Early Quantum Theory
In what may seem like an abrupt change of pace, we will now drop our periodic
table with its focus on practical chemistry of elements and start working on the
background needed to understand and advanced theory called Quantum theory. Don’t
worry, the two are related but you won’t see the tie until the next chapter.
4-1 First ionization energies
Let’s do a quick review of atom structure again
Nucleus - dense and tiny. Surrounded by cloud of electrons. Electrons have
virtually no mass but they are what occupies the volume of the atom.
They are also how one atom interacts chemically with another atoms
SO if you want to understand chemistry, you have to understand electrons
Let’s start with ionization energy - The minimum energy needed to remove and
electron from an atom.
In particular I want to discuss the first ionization energy (I1), the minimum
energy required to remove an electron from a gaseous atom (A) to produce a
positively charged gaseous cation A+
A(g) 6A+(g) + e- (g)
First ionization energy
There are also second, third, fourth, etc ionization energies for the reactions until
you run out of electrons:
A+(g) 6A2+(g) + e- (g)
2nd ionization energy
A2+(g) 6A3+(g) + e- (g)
3rd ionization energy
3+
4+
A (g) 6A (g) + e (g)
4th ionization energy
With each ionization getting harder and harder to do, so the energy gets higher
and higher (Can you explain why?)
Let’s look at the 1st ionization energies of the elements and see if we can dicover
any periodic trends
Figure 4.1
Very distinct trend the group 1 metal has the lowest E1 in any period, and the
inert gas has the highest E1, with atoms in the middle generally following along.
Thinking chemically the electron cloud of the inert gas must be very stable, and
that of the 1A metal very unstable.
The low ionization energy of the group 1 metals ties to their high reactivity
So our first insight is that periodic properties of the elements are tied to the
stability of the electron cloud around the atom
4-2 Ionization Energies and periodicity
Now let’s look at he successive ionization, what happens as you remove more
electrons
Table 4.1
(aJ = attaJ = 10-18 joules/atom)
First let’s get oriented.
Flip back to Figure 4.1
Moderately hard ti ionize H, then really hard to do He, then really easy for Li then
gets harder as you go to Ne, then easy to Na, etc (maybe sketch on board?)
You can see that inthe first column of data in Table 4.1
No go across for Na
Easy for the first
Then a lot harder for the second, then slowly increases up to I9 Then big
jump to I10 (don’t bother with figures, just rough sketch on board)
Figure 4.6 & 4.7 in powerpoint, but just as easy to sketch onboard
Kind of like peeling an onion
first few come off easy (outer skin)
Then a big break and a bunch that come off ~ the same
Then a bigger break and anoth bunch that com off the same
Suggest ‘Shells’ of electrons with similar properties
Now look again at Table 4.1
Always 2 in innermost shell
Then 8 in next shell
The 8 in next
Key Concept:
Call the electrons in the outermost shell the valence electrons
Call the electrons in inner shells the core electrons
Another way to represent valence electrons is with a Lewis dot diagram
Put the name of the atom in the middle and surround it with one dot for each
valence electron
Na.
He:
etc
Similarly we can emphasize the core by writing the [core]
See Table 4.2
Clicker question: Write Lewis diagram for some atoms
4-3 Electromagnetic Spectrum
Now that we are focusing on electrons, we need to better understand how
electrons and matter interact, and to do that we need to understand the electromagnetic
spectrum
EM spectrum - Radio, Microwave, infrared, visible, ultraviolet X-rays and gamma
rays all part of EM spectrum
Figure 4.8
1860's about the time the periodic table was getting nailed down Maxwell
proposed electromagnetic theory of radiation.
Radiation was oscillating electronic and magnetic fields moving through space
Figure 4.9
Magnetic field and electronic fields are perpendicular to each other
Key Concepts/Equations:
Wavelength(ë) - the distance between consecutive peaks or troughs
Frequency (í) - # of crests that pass a point in 1 sec
í×ë = c Speed of light: 2.9979x108 m/s
Exercise
What is the frequency of green light (green ë= 500 nm)?
500 nm = 500x10-9 m = 5x10-7 m
ëí=c
í=c/ë
=3x108/5x10-7 = 6x1014 Hz
What is the wavelength of an FM radio wave with a frequency of 91.1MHz?
91.1 MHz = 91.1x106 Hz = 91.1x10+6 sec-1
ëí=c
ë=c/í
=3x108m/s / 91.1x10+6/s = 3.29 meters
Clicker question more of same? Maybe a set up rather than a calculation?
4-4 Line Spectra of atoms
When you pass while light through a prism you the visible rainbow
400 nm violet 750 nm red
is a continuous spectrum. No gaps
When we excite atoms to emit light (in a flame or arc a spark through a gas)
we get discrete wavelengths - line spectra
If we use a single atom for the source we call this an atomic emission spectrum
Simplest atomic emission spectrum is that of hydrogen
Figure 4.13 & 4.14
Fore years Scientists tried to make sense of this and figure out why we the lines
are where they are
Rydberg figured ou an empirical equation that fit the data
1/ë = Constant (1/4-1/n2) where n = 3,4,5,...
The constant was called the Rydberg constant = 1.097 x 107 m-1)
Every element will give an atomic spectrum
the number of lines and the ë of each line is different
can be used as a fingerprint to identify an element
4-5 Photons
If light is a continuum, why do atomic spectra consist of discrete line?
At this point in time (late 1800's-early 1900's)classical physics, the view that light
is a continuum was having all sorts of problems explaining experiment two key
phenomena
Black Body radiation - Refers to the fact that as heat something up it
emits a continuum of radiation. The hotter it gets the more radiation.
