An electron is accelerated through a potential difference of $10,000 \; V$. Its de Broglie wavelength is,(nearly):
$(m_{e}=9 \times 10^{-31} \; kg)$

  • A
    $12.2 \times 10^{-13} \; m$
  • B
    $12.2 \times 10^{-12} \; m$
  • C
    $12.2 \times 10^{-14} \; m$
  • D
    $12.2 \; nm$

Explore More

Similar Questions

The kinetic energies of an electron,$\alpha$-particle,and a proton are given as $4K, 2K$,and $K$ respectively. The de-Broglie wavelengths associated with the electron $(\lambda_e)$,$\alpha$-particle $(\lambda_\alpha)$,and the proton $(\lambda_p)$ are related as follows:

For which of the following particles will it be most difficult to experimentally verify the de Broglie relationship?

Both an electron and a photon have the same de-Broglie wavelength of $1.2 \ \text{Å}$. The ratio of their energies is nearly

The wavelength of a charged particle of mass $8.0 \times 10^{-31} \ kg$, charge $1.6 \times 10^{-19} \ C$ and kinetic energy $3 \ keV$ will be (Planck constant, $h = 6.4 \times 10^{-34} \ Js$) (in $\text{Å}$)

The wavelength of a very fast-moving electron $(v \approx c)$ is:

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