If l is the angular momentum of the electron in the 3rd orbit of the…

The correct answer is Option (3) → \( 2l \)

Bohr’s second postulate makes the angular momentum of the electron quantised, \( L=nh/2\pi \). So the angular momentum is proportional to the orbit number \( n \).

Comparing the 6th orbit with the 3rd:

    \( \begin{aligned} \frac{L_6}{L_3} &= \frac{6h/2\pi}{3h/2\pi} \\ &= 2 \end{aligned} \)

Since \( L_3=l \), the angular momentum in the 6th orbit is:

    \( L_6=2l \)

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

    \( L=nh/2\pi \qquad (12.5) \)”