An electron of mass m with an initial velocit | vedclass

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

Question

(A) The force acting on the electron due to the electric field is $\vec{F} = q\vec{E} = (-e)(-E_0 \hat{i}) = eE_0 \hat{i}$.The acceleration produced i

Vedclass · Accepted Answer

$\frac{\lambda_0}{\left(1 + \frac{eE_0}{mV_0}t\right)}$

Vedclass · Answer

(A) The force acting on the electron due to the electric field is $\vec{F} = q\vec{E} = (-e)(-E_0 \hat{i}) = eE_0 \hat{i}$.The acceleration produced in the electron is $\vec{a} = \frac{\vec{F}}{m} = \frac{eE_0}{m} \hat{i}$.The velocity of the electron at time $t$ is given by $\vec{v}_t = \vec{v} + \vec{a}t = \left(V_0 + \frac{eE_0}{m}t\right) \hat{i}$.The magnitude of the velocity is $v_t = V_0 + \frac{eE_0}{m}t = V_0 \left(1 + \frac{eE_0}{mV_0}t\right)$.The de-Broglie wavelength at time $t$ is $\lambda_t = \frac{h}{mv_t}$.Substituting $v_t$,we get $\lambda_t = \frac{h}{m V_0 \left(1 + \frac{eE_0}{mV_0}t\right)}$.Since the initial de-Broglie wavelength is $\lambda_0 = \frac{h}{mV_0}$,we have $\lambda_t = \frac{\lambda_0}{\left(1 + \frac{eE_0}{mV_0}t\right)}$.

An electron of mass $m$ with an initial velocity $\vec{V} = V_0 \hat{i} \,(V_0 > 0)$ enters an electric field $\vec{E} = -E_0 \hat{i} \,(E_0 = \text{constant} > 0)$ at $t = 0$. If $\lambda_0$ is its de-Broglie wavelength initially,then its de-Broglie wavelength at time $t$ is:

Similar Questions

An electron beam has a kinetic energy equal to $100\; eV$. Find its wavelength (in $\mathring{A}$) associated with the beam,given the mass of an electron $= 9.1 \times 10^{-31}\; kg$,$1\; eV = 1.6 \times 10^{-19}\; J$,and Planck's constant $= 6.6 \times 10^{-34}\; Js$.

The de Broglie wavelengths associated with a proton and an electron are in the ratio $2: 1$. Their stopping potentials are approximately in the ratio of

The $log-log$ graph between the energy $E$ of an electron and its de-Broglie wavelength $\lambda$ will be

If the momentum of an electron changes by $P$,then the de-Broglie wavelength associated with it changes by $5 \%$. The initial momentum of the electron is: (in $P$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module