Chapter 41:
Wave Mechanics:
In 1924, Louis de Broglie thought that if light has the dual property of waveparticle behavior, why not matter particles such as electrons, protons, and neutrons? He was encouraged by the work of Einstein in 1909 and used a combination of quantum theory and relativity to propose that there is a special wavelength associated with the motion (momentum) of each particle given by
In the Bohr theory of atom, the angular momentum of electron is quantized. This means that
Note that 2πr is the circumference of the orbit of electron around the nucleus and nλ is an integer multiple of wavelength.
This is in agreement with the result that each electron's orbit must be an integer multiple of a certain wavelength!
Electron Diffraction:
In 1926, C. L. Davisson and L. Gerner found out that electrons have also a wavelike behavior as they are bounced off an array of atoms similar to the reflection of X rays from the atomic planes of a crystal. They formed diffraction patterns.
On one hand, the de Broglie wavelength can be determined for an electron that is accelerated and is given speed v inside an electric field of voltage V. Such λ may be calculated as follows: For each electron of mass M and charge q inside a potential difference V, just before collision with a target atom, we may set its P.E. and K.E. equal. P.E. = K.E. qV = (1/2)Mv^{2 } 2MqV = M^{2}v^{2} or Mv = ( 2MqV )^{(1/2)} Since λ = h/Mv (1)

On the other hand, when electrons get reflected from the surface of a metal with an atomic spacing of D, they form diffraction patterns. The positions of diffraction maxima are given by:
D sin φ = n λ. (2)
For electrons accelerated under a 54V potential difference and incident on nickel surface with an atomic spacing of D = 0.215nm, the principal maximum corresponding to n=1 occurs at φ = 50°. Using the above formula (2) yields: λ = 0.165nm while Equation (1) results in λ = 0.167nm, a conclusively close result ! This is another verification of the wavelike motion of electrons.
Example 1: Calculate the de Broglie wavelength for an electron that is accelerated inside a potential difference of 150 volts.
Solution: Using the above formula, we get: λ = 0.1 nm. This is equivalent to 1Å, the interatomic spacing between atoms in a crystal.
Example 2: Calculate the de Broglie wavelength for an electron that is accelerated inside a potential difference of 54 volts.
Solution: Using the above formula, we get: λ = 0.167 nm.
Schrödinger's Wave Equation:
Schrödinger thought that the same way geometric optics is an approximation to wave optics, classical mechanics may be an approximation to wave mechanics. For simplicity, we will use the onedimensional wave equation for the motion of a wave as it moves the particles of the medium according to:
To verify de Broglie's way of thinking on electron motion around the nucleus of atom, let's consider the standing wave solution to the onedimensional wave equation. Note that the "standing wave solution" means "nontransient" that in turn means "timeindependent" or mathematically means "no change with respect to time", and ultimately it translates to (d_{/}dt = 0). It is also called the "steady state" solution. ψ(x) is therefore chosen to determine the shape of the steady state solution and it is a function of position (x) only.
y( x , t ) = ψ(x) sin(ωt).
Substituting this into wave equation (verify), results in:
If U is the potential energy, then the total energy E = p^{2}_{/}2M + U, we may conclude that p^{2} = 2M (E  U), and
and the differential equation becomes (verify):
This is the onedimensional timeindependent Schrödinger wave equation. ψ(x) determines the stationary states of an atomic system when E is a constant in time. When Schrödinger applied this equation to the hydrogen atom for which U = ke^{2}/r, he found that using appropriate boundary conditions naturally lead to discrete energy levels of the Bohr model. Heisenberg at about the same time (1925) developed a different form of quantum mechanics that later on showed equivalent results.
Now, the question is how to interpret the wave associated with the particle (electron in this case)? De Broglie suggested that the wave might represent the particle itself, or might play a guiding role in the particle's motion. Schrödinger suggested that a particle is really a group of waves, a wave packet, somewhat like a fuzzy powderpuff. Max Born guided by Einstein's idea proposed that the intensity of a light wave (that is proportional to the square of the waves amplitude) at a given point, is a measure of the number of photons that arrive at that point. This means that the form of a wave function for electromagnetic field determines the probability of finding a photon. By analogy, Born suggested that the square of the wave function determines the probability per unit volume of finding the particle written as:
ψ^{2}dV = Probability of finding the particle within volume dV
The quantity ψ^{2} is called the probability density. In one dimension, ψ^{2}dx is the probability of finding the particle in the interval between x and x+dx. It is more likely to observe the particle wherever ψ^{2}dx is large and less likely to observe it where ψ^{2}dx is small. Since the particle has to be somewhere the integral of its probability must be equal to one:
A wave function that satisfies this condition is said to be normalized.
The probability interpretation of the wave
function can be shown by a simple experiment. As shown in the middle
column, if a large number of electrons pass through a single slit via a
very narrow beam, say one electron at a time, after a while the majority
will be recorded to concentrate at the center and about the centerline. A
smaller fraction will accumulate fairly evenly at side regions as shown. Eventually after so many electrons go under diffraction at the slit, the familiar diffraction pattern will develop. This means that the probability that the diffracted electrons fall on the centerline is higher than the probability of them falling on the side lines; however, the sum of the probabilities must equal one. The quantity ψ^{2}(x) tells us what fraction of the total number of electrons will fall at position x. 
As another example, look at the patterns that are formed when electrons pass through two slits as shown on the right. As the number of electrons passing through the two slits increases the separation of the bright and dark fringes becomes more apparent.


