An Introduction to Quantum Mechanics (1)

1. The Wave Model of the Atom.

Though useful, especially in terms of identifying spectral lines, Bohr’s model had its limitations. For example, it couldn’t account for how angular momentum is conserved in atoms, nor how electronic transitions originate. More seriously, it wasn’t able to deal with the problem of lost energy and why atoms don’t simply collapse.

    Consider the following dynamical picture:  as the electrons whir about the nucleus they ought to be losing energy, in the context of Bohr’s orbital model. If they lose kinetic energy over time they must spiral into the nucleus, and the atom then ceases to exist. This ought to happen in a very short time, so that most atoms in the universe cease to exist and hence the whole universe. But this isn’t observed. Why?

     The only explanation is that Bohr’s orbital model can’t be correct.  Thus was born the theoretical basis for the wave model which we mostly accept today in modern quantum mechanics.  Unlike the Bohr model, electrons don’t follow defined orbital paths but instead are referenced to regions or volumes in which they will be more or less probable.  The basic allocation of electrons, say for the hydrogen atom, is then confined to “orbitals” or regions of higher probability.

     We now look at the experimental basis provided for this model.

     Around 1926, a young French physicist named Louis de Broglie actually postulated the basis for material particles, such as electrons, acting as waves.  This was experimentally verified in the (1927) Davisson and Germer electron diffraction experiment sketched below:

 

Fig. 1: The Basic Davisson and Germer experiment

    From the experiment, with electrons moving through a potential difference V = 4,000 volts, the kinetic energy gained should be equal to the work done, or:

½ m v²   =   eV

Where m the mass of the electron is: 9.1 x 10 ^-31 kg

And the electron charge e = 1.6 x 10^-19 C

The velocity then is:

v = Ö (2eV/m)

The momentum p = mv = m Ö (2eV/m) = Ö (2eVm)

And the de Broglie wavelength is:

l_D= h/p = h/Ö (2eVm)

For a voltage V = 3,000 V one would find:

l_D =

(6.626 x 10^-34 J-s) / [(3.2 x 10^-19 C) (3000V) ( 9.1 x 10^-31 kg)]^1/2

l_D =  2 x 10^-11 m

Which is the de Broglie wavelength of the electron in this experiment.

    A first step to uncovering the wave model from Bohr’s is to examine his quantized relationship:

m vr  = nh/ 2p  = L

where L is the angular momentum.  Re-arranging:

h/ mv =  2p r/ n = l_D    or  2p r =  n l_D

Showing the circumference is scaled into n (standing) waves of wavelength l_D   as shown below:

Fig. 2. Standing wave model for the Bohr atom.

     This wave-orbiting electron atom still has a radius r,  but with waves each separated by one de Broglie wavelength , l_D.

     Thereby  an integral number of such wavelengths form the circumference of the atomic orbit, as required by  the condition:  2p r =  n l_D.

  For example, in the case of hydrogen the first three of these cloud-wave regions are shown in Fig. 3.

Fig. 3: Electron (“orbital”) clouds-regions in Hydrogen

Let us ‘zoom in’ on the more spherical n= 1 configuration, and the probability for the electron in this space as depicted in Fig. 4 below:

Fig. 4: The n = 1 electron orbital for hydrogen

    This diagram more than any other dispenses with the notion that hydrogen electron occupies a definite position. Instead, it’s confined someplace within a “cloud” or probability space (b) but that probability can be computed as a function of the Bohr radius (a_o = 0.0529 nm).  The probability P_1s for the 1s orbital is itself a result of squaring the “wave function” for the orbital.  If the wave function is defined y (1s) = 1/Öp (Z/ a_o) exp (-Zr/ a_o), and the probability function is expressed:

P = ½y (1s) y (1s) *½

     Where y (1s) * is the complex conjugate, then the graph shown in Fig. 4 is obtained. Inspection shows the probability of finding the electron at the Bohr radius is the greatest, but it can also be found at distances less than or greater than  0.0529 nm.

