De Broglie’s wavelength $\lambda $ of a particle is
(A) Proportional to mass
(B) Proportional to impulse
(C) Inversely proportional to impulse
(D) Doesn’t depend on impulse

Hint We write the de Broglie’s wavelength formula and see how the wavelength varies with mass and impulse to solve this equation. De Broglie’s wavelength is the ratio of Planck's constant to the momentum of the particle. Impulse is nothing but change in momentum.

Complete Step by step solution
De Broglie’s wavelength is the ratio of Planck's constant to the momentum of the particle.
$\lambda = \dfrac{h}{p}$
Here,
Wavelength is represented by $\lambda $
Plank constant is represented by $h$
Momentum of particle is represented by $p$
Momentum is the product of mass and velocity
$p = mv$
Thus, we can write the de Broglie’s wavelength as
$\lambda = \dfrac{h}{{mv}}$
Here,
Mass of particle is represented by $m$
Velocity is particle is represented by $v$
From the first equation we can see that momentum or impulse is inversely proportional to wavelength.
From the second equation we can see that mass is inversely proportional to wavelength.
Hence all the other options expect (c) are wrong

Option (c) inversely proportional to impulse is the correct answer.

Additional information De Broglie’s principle states that any matter can act as a wave as well as a particle. This is called the dual nature of matter. The wavelength of the particle in wave nature is called de Broglie’s wavelength. Although de Broglie’s principle applies for all matter, it is significant only for microscopic matter.

Note One must know what impulse is to solve this question. Impulse is the difference in final and initial momentum. In this question we consider impulse as momentum because we assume the initial momentum to be zero.