Match List I with List II
| LIST I | LIST II | ||
|---|---|---|---|
| A. | Orbital Radius | I. | \( -\displaystyle\frac{e^{2}}{4\pi\varepsilon_0 r} \) |
| B. | Orbital Velocity of electron | II. | \( \left(\displaystyle\frac{e^{2}}{2\varepsilon_0 hn}\right) \) |
| C. | Potential energy of electron | III. | \( \displaystyle\frac{e^{2}}{8\pi\varepsilon_0 r} \) |
| D. | Kinetic energy of electron | IV. | \( \left(\displaystyle\frac{\varepsilon_0h^{2}n^{2}}{\pi me^{2}}\right) \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-IV, B-II, C-I, D-III
These are the standard results of the Bohr model of hydrogen, obtained from the centripetal condition \( \frac{e^{2}}{4\pi\varepsilon_0r^{2}} = \frac{mv^{2}}{r} \) together with the quantisation of angular momentum, \( mvr = \frac{nh}{2\pi} \).
(A) Orbital Radius — this matches (IV) \( \frac{\varepsilon_0h^{2}n^{2}}{\pi me^{2}} \). Combining the two conditions gives \( r_n \propto n^{2} \), with the constant \( \varepsilon_0h^{2}/(\pi me^{2}) \), which is the first Bohr radius.
(B) Orbital Velocity of electron — this matches (II) \( \frac{e^{2}}{2\varepsilon_0hn} \). The speed in the nth orbit is \( v_n = e^{2}/(2\varepsilon_0nh) \), so it decreases as n increases.
(C) Potential energy of electron — this matches (I) \( -\frac{e^{2}}{4\pi\varepsilon_0r} \). It is the electrostatic potential energy of the electron and the nucleus at distance r, and it is negative because the force is attractive.
(D) Kinetic energy of electron — this matches (III) \( \frac{e^{2}}{8\pi\varepsilon_0r} \). From \( mv^{2} = \frac{e^{2}}{4\pi\varepsilon_0r} \), the kinetic energy \( \frac{1}{2}mv^{2} \) is half the magnitude of the potential energy.
Note: Items (A), (C) and (D) of this question are from the latest textbook (2026-27). Item (B), the orbital velocity of the electron, is printed only in the pre-rationalised textbook (2022-23). It was removed from the latest book, but it is still in the CUET 2026 syllabus, so do not skip it.
NCERT (2026-27):
Electron Orbits – “… The kinetic energy (K) and electrostatic potential energy (U) of the electron in hydrogen atom are
\( \displaystyle K = \frac{1}{2}mv^2 = \frac{e^2}{8\pi\varepsilon_0 r} \quad \text{and} \quad U = -\frac{e^2}{4\pi\varepsilon_0 r} \)”
Bohr Model of the Hydrogen Atom – “… The radius of nth possible orbit thus found is
\( \displaystyle r_n = \frac{n^2}{m}\left(\frac{h}{2\pi}\right)^2\frac{4\pi\varepsilon_0}{e^2} \qquad (12.7) \)”
NCERT (Pre-rationalised edition, 2022-23):
Bohr Model of the Hydrogen Atom – “… Combining it with Eq. (12.13), we get the following expressions for vn and rn,
\( \displaystyle v_n = \frac{1}{n}\,\frac{e^2}{4\pi\varepsilon_0}\,\frac{1}{(h/2\pi)} \qquad (12.14) \)”