Match List-I with List-II
| List-I | List-II | ||
| Characteristics of an electron revolving in an orbit of radius \( r \) in a hydrogen atom | Mathematical expressions | ||
| (A) | Kinetic Energy of the electron | (I) | \( [-e^{2}/(4\pi\varepsilon_{0}r)] \) |
| (B) | Velocity of the electron | (II) | \( [-e^{2}/(8\pi\varepsilon_{0}r)] \) |
| (C) | Electrostatic Potential Energy of the electron | (III) | \( [e^{2}/(8\pi\varepsilon_{0}r)] \) |
| (D) | Total Energy of the electron | (IV) | \( [e/(4\pi\varepsilon_{0}mr)^{0.5}] \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
In the Bohr model the electrostatic attraction of the nucleus provides the centripetal force for the electron.
\( \displaystyle \frac{mv^{2}}{r} = \frac{e^{2}}{4\pi\varepsilon_{0}r^{2}} \)
(A) Kinetic Energy of the electron — this matches (III) \( [e^{2}/(8\pi\varepsilon_{0}r)] \). Multiplying the force equation by r/2:
\( \displaystyle K = \frac{1}{2}mv^{2} = \frac{e^{2}}{8\pi\varepsilon_{0}r} \)
(B) Velocity of the electron — this matches (IV) \( [e/(4\pi\varepsilon_{0}mr)^{0.5}] \). Solving the force equation for v:
\( \displaystyle v^{2} = \frac{e^{2}}{4\pi\varepsilon_{0}mr} \ \Rightarrow\ v = \frac{e}{\sqrt{4\pi\varepsilon_{0}mr}} \)
(C) Electrostatic Potential Energy of the electron — this matches (I) \( [-e^{2}/(4\pi\varepsilon_{0}r)] \). The potential energy of charges \( +e \) and \( -e \) at a separation r is negative.
\( \displaystyle U = \frac{1}{4\pi\varepsilon_{0}}\,\frac{(+e)(-e)}{r} = -\frac{e^{2}}{4\pi\varepsilon_{0}r} \)
(D) Total Energy of the electron — this matches (II) \( [-e^{2}/(8\pi\varepsilon_{0}r)] \). The total energy is the sum of the kinetic and potential energies.
\( \displaystyle E = K + U = \frac{e^{2}}{8\pi\varepsilon_{0}r} - \frac{e^{2}}{4\pi\varepsilon_{0}r} = -\frac{e^{2}}{8\pi\varepsilon_{0}r} \)
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
\( \displaystyle E = K + U = \frac{e^2}{8\pi\varepsilon_0 r} - \frac{e^2}{4\pi\varepsilon_0 r} = -\frac{e^2}{8\pi\varepsilon_0 r} \qquad (12.4) \)”