The de Broglie wavelength of a charged partic | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The de Broglie wavelength $\lambda$ of a charged particle accelerated through a potential difference $V$ is given by the formula:$\lambda = \frac{

Vedclass · Accepted Answer

$\frac{10 \lambda}{11}$

Vedclass · Answer

(D) The de Broglie wavelength $\lambda$ of a charged particle accelerated through a potential difference $V$ is given by the formula:$\lambda = \frac{h}{\sqrt{2mqV}}$From this expression,we can see that $\lambda \propto \frac{1}{\sqrt{V}}$.Given that the potential difference is increased by $21 \%$,the new potential difference $V^{\prime}$ is:$V^{\prime} = V + 0.21V = 1.21V$Let the new de Broglie wavelength be $\lambda^{\prime}$. Then:$\frac{\lambda^{\prime}}{\lambda} = \sqrt{\frac{V}{V^{\prime}}} = \sqrt{\frac{V}{1.21V}} = \sqrt{\frac{1}{1.21}} = \frac{1}{1.1} = \frac{10}{11}$Therefore,$\lambda^{\prime} = \frac{10}{11} \lambda$.

The de Broglie wavelength of a charged particle accelerated through a potential difference $V$ is $\lambda$. If the potential difference is increased by $21 \%$,the de Broglie wavelength of the charged particle is

Similar Questions

An electron beam of energy $10 \, KeV$ is incident on a metallic foil. If the interatomic distance is $0.2454 \, \mathring{A}$,then find the angle of diffraction in degrees. $(\sqrt{150} = 12.27)$

What is the de Broglie wavelength of an electron accelerated through $80 \ V$ in $\mathring A$?

If an electron and a photon propagate in the form of waves having the same wavelength,it implies that they have the same

What voltage should be applied to an electron microscope to produce electrons with a wavelength of $\lambda = 1.0 \ \mathring{A}$ (in $V$)?

If the kinetic energy of a free electron is doubled,by what factor does its de Broglie wavelength change?

Vedclass Test Series

Exam Paper Generator

Online Exam Module