An electron of mass m with an initial velocit | vedclass

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Question

(A) The force acting on the electron due to the electric field is $\vec{F} = q\vec{E} = (-e)(-E_0 \hat{i}) = eE_0 \hat{i}$.The acceleration produced i

Vedclass · Accepted Answer

$\frac{\lambda_0}{\left(1 + \frac{eE_0}{mV_0}t\right)}$

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(A) The force acting on the electron due to the electric field is $\vec{F} = q\vec{E} = (-e)(-E_0 \hat{i}) = eE_0 \hat{i}$.The acceleration produced in the electron is $\vec{a} = \frac{\vec{F}}{m} = \frac{eE_0}{m} \hat{i}$.The velocity of the electron at time $t$ is given by $\vec{v}_t = \vec{v} + \vec{a}t = \left(V_0 + \frac{eE_0}{m}t\right) \hat{i}$.The magnitude of the velocity is $v_t = V_0 + \frac{eE_0}{m}t = V_0 \left(1 + \frac{eE_0}{mV_0}t\right)$.The de-Broglie wavelength at time $t$ is $\lambda_t = \frac{h}{mv_t}$.Substituting $v_t$,we get $\lambda_t = \frac{h}{m V_0 \left(1 + \frac{eE_0}{mV_0}t\right)}$.Since the initial de-Broglie wavelength is $\lambda_0 = \frac{h}{mV_0}$,we have $\lambda_t = \frac{\lambda_0}{\left(1 + \frac{eE_0}{mV_0}t\right)}$.

An electron of mass $m$ with an initial velocity $\vec{V} = V_0 \hat{i} \,(V_0 > 0)$ enters an electric field $\vec{E} = -E_0 \hat{i} \,(E_0 = \text{constant} > 0)$ at $t = 0$. If $\lambda_0$ is its de-Broglie wavelength initially,then its de-Broglie wavelength at time $t$ is:

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