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Introduction

In chapters [link] and [link] the so-called wave-particle duality of light is described. This duality states that light displays properties of both waves and of particles, depending on the experiment performed. For example, interference and diffraction of light are properties of its wave nature, while the photoelectric effect is a property of its particle nature. In fact we call a particle of light a photon.

Hopefully you have realised that nature loves symmetry. So, if light which was originally believed to be a wave also has a particle nature, then perhaps particles, also display a wave nature. In other words matter which we originally thought of as particles may also display a wave-particle duality.

De broglie wavelength

Einstein showed that for a photon, its momentum, p , is equal to its energy, E divided by the speed of light, c :

p = E c .

The energy of the photon can also be expressed in terms of the wavelength of the light, λ :

E = h c λ ,

where h is Planck's constant. Combining these two equations we find that the the momentum of the photon is related to its wavelength

p = h c c λ = h λ ,

or equivalently

λ = h p .

In 1923, Louis de Broglie proposed that this equation not only holds for photons, but also holds for particles of matter. This is known as the de Broglie hypothesis.

De Broglie Hypothesis

A particle of mass m moving with velocity v has a wavelength λ related to is momentum p = m v by

λ = h p = h m v

This wavelength, λ , is known as the de Broglie wavelength of the particle (where h is Planck's constant).

Since the value of Planck's constant is incredibly small h = 6 . 63 × 10 - 34 J · s , the wavelike nature of everyday objects is not really observable.

Interesting fact

The de Broglie hypothesis was proposed by French physicist Louis de Broglie (15 August 1892 – 19 March 1987) in 1923 in his PhD thesis. He was awarded the Nobel Prize for Physics in 1929 for this work, which made him the first person to receive a Nobel Prize on a PhD thesis.

A cricket ball has a mass of 0 , 150 kg and is bowled towards a bowler at 40 m · s - 1 . Calculate the de Broglie wavelength of the cricket ball?

  1. We are required to calculate the de Broglie wavelength of a cricket ball given its mass and speed. We can do this by using:

    λ = h m v
  2. We are given:

    • The mass of the cricket ball m = 0 , 150 kg
    • The velocity of the cricket ball v = 40 m · s - 1

    and we know:

    • Planck's constant h = 6 , 63 × 10 - 34 J · s
  3. λ = h m v = 6 , 63 × 10 - 34 J · s ( 0 , 150 kg ) ( 40 m · s - 1 ) = 1 , 11 × 10 - 34 m
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This wavelength is considerably smaller than the diameter of a proton which is approximately 10 - 15 m . Hence the wave-like properties of this cricket ball are too small to be observed.

Calculate the de Broglie wavelength of an electron moving at 40 m · s - 1 .

  1. We are required to calculate the de Broglie wavelength of an electron given its speed. We can do this by using:

    λ = h m v
  2. We are given:

    • The velocity of the electron v = 40 m · s - 1

    and we know:

    • The mass of the electron m e = 9 , 11 × 10 - 31 kg
    • Planck's constant h = 6 , 63 × 10 - 34 J · s
  3. λ = h m v = 6 , 63 × 10 - 34 J · s ( 9 , 11 × 10 - 31 kg ) ( 40 m · s - 1 ) = 1 , 82 × 10 - 5 m = 0 , 0182 mm
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Although the electron and cricket ball in the two previous examples are travelling at the same velocity the de Broglie wavelength of the electron is much larger than that of the cricket ball. This is because the wavelength is inversely proportional to the mass of the particle.

Calculate the de Broglie wavelength of a electron moving at 3 × 10 5 m · s - 1 . ( 1 1000 of the speed of light.)

  1. We are required to calculate the de Broglie wavelength of an electron given its speed. We can do this by using:

    λ = h m v
  2. We are given:

    • The velocity of the electron v = 3 × 10 5 m · s - 1

    and we know:

    • The mass of the electron m = 9 , 11 × 10 - 31 kg
    • Planck's constant h = 6 , 63 × 10 - 34 J · s
  3. λ = h m v = 6 , 63 × 10 - 34 J · s ( 9 , 11 × 10 - 31 kg ) ( 3 × 10 5 m · s - 1 ) = 2 , 43 × 10 - 9 m

    This is the size of an atom. For this reason, electrons moving at high velocities can be used to “probe" the structure of atoms. This is discussed in more detail at the end of this chapter. [link] compares the wavelengths of fast moving electrons to the wavelengths of visible light.