Classical physics is based on the principle of determinism that means: Given the initial position and velocity of a particle as well as the forces acting on it, its future course can accurately be predicted, at least theoretically. The statistical interpretation of the wave function states that one can only predict the probability of finding a particle at a given position. It is not possible to exactly determine the position of a single particle at a given time. Quantum mechanics can correctly predict the average values of physical quantities, not the results of individual measurements.
Heisenberg Uncertainty Principle:
The wavelength of a wave that extends over many cycles is welldefined but its position is not. For example, if you see continuous ripples on a calm swimming pool, you will be able to measure the wavelength fairly accurately; however, you can not exactly measure the position of the moving wave. This is shown in Figure 1. Now, let's look at another scenario. When waves of different wavelengths interfere and add up, they can form a localized wave packet, but the wavelength is not welldefined. An example for this case is when two waves of different wavelengths interfere (waves generated by two moving boats on a calm lake), they generate waveforms that appear stationary or at least appearing at the same location periodically. In this example, the position of the wave packet can be determined, but not the wavelengths. This is shown in Figure 2. 

Based on the above, in order to reduce the uncertainty in the position of a particle, Δx, one can superpose many wavelengths to form a reasonably welllocalized wave packet, as shown in Fig. 2. From λ = h/p, it is clear that a spread in wavelengths (Δλ) results in a spread in momentum (Δp). According to Heisenberg Uncertainty Principle, the uncertainties in position and momentum are related by
Δx Δp ≥ h (_{*})
It is impossible to measure both the position and the momentum of a particle simultaneously to arbitrary precisions. This inability has nothing to do with experimental skill or equipment; it is a fundamental restriction imposed on us by nature. For a wave packet, the uncertainty relation is an intrinsic property independent of the measuring apparatus. Heisenberg arrived at this conclusion in 1927. Here is a simplified version:
Suppose we want to locate an electron by bouncing a photon off of it. We should not expect to measure the position with a precision better than the wavelength of the photon used. Thus the uncertainty in the position of the electron is at least Δx = λ or the precision in the position is at its best equal to λ. The photon may transfer some or all of its linear momentum to the electron depending on the collision and scattering angle. Thus the uncertainty in the momentum of the electron is roughly equal to the original momentum of the photon itself. Δp = h _{/}λ. The product of these uncertainties is
Δx Δp ≈ h.
This is the same as the equation we labeled as (_{*}). To reduce the uncertainty in position Δx, a shorter λ must be used. A shorter λ means a higher f or a higher photon energy hf that means a higher momentum p of the incident photon. A higher p of the incident photon means a greater spread or uncertainty in the momentum measurement of the electron, Δp. Thus, lowering the uncertainty in locating the position (Δx) of electron causes the uncertainty in its momentum measurement (Δp) to increase. The conclusion is that one can measure either the position or the momentum precisely but not both at the same time. In this example, the very act of measurement disturbs the system under study. One cannot speak of the system as an isolated entity. There is always an inevitable interaction between observer and what is being measured.
Test Yourself 1:
1) In the de Broglie formula λ = h _{/}p, p is (a) potential energy (b) kinetic energy (c) momentum. click here.
2) The de Broglie wavelength for an electron moving at 2.00x10^{5}m/s is (a) 3.64Å (b) 3.64nm (c) 1.06Å.
3) The angular momentum of a particle of mass M traveling at angular speed ω around a circle of radius R is (a) MRω (b) MR^{2}ω (c) Mω.
4) The angular momentum of a particle of mass M traveling at angular speed v around a circle of radius R is (a) MR^{2}v (b) MRv (c) Mv. click here.
5) According to Bohr's theory, the angular momentum of electron is quantized. This means that MRv (a) can have any value (b) is always an integer (c) is an integer multiple of h_{/}(2π).
6) The condition 2πR = nλ derived from Bohr's model and de Broglie's wavelength formulation means that (a) any circular path electron takes about the nucleus must be an integer multiple of a certain wavelength (b) the radii at which electron takes on circular wavy motion are discrete and not continuous (c) both a and b.
7) One verification of de Broglie's wavelength formulation is the (a) diffraction pattern formed on a sensitive film as accelerated electrons are bounced off a nickel surface (b) increase in the K.E. of bounced off electrons from a surface (c) decrease in the wavelength of electrons when they bounce off a metal surface.
8) When an electron is accelerated to speed v by imposing a voltage V on it, P.E. is converted to K.E. such that (a) we may set P.E.=K.E. (b) qV=0.5Mv^{2} (c) neither a nor b (d) both a and b click here.
9) Multiplying both sides of qV= 0.5Mv^{2} by mass M , we get (a) qVM=0.5M^{2}v^{2} ,only (b) (2qVM) = M^{2}v^{2} ,only (c) Mv=(2qVM)^{0.5} ,only (d) a, b, and c.
10) If p is the momentum of a particle, according to de Broglie (a) λ = h _{/}p (b) λ = h _{/}(Mv) (c) both a and b.
11) If λ = h _{/}(Mv) then (a) λ = h _{/}(Mv)^{2} as well (b) λ = h _{ /}(2qVM)^{0.5} as well (c) both a and b. click here.
12) In the formula λ = h _{/}(2qVM)^{0.5} (a) h, q, and M are always constant (b) at speeds less than (1/20)c we may treat all factors to be fairly constant (c) neither a nor b.
13) In the formula λ = h _{/}(2qVM)^{0.5} , if h, q, and M are constants, for a reasonable voltage V applied to an electron to accelerate it, the de Broglie wavelength (a) can be easily calculated (b) can be calculated if P.E. is given (c) both a and b. click here.
14) When electrons are reflected from the surface of a metal with an atomic spacing of D, diffraction patterns are formed. The positions of diffraction maxima are given by: (a) D sin φ = λ (b) 3D sin φ = n λ (c) D sin φ = n λ.
15) What voltage V must be applied to an electron such that the wavelength associated with its motion is 0.183nm? (a) 90.0V (b) 25.0V (c) 45.0V. click here.