Physics Topics can be both theoretical and experimental, with scientists using a range of tools and techniques to understand the phenomena they investigate.
What do you Mean by Wave Function?
Davisson and Germer Experiment: us physicists C J Davisson and L H Germer together first established the reality of matter waves. The experiment conducted by them in the year 1927 is described below.
The experiment: The electrons evaporated from the heated filament (F) after coming out of the electron gun are passed through a potential difference V. Thus, they acquire a high velocity and hence an increased kinetic energy. These electrons are then directed through a narrow hole and are thus collimated into a unidirectional beam. The kinetic energy of each single electron of this beam is eV. Arrangements are such that this beam impinges normally on a specially hewn nickel single crystal [Fig.]. Because of this impact, the electrons may be scattered in different directions through different angles ranging from 10° to 90°. The incident and scattered beams are generally referred to as the incident ray and scattered ray. A collector D capable to rotate round the crystal measures the intensity of the rays scattered in different directions. What is actually measured in the process is the number of electrons collected per second.
Observation : Intensity I obtained for the different values of the angle of scattering θ of the scattered ray is expressed by means of a polar graph where the scattered intensity is plotted as a function of the scattering angle. The convention in this respect is this :
1. The point of incidence on the crystal is taken as the origin O of the graph;
2. the direction OP opposite the direction of the incident ray PO is taken as the standard axis of the graph [Fig.]. Suppose, for a stream of particles with a definite energy, which is incident on the crystal, the scattering angles are θ1, θ2, ….. and the corresponding intensities I1, I2,……… . Hence, the line joining the points represented by the polar coordinates (I1, θ1), (I2, θ2), …, etc. will be the polar graph indicating the result of the experiment. Obviously, the polar coordinates of the points A and B in the figure are (I1, θ1) and (I2, θ2); that is, I1 = OA, I2 = OB, and θ1 = ∠AOP, θ2 = ∠BOP. In this figure, the line OABQ joining the points A, B … is the polar graph.
In this experiment, if the kinetic energy of the incident electrons is considerably low [as for the 40 eV electrons in [Fig.(a)], then it is observed that the intensity I corresponding to changes in the angle of scattering from 10° to 90° keeps decreasing continuously, which does not indicate any characteristic property of the electron beam.
After this, Davisson and Germer noticed that when the kinetic energy of the incident electrons is gradually increased, a distinct hump appears at 44 eV at a scattering angle of about 60° [Fig.(b)]. The hump becomes most prominent at 54 eV and scattering angle 50° [Fig.(c)]. At still higher potentials, the hump decreases until it disappears completely at 68 eV [Fig.(e)).
If the nickel crystal is compared with a diffraction grating, it is observed that if X-ray of wavelength 1.65Å is incident normally on the nickel crystal instead of the electron beam, then also, a peak representing maximum intensity will be obtained at 50° angle of scattering. On the other hand, when de Broglie formula for matter wave is applied, the de Broglie wavelength of 54 eV electrons is λ = \(\frac{12.27}{\sqrt{54}}\) = 1.67Å.
This striking similarity of the experimental value to the theoretical value according to de Broglie relation clearly demonstrates the behaviour of the electron stream as matter wave. The experiment is termed electron diffraction experiment as well.
1. The atoms of pure solids are arranged in crystal lattices of definite shapes. If an infinitely large number, say, 1020 or more, of such cyrstals are aligned in a three-dimensional array to form a large piece of matter, then the specimen is called a single crystal. For example, the cubical grains of common salt thus formed are each a single crystal.
2. Davisson and Germer used for their experiment comparatively slow-moving electrons with energy varying between 30 eV and 60 eV. Later on, G.P. Thomson successfully conducted diffraction experiment with even high-energy electron beams; electrons of energy as high as 10 key to 50 keV. In the subsequent years, the wave nature of other elementary particles, like the proton and the neutron, has also been established experimentally.
The nature of matter wave: According to de Broglie’s hypothesis, every moving particle can be represented as a matter wave, which, obviously therefore, has to obey certain conditions:
- Corresponding to a moving particle, a matter wave must also be a moving wave or progressive wave.
- The wave velocity must be equal to the particle velocity.
- Because the particle has a definite position at any time, the matter wave must be such that it can indicate the position of the particle at that time.
