Mathematics WAVE NATURE OF MATTER AND DAVISSON AND GERMER EXPERIMENT

### Topic Covered

color{blue}{star} WAVE NATURE OF MATTER
color{blue}{star} DAVISSON AND GERMER EXPERIMENT

### WAVE NATURE OF MATTER

color{blue} ✍️The dual (wave-particle) nature of light (electromagnetic radiation, in general) comes out clearly from what we have learnt in this and the preceding chapters.

color{blue} ✍️The wave nature of light shows up in the phenomena of interference, diffraction and polarisation. On the other hand, in photoelectric effect and Compton effect which involve energy and momentum transfer, radiation behaves as if it is made up of a bunch of particles – the photons.

color{blue} ✍️Whether a particle or wave description is best suited for understanding an experiment depends on the nature of the experiment. For example, in the familiar phenomenon of seeing an object by our eye, both descriptions are important.

color{blue} ✍️The gathering and focussing mechanism of light by the eye-lens is well described in the wave picture. But its absorption by the rods and cones (of the retina) requires the photon picture of light.

color{blue} ✍️A natural question arises: If radiation has a dual (wave-particle) nature, might not the particles of nature (the electrons, protons, etc.) also exhibit wave-like character? In 1924, the French physicist Louis Victor de Broglie (pronounced as de Broy) put forward the bold hypothesis that moving particles of matter should display wave-like properties under suitable conditions.

color{blue} ✍️He reasoned that nature was symmetrical and that the two basic physical entities – matter and energy, must have symmetrical character. If radiation shows dual aspects, so should matter. De Broglie proposed that the wave length l associated with a particle of momentum p is given as

color{blue}(lamda= h/p = h/(mv))

....................... {11.5}

color{blue} ✍️where m is the mass of the particle and v its speed. Equation (11.5) is known as the de Broglie relation and the wavelength l of the matter wave is called de Broglie wavelength.

color{blue} ✍️The dual aspect of matter is evident in the de Broglie relation. On the left hand side of Eq. (11.5), l is the attribute of a wave while on the right hand side the momentum p is a typical attribute of a particle. Planck’s constant h relates the two attributes.

color{blue} ✍️Equation (11.5) for a material particle is basically a hypothesis whose validity can be tested only by experiment. However, it is interesting to see that it is satisfied also by a photon. For a photon, as we have seen,

color{blue}(p = hnu //c)

........................ (11.6)
Therefore,

color{blue}(h/p = c/v = lambda)

................... (11.7)

color{blue} ✍️That is, the de Broglie wavelength of a photon given by Eq. (11.5) equals the wavelength of electromagnetic radiation of which the photon is a quantum of energy and momentum.

color{blue} ✍️Clearly, from Eq. (11.5 ), lamda is smaller for a heavier particle ( large m) or more energetic particle (large v). For example, the de Broglie wavelength of a ball of mass 0.12 kg moving with a speed of 20 m s–1 is easily calculated:

color{purple}(p = m v = 0.12 kg × 20 m s^(–1) = 2.40 kg m s^(–1))

color{purple}(lambda = h/p = (6.63xx10^(-34)Js)/(2.40kg ms^(-1)) = 2.76 xx 10^(-34)m)

color{blue} ✍️This wavelength is so small that it is beyond any measurement. This is the reason why macroscopic objects in our daily life do not show wave-like properties.

color{blue} ✍️On the other hand, in the sub-atomic domain, the wave character of particles is significant and measurable. Consider an electron (mass m, charge e) accelerated from rest through a potential V. The kinetic energy K of the electron equals the work done (eV ) on it by the electric field:

color{blue}(K = e V)

...........(11.8)

color{blue} ✍️Now, color{purple}(K = 1/2 mv^2 = (p^2)/2m ,) so that

color{blue}(p = sqrt(2 mK) = sqrt(2 m eV)

.....(11.9)

color{blue} ✍️The de Broglie wavelength l of the electron is then

color{blue}(lamda= h/p = h/sqrt(2 mK) = h/(2 meV))

..... (11.10)

color{blue} ✍️Substituting the numerical values of h, m, e, we get

color{blue}(lamda = (1.227)/sqrt(V)nm)

