The angular momentum of the electron in a hydrogen atom is \(4.2\times10^{-34}\ \mathrm{J\,s}\). The total energy of the electron is (Take \(h=6.6\times10^{-34}\ \mathrm{J\,s}\) and \(E_n=-13.6/n^2\ \mathrm{eV}\))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(-0.85\ \mathrm{eV}\)
Bohr’s rule \( L=nh/2\pi \) gives the orbit number from the angular momentum. The energy of that orbit then comes from \( E_n=-13.6/n^2\ \mathrm{eV} \).
\( \begin{aligned} n &= \frac{2\pi L}{h} \\ &= \frac{2\times3.14\times4.2\times10^{-34}}{6.6\times10^{-34}} \\ &= 4 \end{aligned} \)
\( \begin{aligned} E_4 &= -\frac{13.6}{4^2} \\ &= -0.85\ \mathrm{eV} \end{aligned} \)
Therefore, the electron is in the \( n=4 \) orbit and its total energy is \( -0.85\ \mathrm{eV} \).
NCERT (2026-27):
Bohr Model of the Hydrogen Atom – “… This postulate states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of h/2π where h is the Planck’s constant (= 6.6 × 10–34 J s). Thus the angular momentum (L) of the orbiting electron is quantised. That is”
\( \displaystyle L=\frac{nh}{2\pi} \)\( \qquad (12.5) \)
Bohr Model of the Hydrogen Atom – “Atomic energies are often expressed in electron volts (eV) rather than joules. … can be rewritten as”
\( \displaystyle E_n=-\frac{13.6\ \text{eV}}{n^2} \)\( \qquad (12.10) \)