     We thereby see from Figs. 3-4 that Bohr’s original quantizing number, n (the "principal quantum number"), has far more meaning than simply to parse the number of standing waves for a given atom. We already see that it determines the energy of the atom, viz.

E _n =  - 13.6/ n²   

     But it also indicates the average distance of the electron from the nucleus.  Thus, de Broglie’s wavelength provides the basis for the wave-particle duality that lies at the basis of the “smeared” probabilistic atoms peculiar to modern atomic theory.

       At the bottom of Fig. 3 are the quantum numbers: n and ℓ, which are identified as: the principal quantum number, and the angular momentum quantum number, respectively.

    There are two physical meanings attendant on n: i) it determines the energy of an orbital (specifically in the H-atom), and (ii) it indicates the average distance of an electron in a particular orbital, to the nucleus, To fix ideas, I show in the accompanying diagram (Fig. 5)  a sketch of one lobe for an electron orbital associated with the (3, 2, +2) state in the Hydrogen atom. The key point is the orbital denotes an electron density associated with a probability of finding the electron in some defined space.

Fig. 5: one lobe for an electron orbital associated with the (3, 2, +2) state in hydrogen

     In the case shown one must also visualize a symmetrical lobe on the other side (making the whole orbital resemble a dumbbell) to make it complete. As one alters the set of quantum numbers the electron densities change and so do the probabilities associated with the orbit.(See Fig.6)

    Fig. 6: Further hydrogen orbitals with higher quantum numbers

     Describing orbitals using the set of given quantum numbers means knowing the numbering rules applied to each. In the case of the principal quantum number, n, we allow it to have integral (non-zero) values: 1, 2, 3, 4 etc.

     The physical significance of the angular momentum quantum number (ℓ) is to convey the shape of the probability density cloud or orbital. The numbering rule for l is directly contingent on the value for n. Thus, for any given n, then ℓ must be such that it has integral values from 0 to (n -1). This means if n = 2, then ℓ can have (n- 1) = (2 -1) = 1. But if n =1, then ℓ = (n - 1) = 1 - 1 = 0.

       Note that the ℓ -quantum numbers appear more than once for any  orbital with n >1. Thus, for the n = 2 case, we have two values of ℓ occurring: one for l = 0, the other for ℓ = 1. If we go on to n= 3 there are three values of ℓ, for n = 4, four values and so on. One also finds the ℓ -value specified for lettered orbitals: s, p, d, f, g, h. The s-orbital is for ℓ =0, the p for ℓ =1, the d for ℓ = 2 and so on.

     There is no special significance to the letters (apart from the physical meaning we already gave for the angular momentum quantum number, ℓ, and they are mainly of historical import- though still retained, for example, in chemistry. (By extension, one also often hears the term "atomic shell" used in chemistry). A collection of orbitals under the same value of n is called a "shell". Thus, for n = 4, we have ℓ =0, ℓ = 1, ℓ =2, ℓ = 3 so comprising the collection of orbitals: s, p, d and f.

Lastly, there is the magnetic quantum number, usually designated m _ℓ (subscript the same as the angular momentum quantum number) because it is contingent upon it. This quantum number describes the orientation of the orbital in 3-D space. For a given angular momentum quantum number, ℓ, we have integral values specified as follows:

m _ℓ = - ℓ, (-ℓ +1)....0......( ℓ - 1), + ℓ

Note the above set of m _ℓ numbers is given as a SERIES, e.g. starting with (-ℓ) and terminating at + ℓ. Look at the simplest example for ℓ = 0, then:

m _ℓ  = 0.    (Since all terms are zero)

What about ℓ = 1?

Then: m _ℓ = -1, 0, 1

What about ℓ = 2?

We have:

m _ℓ = -2, -1, 0, 1, 2


As a general rule then, we can use the formula:

N(m _ℓ) = {(2 x l) + 1}

to give the total number of m _ℓ   numbers.

