Match List-I with List-II
| List-I | List-II |
| Physical quantity (According to Rutherford Model) | Formula |
| (A) Total energy of the electron in hydrogen atom | (I) \( e^2/(8\pi\varepsilon_0 R) \) |
| (B) Kinetic energy of the electron in hydrogen atom | (II) \( -\,e^2/(8\pi\varepsilon_0 R) \) |
| (C) Radius of the circular orbit around the nucleus | (III) \( -\,e^2/(4\pi\varepsilon_0 R) \) |
| (D) Potential energy of the electron in hydrogen atom | (IV) \( e^2/(4\pi\varepsilon_0 m v^2) \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (II), (B) - (I), (C) - (IV), (D) - (III)
In the Rutherford model the electron of charge \( e \) goes round the nucleus in a circle of radius \( R \) at speed \( v \). The electrostatic attraction provides the centripetal force.
\( \displaystyle \frac{e^{2}}{4\pi\varepsilon_0 R^{2}} = \frac{m v^{2}}{R} \)
(A) Total energy of the electron in hydrogen atom — this matches (II) \( -\,e^2/(8\pi\varepsilon_0 R) \). It is the sum of the kinetic and potential energies.
\( \displaystyle E = K + U = \frac{e^{2}}{8\pi\varepsilon_0 R} - \frac{e^{2}}{4\pi\varepsilon_0 R} = -\frac{e^{2}}{8\pi\varepsilon_0 R} \)
(B) Kinetic energy of the electron in hydrogen atom — this matches (I) \( e^2/(8\pi\varepsilon_0 R) \). From the centripetal condition, \( mv^{2} = e^{2}/(4\pi\varepsilon_0 R) \).
\( \displaystyle K = \frac{1}{2}mv^{2} = \frac{e^{2}}{8\pi\varepsilon_0 R} \)
(C) Radius of the circular orbit around the nucleus — this matches (IV) \( e^2/(4\pi\varepsilon_0 m v^2) \). It follows by solving the same condition for \( R \).
\( \displaystyle R = \frac{e^{2}}{4\pi\varepsilon_0 m v^{2}} \)
(D) Potential energy of the electron in hydrogen atom — this matches (III) \( -\,e^2/(4\pi\varepsilon_0 R) \). The potential energy of the electron and the nucleus is negative because they attract.
\( \displaystyle U = -\frac{e^{2}}{4\pi\varepsilon_0 R} \)
NCERT (2026-27):
Electron Orbits – “… Thus the relation between the orbit radius and the electron velocity is
\( \displaystyle r = \frac{e^2}{4\pi\varepsilon_0 m v^2} \qquad (12.3) \)
The kinetic energy (K) and electrostatic potential energy (U) of the electron in hydrogen atom are
\( \displaystyle K = \frac{1}{2}mv^2 = \frac{e^2}{8\pi\varepsilon_0 r} \quad \text{and} \quad U = -\frac{e^2}{4\pi\varepsilon_0 r} \)
(The negative sign in U signifies that the electrostatic force is in the –r direction.) Thus the total energy E of the electron in a hydrogen atom is
\( \displaystyle E = K + U = \frac{e^2}{8\pi\varepsilon_0 r} - \frac{e^2}{4\pi\varepsilon_0 r} = -\frac{e^2}{8\pi\varepsilon_0 r} \qquad (12.4) \)”