A pure sine curve cannot represent a matter wave, which means that no proper idea can be formed about the true nature of the matter wave from such curves. Let us suppose that a progressive wave moving in the positive x-direction is given by ψ = a sin(ωt – kx + δ) where a = amplitude, ω = angular velocity, k = propagation constant, and δ = phase difference [Fig.].
i) This wave extends from x = -∞ to + ∞ with no dissipation or damping anywhere along the path. Hence, it can never indicate the instantaneous position of the moving particle.
ii) is assumed however that the matter wave is analogous to the pure sine wave, then it can be shown that the velocity of the de Broglie wave turns out to be greater than the speed of light in vacuum, which is impossible contradicting Einstein’s special theory of relativity.
Moreover, a little deliberation will convince that the pure sine wave does not exist in nature. No real wave can possibly extend from -∞ to +∞ without any damping. This implies that what we observe in reality is a wave group. For a practical example, consider the shape of waves formed when a stone is dropped in a pond. A wave group as shown in Fig., comprising just a couple of wave crests and wave troughs is formed in the water and proceeds in circles over the water surface.
Wave group or wave pocket: If there is superposition of two or more sine waves of different frequencies, then the waveform undergoes a change. If many such sine waves of continuously varying frequencies are superposed, the resultant wave that is formed has a general form like the one shown in Fig.
This is what is called wave group or wave packet. Characteristics of wave packet are-
i) This too is a progressive wave moving in a definite direction, in this case, along the +x -axis.
ii) It is a localised wave, which means that it is limited within a rather small interval. Hence, such a wave group can indicate the possible instantaneous location of a moving particle.
iii) It can be shown analytically that the velocity of such wave groups, vg = \(\frac{d \omega}{d k}\). It is called group velocity. Rigorous calculation shows that vg = v; that is, the group velocity is equal to the particle velocity. Thus properly constituted wave group or wave packet alone can correctly represent matter wave.
Wave function: It has to be noted with particular care that matter wave is neither an elastic wave like the sound wave nor an electromagnetic wave like light wave. This is because an elastic wave is associated with the vibration of particles in an elastic medium, whereas an electromagnetic wave involves the simultaneous vibrations of the electric field and the magnetic field vectors.
In analogy, a matter wave is associated with a quantity known as wave function ψ. The origin and propagation of the matter wave are perfectly consistent with the vibrations of this ψ-function with respect to position and time.
It can be noticed that while the moving particle exists at a particular point at a particular moment of its motion, the corresponding matter wave occupies an extent of space at that moment, rather than be limited to a point [Fig.]. It can be inferred from quantum mechanics that this property is an inherent property of matter waves; a wave packet is never concentrated at a single point. The probability of the particle existing at a particular point at a particular time is given by ψ2, the modulus squared of ψ.
Photon wave: We have seen, even before discussing wave particle duality, that electromagnetic radiation has dual nature too; radiation is sometimes represented by waves, at other times in terms of photon beams. Obviously, it is possible also to represent moving photons by a corresponding waveform as is done with a moving particle. A single photon will naturally be represented by a wave group as in Fig. The only difference here is that this (photon) wave packet must be an electro-magnetic wave packet with group velocity in vacuum or air equal to the velocity of light.
Numerical Examples
Example 1.
Find de Broglie wavelength of neutron at 127°C. Given, Boltzmann constant, k = 1.38 × 10-23 J ᐧ mol-1 ᐧ K-1; Planck’s constant, h = 6.626 × 10-34 J ᐧ s; mass of neutron, m = 1.66 × 10-27 kgᐧ
Solution:
Kinetic energy of neutron, E = \(\frac{3}{2}\)kT
We know, E = \(\frac{1}{2}\)mv2 or, 2Em = m2v2
or, mv = \(\sqrt{2 E m}\) = \(\sqrt{2 m \times \frac{3}{2} k T}\) = \(\sqrt{3 m k T}\)
∴ de Broglie wavelength,
λ = \(\frac{h}{m v}\) = \(\frac{h}{\sqrt{3 m k T}}\)
= \(\frac{6.626 \times 10^{-34}}{\sqrt{3 \times 1.66 \times 10^{-27} \times 1.38 \times 10^{-23} \times 400}}\)
= 1.264 × 10-10 m
Example 2.
Under what potential difference should an electron be accelerated to obtain electron waves of λ = 0.6 Å ? Given, mass of electron, m = 9.1 × 10-31 kg; Planck’s constant, h = 6.62 × 10-34 J ᐧ s.
