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Step-by-Step Solution
Step 1: Identify the initial conditions
• The electron has an initial velocity
$ \vec{v} = v_{0}\,\hat{i} $ (with $v_{0} > 0$).
• The electric field is given by
$ \vec{E} = -\,E_{0}\,\hat{i} $ (where $E_{0} > 0$).
• The initial de Broglie wavelength is
$ \lambda_{0} = \frac{h}{m\,v_{0}}. $
Step 2: Find the force acting on the electron
The charge of an electron is $-e$.
Because $ \vec{E} = -E_{0}\,\hat{i} $, the force on the electron is:
$ \vec{F} = q\,\vec{E} = (-e)\bigl(-E_{0}\,\hat{i}\bigr) = e\,E_{0}\,\hat{i}. $
Step 3: Determine the acceleration of the electron
From Newton's second law, $ \vec{F} = m\,\vec{a} $. Hence,
$ \vec{a} = \frac{\vec{F}}{m} = \frac{e\,E_{0}}{m}\,\hat{i}. $
Step 4: Calculate the velocity at time t
The velocity at time $t$ can be found using
$ \vec{v}_{t} = \vec{v}_{0} + \vec{a}\,t. $
Here, $ \vec{v}_{0} = v_{0}\,\hat{i} $, so
$ \vec{v}_{t} = \bigl(v_{0} + \tfrac{e\,E_{0}}{m}t\bigr)\,\hat{i}. $
Thus, its magnitude at time t is
$ |\vec{v}_{t}| = v_{0} + \frac{e\,E_{0}}{m}\,t. $
Step 5: Use the de Broglie wavelength formula
The de Broglie wavelength $ \lambda $ is given by
$ \lambda = \frac{h}{p} = \frac{h}{m\,v}. $
At time $t$, the electron's speed is $v_{t} = v_{0} + \tfrac{e\,E_{0}}{m}\,t$, so
$ \lambda_{t} = \frac{h}{m\,v_{t}}
= \frac{h}{m\Bigl(v_{0} + \tfrac{e\,E_{0}}{m}t\Bigr)}. $
Step 6: Express in terms of the initial wavelength
Recall the initial de Broglie wavelength
$ \lambda_{0} = \frac{h}{m\,v_{0}}. $
Factor this out in the denominator:
$ \lambda_{t}
= \frac{h}{m\,v_{0}\Bigl(1 + \tfrac{e\,E_{0}}{m\,v_{0}}\,t\Bigr)}
= \frac{\lambda_{0}}{1 + \tfrac{e\,E_{0}}{m\,v_{0}}\,t}. $
Hence, the de Broglie wavelength at time $t$ is
$ \boxed{ \lambda_{t} \;=\; \frac{\lambda_{0}}{\,1 + \frac{e\,E_{0}}{m\,v_{0}}\,t\,}\,. } $
