An electron of mass m and charge e initially | vedclass

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Question

(A) The force on the electron is $F = eE$. The acceleration is $a = \frac{eE}{m}$.Since the electron starts from rest,its velocity at time $t$ is $v =

Vedclass · Accepted Answer

$-\frac{h}{eEt^2}$

Vedclass · Answer

(A) The force on the electron is $F = eE$. The acceleration is $a = \frac{eE}{m}$.Since the electron starts from rest,its velocity at time $t$ is $v = at = \frac{eEt}{m}$.The momentum of the electron is $p = mv = m(\frac{eEt}{m}) = eEt$.The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{eEt}$.To find the rate of change of wavelength,we differentiate $\lambda$ with respect to time $t$:$\frac{d\lambda}{dt} = \frac{d}{dt}(\frac{h}{eEt}) = \frac{h}{eE} \frac{d}{dt}(t^{-1}) = \frac{h}{eE} (-t^{-2}) = -\frac{h}{eEt^2}$.

An electron of mass $m$ and charge $e$ initially at rest gets accelerated by a constant electric field $E$. The rate of change of de-Broglie wavelength of the electron at time $t$ is (Ignore relativistic effect) ($h=$ Planck's constant).

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