An electron with mass m and an initial veloci | vedclass

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

Question

(D) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.Since the magnetic force $\vec{F} = q(\vec{v} 	imes \vec{B})$ is alw

Vedclass · Accepted Answer

$\lambda_0$

Vedclass · Answer

(D) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.Since the magnetic force $\vec{F} = q(\vec{v} 	imes \vec{B})$ is always perpendicular to the velocity $\vec{v}$,the magnetic field does no work on the electron.According to the work-energy theorem,the kinetic energy of the electron remains constant.Since the kinetic energy $K = \frac{1}{2}mv^2$ is constant,the speed $v$ of the electron remains constant.As the speed $v$ remains constant,the momentum $p = mv$ remains constant.Therefore,the de Broglie wavelength $\lambda = \frac{h}{p}$ remains constant over time.Thus,the wavelength at time $t$ is equal to the initial wavelength $\lambda_0$.

An electron with mass $m$ and an initial velocity $(t=0)$ $\vec{v} = v_0 \hat{i}$ $(v_0 > 0)$ enters a magnetic field $\vec{B} = B_0 \hat{j}$. If the initial de Broglie wavelength at $t=0$ is $\lambda_0$,then its value after time $t$ would be:

Similar Questions

An electron of mass $m$ and charge $e$ initially at rest gets accelerated by a constant electric field $E$. The rate of change of de-Broglie wavelength of the electron at time $t$ is (Ignore relativistic effect) ($h=$ Planck's constant).

What does the de Broglie equation for an electron represent?

Two points $A$ and $B$ have potentials of $20 \ V$ and $40 \ V$ respectively. An electron is accelerated from rest between them. Find the de Broglie wavelength associated with the electron at point $B$.

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

$A$ particle of mass $1 \, mg$ has the same wavelength as an electron moving with a velocity of $3 \times 10^6 \, m/s$. The velocity of the particle is (mass of electron $= 9.1 \times 10^{-31} \, kg$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module