The potential energy of a particle of mass $m$ is given by $U(x) = \begin{cases} E_0; & 0 \le x \le 1 \\ 0; & x > 1 \end{cases}$. $\lambda_1$ and $\lambda_2$ are the de-Broglie wavelengths of the particle when $0 \le x \le 1$ and $x > 1$ respectively. If the total energy of the particle is $2 E_0$,the ratio $\frac{\lambda_1}{\lambda_2}$ will be:

  • A
    $2$
  • B
    $1$
  • C
    $\sqrt{2}$
  • D
    $\frac{1}{\sqrt{2}}$

Explore More

Similar Questions

Electrons used in an electron microscope are accelerated by a voltage of $25 \ kV$. If the voltage is increased to $100 \ kV$,then the de-Broglie wavelength associated with the electrons would

What is the de Broglie wavelength of an electron accelerated through a potential difference of $ 100 \ V $ (in $\text{Å}$)?

An electron microscope uses electrons of $40 \ keV$. The de Broglie wavelength associated with these electrons is approximately:

$A$ material particle with a rest mass $m_0$ is moving with the velocity of light $C$. Then,the wavelength of the de$-$Broglie wave associated with it is

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

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