An electron (mass m) with an initial velocity | vedclass

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

Question

(C) Initial de Broglie wavelength of the electron is $\lambda_0 = \frac{h}{mv_0}$.Force on the electron in the electric field is $\vec{F} = -e\vec{E}

Vedclass · Accepted Answer

$\frac{\lambda_0}{\sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}$

Vedclass · Answer

(C) Initial de Broglie wavelength of the electron is $\lambda_0 = \frac{h}{mv_0}$.Force on the electron in the electric field is $\vec{F} = -e\vec{E} = -eE_0 \hat{j}$.Acceleration of the electron is $\vec{a} = \frac{\vec{F}}{m} = -\frac{eE_0}{m} \hat{j}$.The initial velocity is $\vec{v}_0 = v_0 \hat{i}$.Velocity at time $t$ is $\vec{v} = \vec{v}_0 + \vec{a}t = v_0 \hat{i} - \frac{eE_0 t}{m} \hat{j}$.The magnitude of velocity at time $t$ is $v = \sqrt{v_0^2 + (\frac{eE_0 t}{m})^2} = v_0 \sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}$.The de Broglie wavelength at time $t$ is $\lambda = \frac{h}{mv} = \frac{h}{m v_0 \sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}$.Substituting $\lambda_0 = \frac{h}{mv_0}$,we get $\lambda = \frac{\lambda_0}{\sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}$.

An electron (mass $m$) with an initial velocity $v = v_0 \hat{i}$ is in an electric field $E = E_0 \hat{j}$. If $\lambda_0 = h/mv_0$,its de Broglie wavelength at time $t$ is given by

Similar Questions

The de-Broglie wavelength of an electron moving with a velocity of $1.5 \times 10^8 \ m/s$ is equal to that of a photon. What is the ratio of the kinetic energy of the electron to that of the photon? (Given: $c = 3 \times 10^8 \ m/s$)

Photons of energy $4.5 \ eV$ are incident on a photosensitive material of work function $3 \ eV$. The de Broglie wavelength associated with the photoelectrons emitted with maximum kinetic energy is nearly (in $Å$)

The energy that should be added to an electron to reduce its de-Broglie wavelength from $1 \,nm$ to $0.5 \,nm$ is

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

If the de Broglie wavelengths of an electron and a proton are equal,then the kinetic energy of the electron is ........

Vedclass Test Series

Exam Paper Generator

Online Exam Module