Solution:
We know, λ = \(\frac{h}{m v}\) ; \(\frac{1}{2}\)mv2 = eV
∴ mv = \(\sqrt{2 m e V}\) or, λ = \(\frac{h}{\sqrt{2 m e V}}\)
or λ2 = \(\frac{h^2}{2 m e V}\) or, V = \(\left(\frac{h}{\lambda}\right)^2 \cdot \frac{1}{2 m e}\)
∴ V = \(\left(\frac{6.62 \times 10^{-34}}{0.6 \times 10^{-10}}\right)^2\) × \(\frac{1}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19}}\)
= 418.04 V
Example 3.
An α-particle and a proton are accelerated from rest through the same potential difference V. Find the ratio of de Broglie wavelengths associated with them.
Solution:
The kinetic energy of a particle of mass m,
E = \(\frac{p^2}{2 m}\), where p = \(\sqrt{2 m E}\)
So, de Broglie wavelength, λ = \(\frac{h}{p}\) = \(\frac{h}{\sqrt{2 m E}}\)
Now, if a particle of charge q is accelerated by applying potential difference V, then
E = qV or, λ = \(\frac{h}{\sqrt{2 m q V}}\)
∴ λ ∝ \(\frac{1}{\sqrt{m q}}\) where V is constant
Hence, \(\frac{\lambda_1}{\lambda_2}\) = \(\sqrt{\frac{m_2 \cdot q_2}{m_1 \cdot q_1}}\)
For proton and α-particle, \(\frac{m_\alpha}{m_p}\) = 4, \(\frac{q_\alpha}{q_p}\) = 2
∴ \(\frac{\lambda_p}{\lambda_\alpha}\) = \(\sqrt{\frac{m_\alpha}{m_p} \cdot \frac{q_\alpha}{q_p}}\) = \(\sqrt{4 \times 2}\) = 2\(\sqrt{2}\)
Hence, the required ratio is 2\(\sqrt{2}\) : 1
Example 4.
For what kinetic energy of neutron will the associated de Broglie wavelength be 1.40 × 10-10 m ? The mass of neutron is 1.675 × 10-27 kg and h = 6.63 × 10-34 J ᐧ s.
Solution:
If m be the mass and K be the kinetic energy of neutron, the de Broglie wavelength associated with it is given by,
λ = \(\frac{h}{\sqrt{2 m K}}\)
or, K = \(\frac{h^2}{2 m \lambda^2}\) = \(\frac{\left(6.63 \times 10^{-34}\right)^2}{2 \times 1.675 \times 10^{-27} \times\left(1.40 \times 10^{-10}\right)^2}\)
= 6.69 × 10-21 J
Example 5.
Find the wavelength of an electron having kinetic energy 10 eV.(h = 6.33 × 10-34J ᐧ s, me = 9 × 10-31 kg) [WBCHSE Sample Question]
Solution:
Kinetic energy of the electron,
E = \(\frac{1}{2}\)mv2 = 10eV = 10 × (1.6 × 10-19)J
∴ m2v2 = 2mE, or, momentum p = mv = \(\sqrt{2 m E}\)
∴ The de Broglie wavelength of the electron,
λ = \(\frac{h}{\sqrt{2 m E}}\) = \(\frac{6.63 \times 10^{-34}}{\left[2 \times\left(9 \times 10^{-31}\right) \times\left(10 \times 1.6 \times 10^{-19}\right)\right]^{1 / 2}}\)
= 3.9 × 10-10m = 3.9 Å
Example 6.
An α-particle moves in a circular path of radius 0.83 cm in the presence of a magnetic field of 0.25 Wb/m2. What is the de Broglie wavelength associated with the particle? [AIPMT ‘12]
Solution:
The radius of a charged particle rotating in a circular path in a magnetic field,
R = \(\frac{m v}{B q}\) or, mv = RBq
The de Broglie wavelength associated with the particle,
λ = \(\frac{h}{m v}\) = \(\frac{h}{R B q}\)
Here, R = 0.83 cm = 0.83 × 10-2 m, B = 0.25 Wb ᐧ m-2, q = 2e = 2 × 1.6 × 10-19C [∵ α-particle]
∴ λ = \(\frac{6.6 \times 10^{-34}}{0.83 \times 10^{-2} \times 0.25 \times 2 \times 1.6 \times 10^{-19}}\) = 0.01 Å