The total energy of the electron in a hydrogen atom is -1.51eVA. Its…

The correct answer is Option (4) → B and C only

For an electron in a Bohr orbit the kinetic energy is positive, \( K = e^{2}/(8\pi\varepsilon_0r) \), and the potential energy is negative and twice as large in magnitude, \( U = -e^{2}/(4\pi\varepsilon_0r) = -2K \). Hence the total energy is \( E = K+U = -K = U/2 \). The level energies are \( E_n = -13.6/n^{2}\,\mathrm{eV} \).

    \( K = -E = +1.51\,\mathrm{eV} \)

    \( U = 2E = -3.02\,\mathrm{eV} \)

    \( \displaystyle n^{2} = \frac{13.6}{1.51} \approx 9 \Rightarrow n = 3 \)

(A) Its Kinetic energy is -1.51eV — Incorrect. Kinetic energy cannot be negative; it is \( +1.51\,\mathrm{eV} \).

(B) Its potential energy is -3.02eV — Correct. \( U = 2E = -3.02\,\mathrm{eV} \).

(C) electron is in second excited stated — Correct. \( n = 3 \): the level \( n=1 \) is the ground state, \( n=2 \) the first excited state and \( n=3 \) the second excited state.

(D) The ground state energy of hydrogen atom is 13.6 eV — Incorrect. The ground state energy is \( -13.6\,\mathrm{eV} \); the value \( 13.6\,\mathrm{eV} \) is the energy needed to ionise the atom from the ground state.

NCERT (2026-27):
Electron Orbits – “… The kinetic energy (K) and electrostatic potential energy (U) of the electron in hydrogen atom are

    \( \displaystyle K = \frac{1}{2}mv^2 = \frac{e^2}{8\pi\varepsilon_0 r} \quad \text{and} \quad U = -\frac{e^2}{4\pi\varepsilon_0 r} \)

… Thus the total energy E of the electron in a hydrogen atom is

    \( \displaystyle E = K + U = \frac{e^2}{8\pi\varepsilon_0 r} - \frac{e^2}{4\pi\varepsilon_0 r} = -\frac{e^2}{8\pi\varepsilon_0 r} \qquad (12.4) \)”

Energy Levels – “… The energy of this state (n = 1), E₁ is –13.6 eV. … Similarly, E₃ = –1.51 eV and E₃ – E₁ = 12.09 eV, or to excite the hydrogen atom from its ground state (n = 1) to second excited state (n = 3), 12.09 eV energy is required, and so on.”