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# Wave Nature Of Matter And De Broglie's Equation - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• Wave nature of matter is considered one the most difficult concept.

• De-broglie wavelength of an electron is considered one of the most asked concept.

• 115 Questions around this concept.

## Solve by difficulty

A charged oil drop is suspended in a uniform field of 3 x 104 V/m so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge = 9.9 × 10-15 kg and g = 10 m/s2)

An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-rays.  It produces continuous as well as characteristic X-rays. If λmin is the smallest possible wavelength of X-ray in the spectrum, the variation of log λmin with log V is correctly represented in :

A particle A of mass m and initial velocity $\dpi{100} v$ collides with a particle B of mass $\dpi{100} \frac{m}{2}$ which is at rest.  The collision is head-on and elastic.  The ratio of the de-Broglie wavelengths λA to λB after the collision is :

De-Broglie wavelength associated with the electron in the n=4 level is :

For which of the following particles will it be most difficult to experimentally verify the de-Broglie relationship?

According to De Broglie's equation, what is the wavelength of a particle with a momentum of $2.5\times 10^{-2}kg\, m/s$?

An electron microscope is designed to study very small objects like viruses, microbes, and the crystal structure of solids. What is the De-Broglie wavelength of the electrons utilized by the microscope, given that they are accelerated by a potential difference of 500 V?

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Assertion: Electrons can exhibit wave-like behavior.

Reason: According to De Broglie’s equation, every particle in motion has a wave-like behavior associated with it.

Determine the de Broglie wavelength of an electron that undergoes acceleration by a potential difference of 200 V.

(Given: $h= 6.626\times 10^{-3}Js,m_{e}= 9.1 \times 10^{-31}kg, e= 1.6\times 10^{-19}C$)

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## Concepts Covered - 2

Wave nature of matter

As we know light behaves both as a wave and particle. Because If you are observing phenomenon like the interference, diffraction or reflection, you will find that light is a wave. However, if you are looking at phenomena like the photoelectric effect, you will find that light has a particle character.

De Broglie’s hypothesis stated that there is symmetry in nature and that if the light behaves as both particles and waves, matter too will have both the particle and wave nature.

i.e if lightwave can behave as a particle then the particle can also behave as waves.

De Broglie’s Equation-

According to De Broglie, A moving material particle can be associated with the wave.

De Broglie proposed that the wavelength $\lambda$ associated with the moving material particle of momentum p is given as

$\lambda =\frac{h}{p}$

where $h= plank's\: constant$ and $h=6.626\times 10^{-34} \ Js$

further, we can write De - Broglie wavelength as

$\lambda = \frac{h}{p}= \frac{h}{mv}= \frac{h}{\sqrt{2mK}}$

where

$h= plank's\: constant$

$m= mass \: of\: particle$

$v= speed \: of\: the \: particle$

$K= Kinetic \: energy \: of \: particle$

So from De Broglie’s Equation, we can conclude that-

• $\lambda \: \alpha\; \frac{1}{m}$

i.e Wavelength associated with a heavier particle is smaller than that with a lighter particle.

• $\lambda\: \alpha \frac{1}{\nu }$

i.e when Particle moves faster, then wavelength will be smaller and vice versa

• if the particle at rest then De - Broglie wavelength will be infinite ($\lambda =\infty$)
• $\lambda\: \alpha \: \frac{1}{p}\: \alpha \frac{1}{v}\: \alpha \frac{1}{\sqrt{K}}$
• De - Broglie wavelength (λ) is independent of charge.

De-broglie wavelength of an electron

De - Broglie wavelength of Electron-

As De Broglie’s Equation is given as $\lambda = \frac{h}{p}= \frac{h}{mv}= \frac{h}{\sqrt{2mK}}$

So for an electron having velocity v attained by it when it is accelerated through a potential difference of V.

then (Kinetic energy gain by the electron)=(work is done on an electron by the electric field)

i.e $K=W_E\Rightarrow \frac{1}{2}m_ev^2=eV$

So De - Broglie wavelength of Electron is given as $\lambda_e = \frac{h}{m_ev}= \frac{h}{\sqrt{2m_eK}}=\frac{h}{\sqrt{2m_e(eV)}}$

using $h=6.626\times 10^{-34} \ Js$ and $m_e=9.1\times 10^{-31} \ kg$ and $e=1.6\times 10^{-19} \ C$

we get  $\lambda _{e }= \frac{12.27}{\sqrt{V}}A^{\circ}$   ( i.e answer will be in $A^0=Angstrom$

Similarly, we can find De - Broglie wavelength associated with charged particle

De - Broglie wavelength with charged particle-

$\lambda = \frac{h}{\sqrt{2mK}}= \frac{h}{\sqrt{2mqV}}$

Where $K\rightarrow kinetic\: energy\: o\! f particle$

$q\rightarrow charged \: particle$

$V\rightarrow potential \: diffenence$

• De - Broglie wavelength of the proton

using $m_p=1.67\times 10^{-27} \ kg$ and $q_p=e=1.6\times 10^{-19} \ C$

we get $\lambda _{proton}= \frac{0.286}{\sqrt{V}}A^{\circ}$

• De - Broglie wavelength of Deuteron

using $m_D=2\times 1.67\times 10^{-27} \ kg$ and $q_D=e=1.6\times 10^{-19} \ C$

we get $\lambda _{deutron}= \frac{0.202}{\sqrt{V}}A^{\circ}$

•  De - Broglie wavelength of an Alpha particle  (He2+)

using $m_ {\alpha ^{2+}}=4\times 1.67\times 10^{-27} \ kg$ and $q_ {\alpha ^{2+}}=2e=2\times 1.6\times 10^{-19} \ C$

we get $\lambda _{\alpha -partical}= \frac{0.101}{\sqrt{V}}A^{\circ}$

• Electron microscope-

An electron microscope is an important application of de-Broglie waves designed to study very minute objects like viruses,
microbes and the crystal structure of the solids. In the electron microscope, by selecting a suitable value of potential difference V ,we can have an electron beam of as small wavelength as desired. And this de-Broglie wavelength is calculated by using the formula $\lambda _{e }= \frac{12.27}{\sqrt{V}}A^{\circ}$.

## Study it with Videos

Wave nature of matter
De-broglie wavelength of an electron

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