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Since the de Broglie wavelength of a particle is inversely proportional to its velocity, the wavelength decreases as the velocity increases. This is confirmed in the last two examples with the electrons. De Broglie's hypothesis was confirmed by Davisson and Germer in 1927 when they observed a beam of electrons being diffracted off a nickel surface. The diffraction means that the moving electrons have a wave nature. They were also able to determine the wavelength of the electrons from the diffraction. To measure a wavelength one needs two or more diffracting centres such as pinholes, slits or atoms. For diffraction to occur the centres must be separated by a distance about the same size as the wavelength. Theoretically, all objects, not just sub-atomic particles, exhibit wave properties according to the de Broglie hypothesis.

The wavelengths of the fast electrons are much smaller than that of visible light.

Questions & Answers

How to calculate velocity
Ibanathi Reply
distance/time
AJITHKUMAR
v=s/t units is metres per second/km per hour
Zodwa
v=displacement/change in time= v=Xf-Xi/t
LUTHO
what happen when you add water on an ester
Rose Reply
 present to produce ethanoic acid and ethano
Renier
whats a functional isomer
Andile
methypropanone
Teekayairs
what's a momentum
Teekayairs Reply
Momentum is the product of mass and velocity P=m×v
Thato
thanks
Teekayairs
this is all about physics only or chemistry can be asked?
Teekayairs
Physics only
Thato
how to calculate the change in momentum of a rocket is initially for 250s at rest with a mass of 2.55×10^5kg of which 1.8×10^5 is fuel.then the rocket engine is fired up for 250s during which 1.2×10^5kg of fuel is consumed.the speed of a rocket after firing is 2.78×10^9 in a forward direction
Teekayairs
wich equation should I use if I'm given the momentum and the change in time..
Alakhe Reply
how to do IUPAC naming ?
Nomsakk Reply
u must first master the functional group of your parent chain
Vunene
how to calculate a single resistor
TEBATSO Reply
one function of diode
TEBATSO Reply
to allow current to flow through one direction
Zodwa
one function of variable resistor
TEBATSO
how to calculate speed
Mahlatsi Reply
Distance /time
Loretta
what's the Conversation about?
Katlego Reply
what is butane
Spirit Reply
butane is a compound that have four carbons
Banele
what about propanoate how many carbon atoms does it have?
Lefa
it has got 3 carbons if I'm not mistaken
Mazwi
yes prop means 3
Katlego
calculatw the kinetic energy of a 500kg car travelling at 7km/h
Thabelo Reply
you first convert km/ to velocity then calculate E=½mv²
Linky
it's Ek= ½×m×v²
Linky
E(kinetic)=1/2mv²
Enow
convert Km/h to m/s for velocity
Enow
what are functional groups
Pheladi Reply
structural formula of the monomer used to form polyethene through addition
Udith Reply
what is the functional group of alcohol
Nduduzo Reply
hydroxyl Group
Tshego
what are esters
Andile
they are unsaturated organic compounds with a carbonyl and an O atom bonded to C atom
Ronewa
is when alcohol react with carboxyl group
Ravele
esters is formed when an alcohol react with a carboxyl acid to bring the process of estirification
Thami
Hi could you please explain to me on condensation polymerization
Zenande
Condensation polymers are any kind of polymers formed through a condensation reaction—where molecules join together—losing small molecules as byproducts such as water or methanol. ... Linear polymers are produced from bifunctional monomers, i.e. compounds with two reactive end groups
Mabaya
Ok thanks I think I get it
Zenande
what are hydrocarbon
Mlondi Reply
carbonic compounds containing carbon and hydrogen ONLY
Kamva

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Source:  OpenStax, Siyavula textbooks: grade 12 physical science. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11244/1.2
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