According to the de-Broglie hypothesis,the wavelength associated with a moving electron of mass $m$ is $\lambda_e$. Using the mass-energy relation and Planck's quantum theory,the wavelength associated with a photon is $\lambda_p$. If the energy $(E)$ of the electron and the photon is the same,then the relation between $\lambda_e$ and $\lambda_p$ is:

  • A
    $\lambda_p \propto \lambda_e$
  • B
    $\lambda_p \propto \lambda_e^2$
  • C
    $\lambda_p \propto \sqrt{\lambda_e}$
  • D
    $\lambda_p \propto \frac{1}{\lambda_e}$

Explore More

Similar Questions

What should be the order of arrangement of de-Broglie wavelength of an electron $(\lambda_{e})$,an $\alpha$-particle $(\lambda_{\alpha})$,and a proton $(\lambda_{p})$,given that all have the same kinetic energy?

Compute the typical de Broglie wavelength of an electron in a metal at $27\,^{\circ} C$ and compare it with the mean separation between two electrons in a metal which is given to be about $2 \times 10^{-10} \; m$.

What is the de Broglie wavelength of a nitrogen molecule in air at $300 \;K$ (in $nm$)? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen $= 14.0076 \;u$)

If $\lambda_0$ is the de-Broglie wavelength for a proton accelerated through a potential difference of $100 \ V$,the de-Broglie wavelength for an $\alpha$-particle accelerated through the same potential difference is

The ratio of momentum of an electron and an alpha particle which are accelerated from rest by a potential difference of $100\,V$ is

Difficult
View Solution

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