Write the quantum condition suggested by Bohr for the angular momentum of the electron.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) According to Bohr's second postulate,the electron revolves around the nucleus only in those orbits for which its angular momentum is an integral multiple of $\frac{h}{2\pi}$.
Mathematically,this condition is expressed as:
$L = mvr = \frac{nh}{2\pi}$
Where:
$L$ is the angular momentum of the electron.
$m$ is the mass of the electron.
$v$ is the velocity of the electron in its orbit.
$r$ is the radius of the orbit.
$n$ is the principal quantum number $(n = 1, 2, 3, ...)$.
$h$ is Planck's constant.

Explore More

Similar Questions

When the electron orbiting in a hydrogen atom in its ground state moves to the third excited state,the de-Broglie wavelength associated with it

In the Bohr model,an electron of mass $m$ moves in a circular orbit around the proton. Considering the orbiting electron as a circular current loop,find the magnetic moment of the hydrogen atom when the electron is in the $n$th orbit. (Assume $h$ is Planck's constant)

According to Bohr's theory,the time-averaged magnetic field at the centre (i.e.,nucleus) of a hydrogen atom due to the motion of electrons in the $n^{th}$ orbit is proportional to ($n =$ principal quantum number).

The de-Broglie wavelength of an electron in the ground state of the hydrogen atom is

The ratio of radii of $3^{\text{rd}}$ and $6^{\text{th}}$ Bohr's orbit in a hydrogen atom is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo