The time of revolution of an electron around a nucleus of charge $Ze$ in the $n^{th}$ Bohr orbit is directly proportional to

  • A
    $n$
  • B
    $\frac{n^3}{Z^2}$
  • C
    $\frac{n^2}{Z}$
  • D
    $\frac{Z}{n}$

Explore More

Similar Questions

Electrons in a certain energy level $n=n_{1}$ can emit $3$ spectral lines. When they are in another energy level $n=n_{2}$,they can emit $6$ spectral lines. The orbital speed of the electrons in these orbits are in the ratio:

An electron in the hydrogen atom initially in the fourth excited state makes a transition to $n^{\text{th}}$ energy state by emitting a photon of energy $2.86 \ eV$. The integer value of $n$ will be . . . . . . .

The ionization potential of a hydrogen atom is $13.6 \ eV$. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy $12.1 \ eV$. According to Bohr's theory,the number of spectral lines emitted by the hydrogen atoms will be:

Using Bohr's atomic model,derive an equation for the radius of the $n^{th}$ orbit of an electron.

Which of the following graphs in the figure shows the speed $(v)$ of an electron in a hydrogen atom as a function of the principal quantum number $(n)$?

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