In any problems to do with identifying electron shell structure- configuration, it is well to bear in mind the Pauli Exclusion Principle to make sure the electrons are distributed so that no two electrons have the same set of quantum numbers. It is instructive to study the table below to see how the principal quantum number, n, changes with ℓ, and the subshell as well as the symbol for the primary electron shell.  The  Table shown below shows the respective Shell and Subshell Symbols and associated Quantum Numbers.

n  =	SHELL	ℓ   =	Subshell
1	K	0	s
2	L	1	p
3	M	2	d
4	N	3	f
5	O	4	g
6	P	5	h

In working out assorted problems, preparation of a schematic energy level diagram associated with the state of the system can also be of immense value.

Problems:

(1) Find the de Broglie wavelength  l_D  of a proton subject to 4000 V in a Davisson-Germer type experiment.

 (Take m _p = 1.7  x 10 ^-27  kg)

(2) Write out the electron configuration for oxygen (O16), then write out the values for the set of quantum numbers : n, ℓ, m _ℓ, m _s  for each of the electrons in O16.

(Hint: The  m _s or spin quantum number, which we will encounter in the next section, has either +1/2 or -1/2 value. )

Looking at Basic Atomic Physics (1)

1.The Rutherford Model of the Atom.

 What may be called the first foray into basic atomic physics by which further theory could be built upon, commenced with the Geiger and Marsden experiment – first suggested by Lord Rutherford in 1909. The basic setup is depicted in the rough sketch below:

Fig. 1: Basic Layout of the Geiger-Marsden Experiment

From the Rutherford experiment design, Geiger and Marsden made use of a source of alpha particles to bombard a thin metal foil, on the other side of which was a detecting zinc sulfide screen. They found that while most alpha particles arrived at A, in the direction shown, a few also scattered to positions at B and C which could be detected when the screen at A was moved to the other positions.  The nature of the scattering and deflections (especially some alpha particles at very large angles) was such that there had to be a highly concentrated charge or “nucleus” at the center of the atom. Since the alpha particles are relatively massive (at about 4.002 amu each) the deflections at wide angles meant nearly all the atomic mass was concentrated in the center of the atom and electrons were in the distant outer regions.

Rutherford thereby proposed a model of the atom in which nearly all the mass was concentrated in a very small nucleus while the electrons were scattered at some distance away. This is depicted below in Fig. 2.

Fig. 2: The Rutherford Model of the atom The key consequence was that the Rutherford experiment, carried out by Geiger and Marsden, showed that the “pudding pie” model of J.J. Thomson was incorrect. If Thomson’s model was correct, then the expected deflection could be no larger than 0.0001 radians or less than a degree. Since the observed deflections were in some cases more than 100 degrees, it failed the experimental test. Despite this success, Rutherford’s model still hadn’t won the day. It was largely accepted because it could quantitatively alpha-scattering by thin foils. His model could not: 1) explain line spectra in atoms, including both absorption and emission lines, 2) account for the stability of atoms and could only account for half the nuclear mass. 2. The Bohr  Model of the Atom. The Bohr Model of the atom, proposed by Neils Bohr, directly challenged the Rutherford model by showing how the observed emission and absorption lines of spectra could be explained. At the heart of Bohr’s model was simplicity, with the hydrogen atom – for example – configured to a miniature solar system with the nucleus at the center and the electron in orbit around it. Fig. 3: The Bohr Model of the atom. From the diagram the electron (e) orbits at a radius r from the central nucleus of charge Ze. As with the planets, a centripetal (inner directed force) F acts toward the center.  Bohr’s major concept was to quantize the electron orbits. He proceeded by first quantizing the angular momentum of the orbit: m vr  = nh/ 2p  = n ħ where  ħ  =  h/ 2p   is the Planck constant divided by 2p. The Planck constant, first proposed by Max Planck, is: h = 6.626069 x 10- 34 J-s Then the value of ħ  = 1.0546 x 10- 34 J-s Next: both sides are squared: (m vr ) 2 =  (n ħ)2 So:  m2 v2 r 2 =  n2 ħ2 And:  v2  =  n2 ħ2     / m2  r 2 Now, Bohr looked at the total energy of the H-atom in terms of it kinetic (K) and potential (V) contributions, so: E = K + V  =    ½ m v2  -  k e2  / r E = K + V  =   k e2  /  2r    - k e2  / r =  - k e2  /  2r     (Since  ½ m v2   =  k e2  /  2r ) Now solve for r (actually the quantized r n ): r n  = [  n2 ħ2/ m2 v2 ] ½ But  from the kinetic energy equivalence: v2 =  k e2  /  mr =  n2 ħ2   m2  r 2 \    r n  = [n2 ħ2   / m k e2  ] The Bohr radius is just the value when the principal quantum number n = 1, so : r o  = [ ħ2   / m k e2  ]   = 0.0529 nm = 5.2917 ×10−11 m This is just the most probable radius, i.e. distance between proton and electron, in the hydrogen ground state. Now, to obtain the quantized energy (E n) we substitute the value for r n  into the total energy equation: E =  - k e2  /  2r     =  - k e2  /  2[n2 ħ2   / m k e2 ]    E = - m k 2 e4  /  2n2 ħ2   =  - m k 2 e4  /  2 ħ2     [1/ n2] = - 13.6/ n2    Where the last quantity  is in eV, or electron volts. Here the n refers to the energy level, ground state is n = 1, so can allow the computation of energy for a given level. Or, the energy for a photon emitted from an atom when an electron makes a transition – say from n = 2 to n = 3. Such a situation is shown below:

