An electron of mass with an initial velocity enters an electric field at . If is its de-Broglie wavelength initially, then its de-Broglie wavelength at time is [2018]

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Solution

Q.
An electron of mass $m$ with an initial velocity $\vec{v} = v_{0} \hat{i}$ $(v_{0} > 0)$ enters an electric field $\vec{E} = - E_{0} \hat{i}$ $(E_{0} = constant > 0)$ at $t = 0$ . If $λ_{0}$ is its de-Broglie wavelength initially, then its de-Broglie wavelength at time $t$ is [2018]

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

2 $λ_{0} (1 + \frac{e E_{0}}{m v_{0}} t)$

3 $λ_{0} t$

4 $λ_{0}$

Ans.
(1)

Here, $\vec{E} = - E_{0} \hat{i}$ ; initial velocity $\vec{v} = v_{0} \hat{i}$

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

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

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