...... (11.11)

color{blue} ✍️where V is the magnitude of accelerating potential in volts. For a 120 V accelerating potential, Eq. (11.11) gives lamda = 0.112 nm. This wavelength is of the same order as the spacing between the atomic planes in crystals. This suggests that matter waves associated with an electron could be verified by crystal diffraction experiments analogous to X-ray diffraction. We describe the experimental verification of the de Broglie hypothesis in the next section. In 1929, de Broglie was awarded the Nobel Prize in Physics
for his discovery of the wave nature of electrons.

color{blue} ✍️The matter–wave picture elegantly incorporated the Heisenberg’s uncertainty principle. According to the principle, it is not possible to measure both the position and momentum of an electron (or any other particle) at the same time exactly. There is always some uncertainty (D x) in the specification of position and some uncertainty (Dp) in the specification of momentum. The product of Deltax and Deltap is of the order of h* (with h= h//2p), i.e.,

color{blue}(Deltax Deltap approx h)

.......(11.12)

color{blue} ✍️Now, if an electron has a definite momentum p, (i.e. Deltap = 0), by the de Broglie relation, it has a definite wavelength l. A wave of definite (single)
wavelength extends all over space. By Born’s probability interpretation this means that the electron is not localised in any finite region of space. That is, its position uncertainty is infinite (Delta->infty ), which is consistent with the uncertainty principle

color{blue} ✍️In general, the matter wave associated with the electron is not extended all over space. It is a wave packet extending over some finite region of space. In that case Deltax is not infinite but has some finite value depending on the extension of the wave packet.

color{blue} ✍️Also, you must appreciate that a wave packet of finite extension does not have a single wavelength. It is built up of wavelengths spread around some central wavelength. By de Broglie’s relation, then, the momentum of the electron will also have a spread – an uncertainty Deltap.

color{blue} ✍️ This is as expected from the uncertainty principle. It can be shown that the wave packet description together with de Broglie relation and Born’s probability interpretation reproduce the Heisenberg’s uncertainty principle exactly.

In Chapter 12, the de Broglie relation will be seen to justify Bohr’s postulate on quantisation of angular momentum of electron in an color{blue} ✍️atom. Figure 11.6 shows a schematic diagram of (a) a localised wave packet, and (b) an extended wave with fixed wavelength.

Q 3149656513

What is the de Broglie wavelength associated with (a) an electron moving with a speed of 5.4×106 m/s, and (b) a ball of mass 150 g travelling at 30.0 m/s?
Class 12 Chapter 11 Example 4
Solution:

(a) For the electron:
Mass m = 9.11×10^(–31) kg, speed v = 5.4×10^6 m//s. Then, momentum
p = m v = 9.11×10^(–31) (kg) × 5.4 × 10^6 (m//s)
p = 4.92 × 10^(–24) kg m//s
de Broglie wavelength, lamda = h//p

= (6.63 xx 10^(-34) Js)/(4.92xx10^(-24) kg m//s)
lamda= 0.135 nm

(b) For the ball:
Mass m’ = 0.150 kg, speed v ’ = 30.0 m//s.
Then momentum p’ = m’ v’ = 0.150 (kg) × 30.0 (m//s)
p’= 4.50 kg m//s
de Broglie wavelength lamda = h//p’.

= (6.63 xx 10^(-34) Js)/(4.50xxkgm//s)
lamda = 1.47 xx10^(-34)m
The de Broglie wavelength of electron is comparable with X-ray wavelengths. However, for the ball it is about 10–19 times the size of the proton, quite beyond experimental measurement.
Q 3169656515

An electron, an a-particle, and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength?
Class 12 Chapter 11 Example 5
Solution:

For a particle, de Broglie wavelength, lamda = h//p
Kinetic energy, K = p^2//2m

Then, lamda=h // sqrt(2mK)

For the same kinetic energy K, the de Broglie wavelength associated with the particle is inversely proportional to the square root of their masses. A proton (text()_1^1H) is 1836 times massive than an electron and an alpha-particle (text()_4^2H) four times that of a proton.

