Question

Minimum de-Broglie wavelength is associated with
(A) electron
(B) proton
(C) $ C{O_2} $ molecule
(D) $ S{O_2} $ molecule

Answer 1

Hint: In order to answer this question, first we will choose the correct option which is associated with the minimum de-Broglie wavelength and then we will explain the reason behind the correct option and we will also explain the de-Broglie equation.

Complete step by step answer:
Among the given particles, Proton will have the minimum de Broglie wavelength.
This is because de Broglie's wavelength is determined by a particle's mass and velocity, and if all particles have the same velocity, the particle's mass will determine its wavelength.
De Broglie’s wavelength: $ \lambda = \dfrac{h}{{mv}} $
where, $ h $ is the Planck’s constant,
 $ m $ is the mass and
 $ v $ is the velocity of the particle.
The molecular mass of $ SO $ is greater than that of $ CO $ and hence it will have the minimum de Broglie’s wavelength.
Because the electron has the lightest mass, it has the longest wavelength, followed by the proton and $ CO $ .
One of the equations widely used to define the wave characteristics of matter is the de Broglie equation. It essentially describes the electron's wave nature. Electromagnetic radiation has the properties of both a particle (with momentum) and a wave (expressed in frequency, wavelength).
Hence, the correct option is (B) proton.

Note:
Matter, it is thought, has a dual nature of wave-particles. The quality of a material item that varies in time or space while behaving like waves, known as de Broglie waves, is named after the discoverer Louis de Broglie. It's also known as matter-waves. It is very comparable to the empirically proved dual nature of light, which operates as both a particle and a wave.