An electron in the ground state goes to the first excited state in the hydrogen atom. Arrange the following energies in decreasing order
(A) Total energy in ground state
(B) Kinetic energy in ground state
(C) Potential energy in ground state
(D) Total energy in first excited state
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (B), (D), (A), (C)
In the Bohr model of hydrogen the total energy of level \( n \) is \( E_n = -13.6/n^{2}\,\mathrm{eV} \). The kinetic energy is the negative of the total energy, and the potential energy is twice the total energy.
(B) Kinetic energy in ground state — \( K = +13.6\,\mathrm{eV} \)
(A) Total energy in ground state — \( E_1 = -13.6\,\mathrm{eV} \)
(C) Potential energy in ground state — \( U = 2E_1 = -27.2\,\mathrm{eV} \)
(D) Total energy in first excited state — \( \displaystyle E_2 = -\frac{13.6}{2^{2}} = -3.4\,\mathrm{eV} \)
Arranging from the largest to the smallest value, \( +13.6 \gt -3.4 \gt -13.6 \gt -27.2 \), that is (B), (D), (A), (C).
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 – “… Atomic energies are often expressed in electron volts (eV) rather than joules. Since 1 eV = 1.6 × 10–19 J, Eq. (12.9) can be rewritten as
\( \displaystyle E_n = -\frac{13.6\ \mathrm{eV}}{n^2} \qquad (12.10) \)”