An electron of mass m with an initial velocity $\vec{v} = v_{0} \hat{i} (v_{0} > 0)$ enters an electric field $\vec{E} = - {\vec{E}}_{0} \hat{i} (E_{0} = constant > 0) at t = 0 .$
If $λ_{0}$ is its de-Broglie wavelength initially, then its de- Broglie wavelength at time t is

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$\frac{λ_{0}}{(1 + \frac{e E_{0}}{m v_{0}} t)}$

$λ_{0} (1 + \frac{e E_{0}}{m v_{0}} t)$

$λ_{0} t$

$λ_{0}$

answer is A.

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Detailed Solution

Here, = $\frac{λ_{0}}{(1 + \frac{e E_{0}}{m v_{0}} t)}$ initial; velocity $\vec{v} = v_{0} \hat{i}$

Force action on electron due to electric field

$\vec{F} = (- e) (- E_{0} \hat{i}) = e E_{0} \hat{i}$

$Acceleration produced in the electron, \vec{a} = \frac{\vec{F}}{m} = \frac{e E_{0}}{m} \hat{i}$

Now, velocity of electron after time t

${\vec{v}}_{t} = \vec{v} + \vec{a} t = (v_{0} + \frac{e E_{0} t}{m}) \hat{i} or |{\vec{v}}_{t}| = v_{0} + \frac{e E_{0} t}{m}$

$Now, λ_{t} = \frac{h}{m v_{t}} = \frac{h}{m (v_{0} + \frac{e E_{0} t}{m})} = \frac{h}{m v_{0} (1 + \frac{e E_{0} t}{m v_{0}})}$

$= \frac{λ_{0}}{(1 + \frac{e E_{0} t}{m v_{0}})} (∵ λ_{0} = \frac{h}{m v_{0}})$

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