The de-Broglie wavelength of an electron movi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) According to Bohr's quantization condition, the angular momentum of an electron in the $n^{th}$ orbit is given by $mvr = \frac{nh}{2\pi}$.From the

Vedclass · Accepted Answer

$2\pi r/n$

Vedclass · Answer

(C) According to Bohr's quantization condition, the angular momentum of an electron in the $n^{th}$ orbit is given by $mvr = \frac{nh}{2\pi}$.From the de-Broglie relation, the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.Rearranging the Bohr quantization condition, we get $mv = \frac{nh}{2\pi r}$.Substituting this into the de-Broglie wavelength formula: $\lambda = \frac{h}{(nh / 2\pi r)}$.Simplifying the expression, we get $\lambda = \frac{2\pi r}{n}$.

The de-Broglie wavelength of an electron moving in the $n^{th}$ Bohr orbit of radius $r$ is

Similar Questions

The ratio of angular momentum of an electron in $n^{\text{th}}$ orbit of a hydrogen atom to the velocity of the electron in the $n^{\text{th}}$ orbit is proportional to

The number of absorption transitions between the first and the fourth energy states of a hydrogen atom is $3$. The number of emission transitions between these states will be

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

The ratio of the velocity of the electron in the first Bohr orbit to that in the second Bohr orbit of a hydrogen atom is

The ionisation energy of a hydrogen atom is $13.6 \; eV$. The ionisation energy of a singly ionised helium atom would be ....... $eV$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module