrostatic force between the electron and proton and is negative because of the attractive nature of the force.

Answer 1

The Correct Option is B

In a hydrogen atom, the electron orbits the nucleus (proton) in a circular path under the influence of the electrostatic force between the electron and proton. The Coulomb force provides the centripetal force for the electron's motion.
The electrostatic force between the electron and proton is given by Coulomb’s law:
\[ F = \frac{1}{4\pi \epsilon_0} . \frac{e^2}{r^2} \] Where: - \( e \) is the charge of the electron, - \( \epsilon_0 \) is the permittivity of free space, - \( r \) is the radius of the electron's orbit.
The kinetic energy \( K.E. \) of the electron is given by the relation:
\[ K.E. = \frac{1}{2} m v^2 \] Using the fact that the centripetal force \( F \) is equal to \( \frac{m v^2}{r} \), and equating the electrostatic force to the centripetal force, we find:
\[ K.E. = - \frac{1}{4\pi \epsilon_0} . \frac{e^2}{r} \] Thus, the correct expression for the kinetic energy of the electron in the hydrogen atom is:
\[ K.E. = - \frac{1}{4\pi \epsilon_0} . \frac{e^2}{r} \] Hence, the correct answer is option (B) \( -\frac{1}{4\pi \epsilon_0} . \frac{e^2}{r} \).