The orbital acceleration of an electron is .........
- A$\frac{n^2 h^2}{4 \pi^2 m^2 r^3}$
- B$\frac{n^2 h^2}{2 n^2 r^3}$
- C$\frac{4 n^2 h^2}{\pi^2 m^2 r^3}$
- D$\frac{n^2 h^2}{\pi^2 m^2 r^3}$
Explore More
Similar Questions
For an electron moving in the $n^{th}$ orbit of an $H$-atom,the angular velocity is proportional to:
Medium
View SolutionThe approximate value of the principal quantum number for a circular orbit of a hydrogen atom of radius $530 \ nm$ is
Experimentally it is found that $12.8 \, eV$ energy is required to separate a hydrogen atom into a proton and an electron. The orbital radius of the electron in this hydrogen atom is $\frac{9}{x} \times 10^{-10} \, m$. The value of $x$ is (Given: $1 \, eV = 1.6 \times 10^{-19} \, J$,$\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, Nm^2/C^2$,and electronic charge $e = 1.6 \times 10^{-19} \, C$)
The orbital acceleration of an electron in a hydrogen atom is given by:
The acceleration of an electron in the first orbit of the hydrogen atom $(Z = 1)$ is
DifficultJEE MAIN 2017
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo