Matter waves, de-Broglie wavelength, physical interpretation, practice problems, FAQs

Do you know that the ball you usually use to play cricket also has the wave-like characteristics (ex-wavelength)? Although the hypothesis was given by Louis Victor de-Broglie, it was confirmed by the famous Davisson and Germer experiment that particles of matter also behave like waves.

Table of contents

Matter waves
de-Broglie wavelength of an electron
Physical interpretation of matter waves
Practice problems
FAQs

Matter waves

In 1924, the French physicist Louis Victor de-Broglie put forward the bold hypothesis that moving particles of matter should display wave-like properties under suitable conditions.
Matter waves or de-Broglie waves: The waves associated with moving particles.
de-Broglie wavelength is the wavelength which is associated with a moving particle.

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In ordinary situations, the de-Broglie wavelength is very small and the wave nature of matter can be ignored. For example: for a baseball of mass 46 g moving at 30 m/s, if you will calculate the de-Broglie wavelength, it comes out to be 4.810-34 m which is too small to be observed.
For macroscopic objects wave characteristics can not be observed.
For microscopic particles wave characteristics can be observed.

de-Broglie wavelength

de-Broglie proposed that the wavelength associated with a particle of matter having momentum p can be related as:

$λ = \frac{h}{p} = \frac{h}{m v}$ where p= momentum

$K E = \frac{p^{2}}{2 m} \Rightarrow p = \sqrt{2 m K E}$

$λ = \frac{h}{\sqrt{2 m K E}}$

When charge 𝑞 accelerated through a potential difference 𝑉 from rest.

Work done, 𝑊=𝐾.𝐸.=𝑞𝑉

$λ = \frac{h}{\sqrt{2 m q V}}$

de-Broglie wavelength of an electron

Mass of an electron, m=9.110-31 kg

Charge on an electron, q=1.610-19 C

$λ = \frac{h}{\sqrt{2 m q V}} = \frac{6.64 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 1.6 \times 10^{- 19} V}}$

$\Rightarrow λ = \frac{12.27}{\sqrt{V}} Å$

Physical interpretation of matter waves

The probability of finding a particle in space is represented by matter waves.
The intensity (square of the amplitude) of the matter wave at a point determines the probability density (probability per unit volume) of the particle at that point

Probability of finding a particle in space, P=IV

Where I=Intensity of matter wave

V= small volume

Practice problems

Q. A proton and an alpha particle are accelerating by the same potential difference. Find the ratio of their de-Broglie wavelength?

$(c h a r g e q_{α} = + 2 e, q_{p r o t o n} = + e, m_{α} = 4 m_{p r o t o n})$

A. Since the proton and alpha particles are accelerated by the same potential difference therefore, their KE will be

K.E = qV

$\Rightarrow \frac{p^{2}}{2 m} = q V$

$\Rightarrow p = \sqrt{2 m K E}$ (Since V is same for both)

$\frac{p_{P}}{p_{α}} = \sqrt{\frac{q_{P} m_{P}}{q_{α} m_{α}}}$

de-Broglie wavelength is given by,

$λ = \frac{h}{p}$

$\Rightarrow λ \propto \frac{1}{p}$

$\frac{λ_{P}}{λ_{α}} = \frac{p_{α}}{p_{P}} = \sqrt{\frac{q_{α} m_{α}}{q_{P} m_{P}}}$

$S i n c e q_{α} = + 2 e, q_{p r o t o n} = + e, m_{α} = 4 m_{p r o t o n}$

$∴ \frac{λ_{P}}{λ_{α}} = \sqrt{\frac{+ 2 e \times 4 m_{P}}{e \times m_{P}}} = \sqrt{8}$

Q. Ultraviolet light of wavelength 99 nm falls on a metal plate of work function 1 eV Find the wavelength of the fastest photoelectron emitted.melectron=9.110-31 kg h=6.6410-34 Js.

