Question

A particle is dropped from a height $H$ . The de-Broglie wavelength of the particle as a function of height is proportional to:
A. $H$
B. ${H^{\dfrac{1}{2}}}$
C. ${H^0}$
D. ${H^{\dfrac{{ - 1}}{2}}}$

Answer 1

First of all, we will find an expression for velocity of the particle with the initial velocity zero. We will substitute this expression in the de-Broglie’s equation and manipulate it accordingly.

Complete step by step answer:
In the given problem, the data supplied is:
The object is dropped from a height $H$ .
It means that during the process it has travelled a distance of $H$ .

From de-Broglie’s equation, which relates wavelength and momentum, is given below:
$\lambda = \dfrac{h}{{mv}}$ …… (1)
Where,
$\lambda$ indicates wavelength.
$h$ indicates Planck’s constant.
$m$ indicates mass of the particle.
$v$ indicates velocity of the particle.

Again, we have an expression for velocity possessed by a particle which is falling from a height of $H$ is given below:
$v = \sqrt {2gH}$ …… (2)
Where,
$v$ indicates velocity of the particle.
$g$ indicates acceleration due to gravity.
$H$ indicates the height from which it is falling.

Now, we will use equation (2) in equation (1):
\lambda =\dfrac{h}{{mv}} \\\
\implies \lambda =\dfrac{h}{{m\sqrt {2gH} }} \\\
\implies \lambda =\dfrac{h}{{m\sqrt {2g} \times \sqrt H }} \\\
\implies \lambda =\dfrac{h}{{m\sqrt {2g} \times {H^{\dfrac{1}{2}}}}} \\\
Simplifying the above expression further, we get:
$\lambda = \dfrac{h}{{m\sqrt {2g} }} \times {H^{\dfrac{{ - 1}}{2}}}$ …… (3)
Since, $h$ ,$m$ and $g$ are constants for a given particle. So, equation (3) can be written as:
$\lambda \propto {H^{\dfrac{{ - 1}}{2}}}$
Hence, the de-Broglie wavelength of the particle as a function of height is proportional to ${H^{\dfrac{{ - 1}}{2}}}$ .

So, the correct answer is “Option D”.

Additional Information:
The De Broglie wavelength is a wavelength expressed in all objects in quantum mechanics, according to wave-particle duality, which specifies the probability density of locating the object at a particular moment in the space of the specification. A particle's de Broglie wavelength is inversely proportional to its momentum.

Note:
In de-Broglie's equation it is clear that, greater the mass of the particle, negligible is the wavelength. Lighter the particle is, the higher the significance of its wavelength. If a small stone is thrown up in the air, it also possesses wavelength, but the wavelength is so small that it has no significance because the mass of the stone is pretty much greater than that of an electron or other subatomic particles.