(A) The acceleration of the electron is $a = \frac{|e|E}{m}$.Since the electron starts from rest $(u = 0)$,its velocity at time $t$ is $v = at = \frac

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(A) The acceleration of the electron is $a = \frac{|e|E}{m}$.Since the electron starts from rest $(u = 0)$,its velocity at time $t$ is $v = at = \frac{|e|E}{m}t$.The de-Broglie wavelength is given by $\lambda = \frac{h}{mv} = \frac{h}{m(\frac{|e|E}{m}t)} = \frac{h}{|e|Et}$.To find the rate of change of wavelength,we differentiate $\lambda$ with respect to $t$:$\frac{d\lambda}{dt} = \frac{d}{dt} (\frac{h}{|e|Et}) = \frac{h}{|e|E} \cdot \frac{d}{dt}(t^{-1}) = \frac{h}{|e|E} (-t^{-2}) = -\frac{h}{|e|Et^2}$.

Class 12 Physics Dual Nature of Radiation and matter Matter Waves and de Broglie Wavelength

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An electron of mass $m$ and magnitude of charge $|e|$ initially at rest gets accelerated by a constant electric field $E$. The rate of change of de-Broglie wavelength of this electron at time $t$ ignoring relativistic effects is:

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A
$\frac{-h}{|e| Et^{2}}$
B
$\frac{|e| Et}{h}$
C
$-\frac{h}{|e| E \sqrt{t}}$
D
$-\frac{h}{|e| Et}$

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Dual Nature of Radiation and matter

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Matter Waves and de Broglie Wavelength

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An electron of mass $m$ and magnitude of charge $|e|$ initially at rest gets accelerated by a constant electric field $E$. The rate of change of de-Broglie wavelength of this electron at time $t$ ignoring relativistic effects is:

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