Fig. 4: A few energy (electron) transitions made in Hydrogen

     An important point is that the quantized angular momentum postulate (m vr  = n ħ) restricts the possible circular orbits to defines sizes according to the quantized radii (r _n  etc.). Thus the normal state of the atom, say hydrogen, will be that for which it has the least energy or the ground state – corresponding in the case of hydrogen to the Bohr radius. Some transitions for different spectral series are shown below:

Fig. 5. Some Energy transitions in the Hydrogen Bohr atom

     As shown in Fig. 4, emission occurs when an electron in the atom, say hydrogen, makes a transition from a higher to a lower energy level, accompanied by the emission of a photon with a defined energy E = hf = h (c/ l). Consider for example, a transition from the n = 2 to the n = 1 level, as depicted in the lower right of Fig. 4 and in the first line of the Lyman series of Fig. 5.

The energy at the n= 2 level is:

E(n=2) =  - 13.6/ n²   = - 13.6/ (2)²      =  - 13.6/4  (eV)

Now, 1 eV =  1.6 x 10^-19 J  so:

E(n=2) =  - 13.6/4  (eV) =  -(3.4) x 1.6 x 10^-19 J  =

 -5.4 x 1.6 x 10^-19 J 

The n= 1 level has energy:

E(n=1) =  - 13.6/ n²   = - 13.6/ (1)²      =  - 13.6  (eV)

E(n=1) =  -(13.6)  x 1.6 x 10^-19 J  = -21.8 x 10^-19 J 

Then the energy difference is:

E2 – E1 = [- 5.4 – (-21.8)]  x 10^-19 J  = 16.4 x 10 ^-19 J 

From this, the wavelength of the photon emitted can be found. Since E = hf = h (c/ l):

l =   hc/ (E2 – E1) 

l =    (6.626069 x 10^{- 34} J-s)(3 x 10 ⁸ m/s)/ 16.4 x 10^-19 J 

l =    1.21 x 10 ^-7 m 

The frequency can be found from:

f = (c/ l) = (3 x 10 ⁸ m/s) / 1.21 x 10 ^-7 m  = 2.47 x 10 ¹⁵ Hz

Insight problem:  Using Fig. 5 as a basis, compute the energies and wavelengths of the photons emitted when the electron in the hydrogen atom makes the 1^st, 2^nd and 3^rd Balmer transitions.