The angular momentum (in Js units) of an electron in the 3rd orbit of hydrogen atom is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 3.16\times10^{-34} \)
By Bohr’s quantisation condition, the angular momentum of the electron in the \( n \)th orbit is an integral multiple of \( h/2\pi \): \( L = nh/2\pi \). It does not depend on the nuclear charge, so only the orbit number \( n = 3 \) is needed.
\( \displaystyle L = \frac{3h}{2\pi} = \frac{3\times6.626\times10^{-34}\,\mathrm{J\,s}}{2\times3.1416} \)
\( L \approx 3.16\times10^{-34}\,\mathrm{J\,s} \)