Physics-de Broglie Wavelength

NEET Physics de Broglie Wavelength MCQ Question

Type: MCQ-diagram based-Easy-Class 12

Consider a diagram showing a particle with mass mm and velocity vv. If it exhibits wave-like properties according to de Broglie's hypothesis, what is the expression for its wavelength λ\lambda?

Question diagram
A

λ=hmv\lambda = \frac{h}{mv}

B

λ=mvh\lambda = \frac{mv}{h}

C

λ=hv\lambda = \frac{h}{v}

D

λ=hmv\lambda = hmv

Correct Answer

Option A

Detailed Explanation

According to de Broglie's hypothesis, the wavelength λ\lambda associated with a particle is given by λ=hmv\lambda = \frac{h}{mv}, where hh is Planck's constant and mvmv is the momentum of the particle.

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