Question:
The de-Broglie wavelength \((\lambda)\) of a particle
The de-Broglie wavelength \((\lambda)\) of a particle
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The de Broglie wavelength is Planck constant divided by momentum.
Updated On: Nov 13, 2026
- is inversely proportional to impulse.
- does not depend on impulse.
- is proportional to mass.
- is proportional to impulse.
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
The de Broglie wavelength is \(\lambda=\dfrac{h}{p}\), where \(p\) is the momentum.
Step 2: Impulse
Impulse of a particle starting from rest equals its momentum, \(J=\Delta p\).
Step 3: Result
So \(\lambda\propto\dfrac{1}{p}\): inversely proportional to impulse, option (A).
Final Answer:
The wavelength is inversely proportional to impulse. \[ \boxed{\text{(A) }\text{inversely proportional to impulse}} \]
The de Broglie wavelength is \(\lambda=\dfrac{h}{p}\), where \(p\) is the momentum.
Step 2: Impulse
Impulse of a particle starting from rest equals its momentum, \(J=\Delta p\).
Step 3: Result
So \(\lambda\propto\dfrac{1}{p}\): inversely proportional to impulse, option (A).
Final Answer:
The wavelength is inversely proportional to impulse. \[ \boxed{\text{(A) }\text{inversely proportional to impulse}} \]
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