Question:
The de-Broglie wavelength \(\lambda\) of a particle
The de-Broglie wavelength \(\lambda\) of a particle
Show Hint
Use lambda = h/p, which shows the wavelength varies inversely with momentum.
Updated On: Nov 12, 2026
- is proportional to mass
- is proportional to momentum
- is inversely proportional to momentum
- does not depend on momentum
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
de Broglie proposed that a moving particle has a wavelength \(\lambda=\dfrac{h}{p}\).
Step 2: Analyse the relation
Here \(h\) is Planck's constant, so \(\lambda\propto\dfrac{1}{p}\).
Step 3: Compare the options
The wavelength is inversely proportional to momentum, which is option (C).
Final Answer:
The de Broglie wavelength is inversely proportional to the momentum of the particle. \[ \boxed{\text{(C) }\text{is inversely proportional to momentum}} \]
de Broglie proposed that a moving particle has a wavelength \(\lambda=\dfrac{h}{p}\).
Step 2: Analyse the relation
Here \(h\) is Planck's constant, so \(\lambda\propto\dfrac{1}{p}\).
Step 3: Compare the options
The wavelength is inversely proportional to momentum, which is option (C).
Final Answer:
The de Broglie wavelength is inversely proportional to the momentum of the particle. \[ \boxed{\text{(C) }\text{is inversely proportional to momentum}} \]
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