An electron moves in a Bohr orbit. The magnetic field at the centre is proportional to

  • A
    $n^{-2}$
  • B
    $n^{-3}$
  • C
    $n^{-4}$
  • D
    $n^{-5}$

Explore More

Similar Questions

The de Broglie wavelength of an electron revolving in the first Bohr orbit of circumference $2\pi r$ is ............

Hydrogen $(H)$,deuterium $(D)$,singly ionized helium $(He^+)$ and doubly ionized lithium $(Li^{2+})$ all have one electron around the nucleus. Consider the $n = 2$ to $n = 1$ transition. The wavelengths of emitted radiations are $\lambda_1, \lambda_2, \lambda_3$ and $\lambda_4$ respectively. Then approximately:

Difficult
View Solution

The radius of the $2^{nd}$ orbit of $He^{+}$ in Bohr's model is $r_1$ and that of the fourth orbit of $Be^{3+}$ is represented as $r_2$. If the ratio $\frac{r_2}{r_1}$ is $x : 1$,then the value of $x$ is .........

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

Assertion $(A)$: The magnetic moment $(\mu)$ of an electron revolving around the nucleus decreases with increasing principal quantum number $(n)$.
Reason $(R)$: Magnetic moment of the revolving electron,$\mu \propto n$.

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