Hence, alpha - particle has the shortest de Broglie wavelength.
Q 3129756611

A particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is 1.813 × 10^(–4). Calculate the particle’s mass and identify the particle.
Class 12 Chapter 11 Example 6
Solution:

de Broglie wavelength of a moving particle, having mass m and velocity v:

lamda = h/p = h/(mv)

Mass, m = lamda/lamdav
For an electron, mass m_e = h/lamda_e v_e
Now, we have v//v_e = 3 and

lamda //lamda_e 1.813 xx10^(-4)
Then, mass of the particle, m = m_e (lamda_e)/(lamda) (v_e)/v

m = (9.11×10^(–31) kg) × (1//3) × (1//1.813 × 10^(–4))
m = 1.675 × 10^(–27) kg.
Thus, the particle, with this mass could be a proton or a neutron.
Q 3169756615

What is the de Broglie wavelength associated with an electron, accelerated through a potential differnece of 100 volts?
Class 12 Chapter 11 Example 7
Solution:

Accelerating potential V = 100 V. The de Broglie wavelength lamda is

lamda = h//p = (1.227)/(sqrtV) nm

lamda= (1.227)/(sqrt100) nm = 0.123nm

The de Broglie wavelength associated with an electron in this case is of the order of X-ray wavelengths.

### DAVISSON AND GERMER EXPERIMENT

color{blue} ✍️The wave nature of electrons was first experimentally verified by C.J. Davisson,who observed diffraction effects with beams of electrons scattered by crystals.

color{blue} ✍️The experimental arrangement used by Davisson and Germer is schematically shown in Fig. 11.7.

color{blue} ✍️It consists of an electron gun which comprises of a tungsten filament F, coated with barium oxide and heated by a low voltage power supply (L.T. or battery).

color{blue} ✍️Electrons emitted by the filament are accelerated to a desired velocity by applying suitable potential/voltage from a high voltage power supply (H.T. or battery). They are made to pass through a cylinder with fine holes along its axis, producing a fine collimated beam.

color{blue} ✍️The beam is made to fall on the surface of a nickel crystal. The electrons are scattered in all directions by the atoms of the crystal. The intensity of the electron beam, scattered in a given direction, is measured by the electron detector (collector).

color{blue} ✍️The detector can be moved on a circular scale and is connected to a sensitive galvanometer, which records the current. The deflection of the galvanometer is proportional to the intensity of the electron beam entering the collector.

color{blue} ✍️The apparatus is enclosed in an evacuated chamber. By moving the detector on the circular scale at different positions, the intensity of the scattered electron beam is measured for different values of angle of scattering q which is the angle between the incident and the scattered electron beams. The variation of the intensity (I ) of the scattered electrons with the angle of scattering q is obtained for different accelerating voltages.

color{blue} ✍️The experiment was performed by varying the accelarating voltage from 44 V to 68 V. It was noticed that a strong peak appeared in the intensity (I ) of the scattered electron for an accelarating voltage of 54V at a scattering angle q = 50º

color{blue} ✍️The appearance of the peak in a particular direction is due to the constructive interference of electrons scattered from different layers of the regularly spaced atoms of the crystals. From the electron diffraction measurements, the wavelength of matter waves was found to be 0.165 nm.

color{blue} ✍️The de Broglie wavelength lamda associated with electrons, using Eq. (11.11), for V = 54 V is given by

color{purple}(lamda = h//p = (1.227)/sqrt(V)nm)
color{purple}(lamda = (1.227)/sqrt(54) nm = 0.167nm)

color{blue} ✍️Thus, there is an excellent agreement between the theoretical value and the experimentally obtained value of de Broglie wavelength. Davisson- Germer experiment thus strikingly confirms the wave nature of electrons and the de Broglie relation.

color{blue} ✍️More recently, in 1989, the wave nature of a beam of electrons was experimentally demonstrated in a double-slit experiment, similar to that used for the wave nature of light. Also, in an experiment in 1994, interference fringes were obtained with the beams of iodine molecules, which are about a million times more massive than electrons.

color{blue} ✍️The de Broglie hypothesis has been basic to the development of modern quantum mechanics. It has also led to the field of electron optics. The wave properties of electrons have been utilised in the design of electron microscope which is a great improvement, with higher resolution, over the optical microscope.