Identify the correct statements in the following:
(A) The total energy of an electron in nth orbit in hydrogen atom is (−13.6/n) eV.
(B) As per Bohr’s theory, the stationary orbits of electrons around the nucleus are equally spaced.
(C) Orbits of electrons around nucleus as per Bohr’s theory are circular.
(D) The potential energy of electron in orbit in the hydrogen atom is twice of its total energy.
Choose the correct answer from the given options.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (C) and (D) only
In the Bohr model of hydrogen the electron moves in circular orbits with \( r_n \propto n^{2} \) and total energy \( E_n = -13.6/n^{2} \) eV.
(A) The total energy of an electron in nth orbit in hydrogen atom is (−13.6/n) eV. — Incorrect. The energy varies as \( 1/n^{2} \), not \( 1/n \).
\( \displaystyle E_n = -\frac{13.6}{n^{2}}\,\mathrm{eV} \)
(B) As per Bohr’s theory, the stationary orbits of electrons around the nucleus are equally spaced. — Incorrect. The radii are \( r_1, 4r_1, 9r_1, \ldots \) since \( r_n \propto n^{2} \), so the gaps grow with n.
(C) Orbits of electrons around nucleus as per Bohr’s theory are circular. — Correct. Bohr assumed circular orbits, with the Coulomb force providing the centripetal force.
(D) The potential energy of electron in orbit in the hydrogen atom is twice of its total energy. — Correct. For an inverse-square force, \( K = -E \) and \( U = 2E \).
\( \displaystyle K = \frac{ke^{2}}{2r},\quad U = -\frac{ke^{2}}{r},\quad E = K + U = -\frac{ke^{2}}{2r} \)
NCERT (2026-27):
Electron Orbits – “… The kinetic energy (K) and electrostatic potential energy (U) of the electron in hydrogen atom are
\( \displaystyle K = \frac{1}{2}mv^2 = \frac{e^2}{8\pi\varepsilon_0 r} \quad \text{and} \quad U = -\frac{e^2}{4\pi\varepsilon_0 r} \)
(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
\( \displaystyle E = K + U = \frac{e^2}{8\pi\varepsilon_0 r} - \frac{e^2}{4\pi\varepsilon_0 r} = -\frac{e^2}{8\pi\varepsilon_0 r} \qquad (12.4) \)”
Bohr Model of the Hydrogen Atom – “… Since 1 eV = 1.6 × 10–19 J, Eq. (12.9) can be rewritten as
\( \displaystyle E_n = -\frac{13.6\ \mathrm{eV}}{n^2} \qquad (12.10) \)”