The energy of an electron revolving in the $n^{th}$ Bohr's orbit of an atom is given by the expression

  • A
    $E_n = - \frac{2\pi^2 m^4 e^2 z^2}{n^2 h^2}$
  • B
    $E_n = - \frac{2\pi^2 m e^2 z^2}{n^2 h^2}$
  • C
    $E_n = - \frac{2\pi^2 m e^4 z^2}{n^2 h^2}$
  • D
    $E_n = - \frac{2\pi m^2 e^2 z^4}{n^2 h^2}$

Explore More

Similar Questions

If $\lambda_0$ and $\lambda$ are respectively the threshold wavelength and wavelength of incident light,the velocity of photoelectrons ejected from the metal surface is

The energy of the second orbit of a hydrogen atom is $-5.45 \times 10^{-19} \ J$. What is the energy of the first orbit of $Li^{2+}$ ion (in $J$)?

Which of the following equations gives the angular momentum of an electron in a stationary orbit?

How does the energy of an orbit change as we move away from the nucleus in a hydrogen-like atom?

If the energies of two light radiations $E_1$ and $E_2$ are $25 \ eV$ and $100 \ eV$ respectively,then their respective wavelengths $\lambda_1$ and $\lambda_2$ would be in the ratio $\lambda_1: \lambda_2=$

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