What is the angular momentum of the electron in a hydrogen atom in its ground state?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( h/2\pi \)
By Bohr’s second postulate, the electron revolves only in orbits where its angular momentum is an integral multiple of \( h/2\pi \). So \( L=nh/2\pi \). The ground state is the orbit with \( n=1 \).
\( \begin{aligned} L &= \frac{nh}{2\pi} \\ &= \frac{1\times h}{2\pi} \\ &= \frac{h}{2\pi} \end{aligned} \)
NCERT (2026-27):
Bohr Model of the Hydrogen Atom – “(ii) Bohr’s second postulate defines these stable orbits. 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
\( L=nh/2\pi \qquad (12.5) \)”
Summary – “… The n = 1 state is called ground state. In hydrogen atom the ground state energy is –13.6 eV.”