CK-12-Physics - Intermediate

23.3. Wave-Particle Duality http://www.ck12.org

Thus, De Broglie assumed that the wavelength of a matter wave would obey the same relationship as the wavelength
of light wave.

Substitutingp=mvinto Equation A, we have:

mv=hλ→λ=mvh

Since De Broglie’s supplied the above equation in his doctoral thesis, it was then a matter of experimentation to see
if his prediction was true.

It was suggested a year later that a crystalline solid may act as a diffraction grating for a beam of electrons. Thus,
if electrons had an associated wave, a diffraction pattern would emerge similar to a diffraction pattern that would be
produced by the light passing through the crystal. In fact, that was the genesis of the idea. It had already been shown
that x-rays passing through a crystal yielded diffraction patterns permitting physicists to measure the wavelength
of x-rays. De Broglie’s prediction for the wavelength of electron matter waves could be tested at wavelengths
comparable to those of x-rays. Such an experiment could confirm if De Broglie’s hypothesis was true.

The experiment was performed by C.J Davisson (1881-1958) and L.H. Germer (1896-1971) and fully confirmed
de Broglie’s hypothesis. It bears restating that De Broglie’s description is valid for all particles– whether they are
charged or uncharged, small or large, all material objects have associated matter waves. In 1929, De Broglie was
awarded the Nobel Prize in physics for his work.

Check your understanding

What is the associated wave for a baseball of mass 145gtraveling with velocity 40. 0 ms?

Solution:

λ=mvh =^6.^626 ×^10

− (^34) J−s
( 0. 145 kg)( 40. 0 ms)=^1.^14 ×^10
− (^34) m
Small wonder that the associated waves of macroscopic objects are not observable experimentally!
The fact that Planck’s constant is so tiny ensures that macroscopic events remain easily within the domain of
Newtonian (classical) physics.
Illustrative Example 23.3.1
What wavelength is associated with an electron beam of kinetic energy 95eV?
(The mass of an electron isme= 9. 11 × 1031 kg)
Solution:
We first must determine the velocity of the electrons:
95 eV= ( 95 eV)

(

1. 60 × 10 −^19

J

eV

)

= 1. 52 × 10 −^17 J→

KE=

1

2

mv^2 → 1. 52 × 10 −^17 J=

1

2

( 9. 11 × 10 −^31 kg)v^2 →

v=

√

2 ( 1. 52 × 10 −^17 )

9. 11 × 10 −^31 kg

= 5. 776 × 106

m

s

Thus, the wavelength is:

λ=mvh =^6.^626 ×^10

− (^34) J−s
( 9. 11 × 10 −^31 kg)( 5. 776 × 106 ms)=^1.^259 ×^10
− (^10) m= 0. 126 nm