A. Given, =99 nm o=1.0 eV

From, Einstein's photoelectric equation,

$K E_{m a x} = \frac{h c}{λ} - ϕ_{o}$

$= \frac{6.64 \times 10^{- 34} \times 3 \times 10^{8}}{99 \times 10^{- 9}} - 1.0 \times 1.6 \times 10^{- 19}$

$K E_{m a x} = 18.4 \times 10^{- 19} J$

de-Broglie wavelength of the particles will be,

$λ = \frac{h}{\sqrt{2 m K E}}$

$= \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 18.4 \times 10^{- 19}}}$

𝜆=0.36 nm

Q. If one wishes to see inside the atom of diameter 100 pm, one must be able to resolve to a width of say 10 pm. If an electron-microscope is used, find the minimum electron energy required. (Assume the wavelength of light used in an electron microscope is nearly equal to the resolving power of the electron microscope.)

A. The wavelength of light used in the electron microscope is almost equal to the resolving power of the electron-microscope.

=1010-12 m

Hence, from the de-Broglie wavelength,

$λ = \frac{h}{p} = \frac{h}{m v}$

$\Rightarrow v = \frac{h}{λ m} = \frac{6.6 \times 10^{- 34}}{1 \times 10^{- 11} \times 9.1 \times 10^{- 31}}$

v=7.25107 m/s

Minimum electron energy,

$K E = \frac{1}{2} m v^{2} = \frac{9.1 \times 10^{- 31} \times {(7.25 \times 10^{7})}^{2}}{2 \times 1.6 \times 10^{- 19}} = 15 k e V$

Q. A proton, when accelerated through a potential difference of V Volts has a wavelength associated with it. If an - particle is to have the same wavelength it must be accelerated through a potential difference of V/n Volts. Find the value of n.

A. Given, p==

The de-Broglie wavelength associated with a particle of charge q and mass m accelerated through a potential difference V is given by,

$λ = \frac{h}{\sqrt{2 m q V}}$

$V = \frac{h^{2}}{2 m q λ^{2}}$ [Since is same for both]

$\Rightarrow V \propto \frac{1}{m q}$

$∵ m_{α} = 4 m_{p} a n d q_{α} = 2 q_{p}$

$∴ \frac{V_{α}}{V_{p}} = \frac{m_{p}}{m_{α}} \times \frac{q_{p}}{q_{α}} = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$

$∴ V_{α} = \frac{V}{8}$

Hence, n=8

FAQs

Q. When the de-Broglie wavelength of neutron and electron are same, what will be their corresponding momenta?

A neutron has higher momentum than an electron.
They both have equal momentum
Electrons have higher momentum than neutrons.
Information is insufficient to answer the question.

The de-Broglie wavelength is given by,

$λ = \frac{h}{p}$

Here, h is constant.

As $λ_{1} = λ_{2} \Rightarrow p_{1} = p_{2}$

So, the momentum of the neutron and electron will be the same.

Q. An electron, an alpha particle, a proton and a neutron have the same de-Broglie wavelength. Give order of their kinetic energies.
A. Since 𝜆 is same for all particles, they will also have the same momentum.

$p = \frac{h}{λ}$

Now, Kinetic Energy is given by, $K E = \frac{p^{2}}{2 m}$

As p is same for all, therefore KE will be inversely proportional to mass of the particle.

The order of the masses of particles is

$m_{α} > m_{n} > m_{p} > m_{e}$

Their KE will be in the following order.

$K E_{e} > K E_{p} > K E_{n} > K E_{α}$

Q. Which of the following options is correct with regard to the validation of the equation E= pc ?

Valid for both an electron and a photon.
Valid for an electron but not valid for a photon.
Valid for a photon but not valid for an electron.
Not valid for both electron and photon.

A. (c)

Energy of a photon,

$E = h ν = \frac{h c}{λ} . . . . . . (1)$

de-Broglie wavelength is given by,

$λ = \frac{h}{p} \Rightarrow p = \frac{h}{λ}$

Putting this in (1) we get, the energy of a photon as E=pc

The equation E=pc is valid only for a particle with zero rest mass. The rest mass of a photon is zero, but the rest mass of an electron is 9.110-31 kg.

Hence, the equation E=pc is valid for a photon but not for an electron.

Q. Can we use the de-Broglie hypothesis for macroscopic matter?
A. Since the wavelength associated with the moving macroscopic particles is too small to be detected, therefore its wave nature is not observable. Hence although we can use the hypothesis, the de-Broglie wavelength associated with it is too small.