The electron belongs to the angular momentum of an $e^{-}$ in a Bohr's orbit of H atom is $4.2178 \times 10^{- 34} K g . m^{2} . S e c^{- 1}$ .

K - shell

No worries! We‘ve got your back. Try BYJU‘S free classes today!

L - shell

No worries! We‘ve got your back. Try BYJU‘S free classes today!

N - shell

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

M - shell

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
N - shell

$\frac{n h}{2 π}$ = $4.2178 \times 10^{- 34}$

$n$ = $\frac{2 π \times 4.2178 \times 10^{- 34}}{h}$

= $\frac{2 \times 3.14 \times 4.2178 \times 10^{- 34}}{6.625 \times 10 - 34}$

$n$ = $4$

N-shell

Suggest Corrections

Similar questions

The electron belongs to the angular momentum of an $e^{-}$ in a Bohr's orbit of H atom is $4.2178 \times 10^{- 34} K g . m^{2} . S e c^{- 1}$ .

The angular momentum of an electron in a Bohr's orbit of H-atom is $4.2178 \times 10^{- 34} K g m^{2} / s e c$ . Then electron belongs to

The electron belongs to the angular momentum of an $e^{-}$ in a Bohr's orbit of H atom is $4.2178 \times 10^{- 34} K g . m^{2} . S e c^{- 1}$ .

4.2178 \times 10^{- 34} {k g m}^{2} / s e c

H

4.2178 \times 10^{- 34} k g m^{2} / s e c

Join BYJU'S Learning Program