Question:
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
(\(h\) = Planck's constant)
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
(\(h\) = Planck's constant)
(\(h\) = Planck's constant)
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Find the momentum from force times time, then differentiate the wavelength with respect to time.
Updated On: Nov 12, 2026
- \(-\frac{h}{eEt^{2}}\)
- \(-\frac{eEt}{h}\)
- \(-\frac{mh}{eEt^{2}}\)
- \(-\frac{h}{eE}\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
A constant force \(F=eE\) acts on an electron starting from rest, so its momentum grows with time.
Step 2: Momentum
\(p=Ft=eEt\)
Step 3: de Broglie Wavelength
\(\lambda=\dfrac{h}{p}=\dfrac{h}{eEt}\)
Step 4: Differentiate
\(\dfrac{d\lambda}{dt}=-\dfrac{h}{eEt^{2}}\)
Final Answer:
The rate of change of wavelength is minus h divided by eE t squared. \[ \boxed{\text{(A) }-\dfrac{h}{eEt^{2}}} \]
A constant force \(F=eE\) acts on an electron starting from rest, so its momentum grows with time.
Step 2: Momentum
\(p=Ft=eEt\)
Step 3: de Broglie Wavelength
\(\lambda=\dfrac{h}{p}=\dfrac{h}{eEt}\)
Step 4: Differentiate
\(\dfrac{d\lambda}{dt}=-\dfrac{h}{eEt^{2}}\)
Final Answer:
The rate of change of wavelength is minus h divided by eE t squared. \[ \boxed{\text{(A) }-\dfrac{h}{eEt^{2}}} \]
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