Question

An electron (mass m) with an initial velocity $\overset{\rightarrow}{v} = v_{0}\widehat{i}(v_{0} > 0)$ is in an electric field $\overset{\rightarrow}{E} = - E_{0}\text{ }\widehat{\text{i}}$ ($E_{0} =$ constant > 0) It’s de Broglie wavelength at time t is given by

• A. $\frac{\lambda_{0}}{\left( 1 + \frac{eE_{0}t}{mv_{0}} \right)}$
• B. $\lambda_{0}\left( 1 + \frac{eE_{0}t}{mv_{0}} \right)$
• C. $\lambda_{0}$
• D. $\lambda_{0}t$

Answer 1

Answer: (A)

$\frac{\lambda_{0}}{\left( 1 + \frac{eE_{0}t}{mv_{0}} \right)}$

Answer 2

: Here $\overrightarrow{E} = - E_{0}î;initialvelocity\overrightarrow{v} = v_{0}î$

Force action on electron due to electric field

$\overrightarrow{F} = ( - e)( - E_{0}î) = eE_{0}î$

Acceleration produced in the electron,$\overrightarrow{a} = \frac{\overrightarrow{F}}{m} = \frac{eE_{0}}{m}î$

Now velocity of electron after time t, $\overset{\rightarrow}{v_{t}} = \overset{\rightarrow}{v_{}} + \overset{\rightarrow}{a}t$

$= \left( v_{0} + \frac{eE_{0}t}{m} \right)î$

Or $\left| \overset{\rightarrow}{v_{t}} \right| = v_{0} + \frac{eE_{0}t}{m}$

Now $\lambda_{t} = \frac{h}{mv_{t}} = \frac{h}{m\left( v_{0} + \frac{eE_{0}t}{m} \right)} = \frac{h}{mv_{0}\left( 1 + \frac{eE_{0}t}{mv_{0}} \right)}$

$= \frac{\lambda_{0}}{\left( 1 + \frac{eE_{0}t}{mv_{0}} \right)}\left( \because\lambda_{0} = \frac{h}{mv_{0}} \right)$