Question

An electron (mass m) with an initial velocity $\overset{\rightarrow}{v} = v_{0}\widehat{i}$ is in an electric field $\overset{\rightarrow}{E} = E_{0}\ \widehat{j}.$ if $\lambda_{0} = \frac{h}{mv_{0}},$ it’s de Broglie wavelength at time t is given by

• A. $\lambda_{0}$
• B. $\lambda_{0}\sqrt{1 + \frac{e^{2}E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}}}$
• C. $\frac{\lambda_{0}}{\sqrt{1 + \frac{e^{2}E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}}}}$
• D. $\frac{\lambda_{0}}{\left( 1 + \frac{e^{2}E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}} \right)}$

Answer 1

Answer: (C)

$\frac{\lambda_{0}}{\sqrt{1 + \frac{e^{2}E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}}}}$

Answer 2

: Initial de Broglie wavelength of electron, $\lambda_{0} = \frac{h}{mv_{0}}$

Force on electron in electric field, $\overrightarrow{F} = - e\overset{\rightarrow}{E} = - eE_{0}\widehat{j}$

Acceleration of electron, $\overrightarrow{a} = \frac{\overrightarrow{F}}{m} = - \frac{eE_{0}}{m}\widehat{j}$

It is acting along negative y-axis

The initial velocity of electron along x-axis $\overset{\rightarrow}{v_{x_{0}}} = v_{0}î$

Initial velocity of electron along y- axis $\overset{\rightarrow}{v_{y_{0}}} = 0$

Velocity of electron after time t along x-axis $\overset{\rightarrow}{v_{x}} = v_{0}î$

($\therefore$There is no acceleration of electron along x-axis velocity of electron after time t along y-axis)

$\overset{\rightarrow}{v_{y}} = 0 + \left( - \frac{eE_{0}}{m}\widehat{j} \right)t = - \frac{eE_{0}}{m}t\widehat{j}$

Magnitude of velocity of electrons after time t is

$\left| \overrightarrow{v} \right| = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{v_{0}^{2} + \left( \frac{- eE_{0}}{m}t \right)^{2}} = v_{0}\sqrt{1 + \frac{e^{2}E_{0}^{2}t^{2}}{(m^{2}v_{0}^{2})}}$

De Broglie wavelength associated with electron at time t is

$\lambda = \frac{h}{mv} = \frac{h}{mv_{0}\sqrt{1 + \frac{e^{2}E_{0}^{2}t^{2}}{(m^{2}v_{0}^{2})}}} = \frac{\lambda_{0}}{\sqrt{1 + \frac{e^{2}E_{0}^{2}t^{2}}{(m^{2}v_{0}^{2})}}}$