Practical example heating iron from red hot to white hot.
It is emitting a continuum of radiation, so you would think that it would fit
the classical (continuum ) theory. But it was emitting the wrong
continuum! According to the continuum theory it should give off lots of
light at high frequencies (short wavelengths) at all temperature. According
to experiment it only did this at very high temperatures
The Photo-electric effect - refers to light hitting a metal in a vacuum tube
as shown in Figure 4.17 & 4.18. Here light had to have a certain minimum
frequency (called threshold frequency) before electrons would pass from
one electrode to the other
It took the brilliance of Einstein to figure out that the two things were related.
Key Concept:
The explanation was that the energy of light is NOT a continuum, but was tied up
in small discrete packets called photons (just as matter is tied up in small
discrete packets called atoms)
Key Equation:
The energy of a single photon of light energy is given by:
E=hí (if you have the í of light)
-orE=hc/ë (if you have the ë of light)
Practice calculation:
Going back to our green light.
What is the energy of a single photon of green light
What is the frequency of green light (green ë= 500 nm)?
In our previous work on this problem
500 nm = 500x10-9 m = 5x10-7 m
ëí=c
í=c/ë
=3x108/5x10-7 = 6x1014 Hz
E=hí
=6.63x10-34 Js x 6x1014 sec-1
=3.98 x10-19 J
or E=hc/ë
=(6.63x10-34 Js 3x108m/s) /5x10-7 m
=3.98 x10-19 J
Clicker question, set up of more of the same
4-6 De Broglie Wavelength
So the explanation seems to be that light displays wave-like properties under
certain conditions, and particle like properties under other conditions. This is
called the wave-particle duality of light
In 1924 de Broglie proposed a really radical idea. If the wave of light can display
particle like properties under certain conditions, then why can’t particles of matter
display wavelike properties?
Using Einstein’s theory of relativity he proposed that both light and matter obey
the equation
Key Equation:
ë=h/mv
Where ë is the wavelength, m is the mass of the particle at rest and v is the
velocity of the particle
Practice problems
What is the deBroglie wavelength of an electron traveling at 1% of the speed of
light
mass of electron 9.11x10-31 kg
v = .01 x c = .01 x 3x108m/s = 3x106 m/s
ë=h/mv = 6.626x10-34 J@sec/(9.11 kg x 3x106 m/s)
Note 1 J = 1 kg m2/s2
so ë = 2.43x10-10 = = .243 nm
this is the wavelength of an X-ray
What is the mass of a photon of UV light (ë=200 nm)
ë=h/mv
më= h/v
m = h/ëv
ë in m = 2x10-7
Photons travel at the speed of light so v = c =3x108 m/s
=6.626x10-34 / (2x10-7 x 3x108)
=1.10x10-35 Kg
which is about 10,000 less mass than an electron!
Clicker question on De broglie equation
4-7 Wave-Particle Duality
Proof of de Broglie’s hypothesis came just a few years latter when 2 Scientists
showed that you could point a stream of electrons at a crystal, and the electrons
would diffract, Figure 4.22 Something that only wave can do!
So we can regard light as wave or a particle depending on the experiment
Similarly we can regard matter as particles or waves depending on the
experiment!
This is called wave-particle duality.
(Interesting aside in text: JJ Thomson (and others) got Nobel Prize in 1906 for
demonstrating that electron was a particle. His Son GP Thomson (and others)
got Nobel prize in 1937 for showing it was a wave!)
4-8 Quantization
Getting back to our line spectra for hydrogen
In 1913 Niels Bohr had looked at the line spectra for H and the Rydberg
equations, and had proposed that the reason the E was quantized was that tha
the electrons in H were restricted to certain circular orbitals
10 years later, when deBroglie proposed that electrons could act like waves, it
was possible to rationalize why the electrons were restricted to only certain
orbitals.
Figure 4.24
It was proposed that the orbit could only be stable if the wave took an integral
number of periods to get around the nucleus so it would return to its starting
point after 1 revolution, otherwise it would cancel itself out like waves do in a
diffraction pattern
Putting a little math together, saying the circumference of the orbit is 2ðr, and
this has to equal në. Throwing in the force of attraction between the nucleus and
the electron he came up with the equation
En = -2.18x10-18J/n2
This quantized the energy of the orbitals to distinct values or energy states
These quantized states are called stationary states
The lowest stationary state (n=1) is called the ground state
All the higher states (n>1) are called excited states
n=2 is called the first excited state
Note that the energy of an orbital is negative. This means that energy is
released as an electron ‘falls’ into this state. And then you have to supply energy
(ionization energy) to release the electron from this orbital
4-9 Electronic Transitions
When an electron is in a stationary state it cannot emit or absorb energy. This
only happens when en electron moves from on e state to another
So if energy is going to be released from an H atom to make an emission line,
and electron has to drop from a high stationary state to a lower stationary state
so some energy can be released
E of photon = Ei - Ef
Where Ei = E of initial state and Ef = E of final state
Key Equation:
Ephoton = 2.18x10-18J (1/nf2 - 1/ni2)
Practice problems
What is the energy released in J, and the wavelength of the photon carrying that
energy when an electron in a hydrogen atom falls from the n=5 state to the n=2 state?
E = 2.18x10-18J (1/nf2 - 1/ni2)
=2.18x10-18 J (1/22-1/52)
=2.18x10-18 J (1/4-1/25)
=2.18x10-18 J (.25-.04)
=2.18x10-18 J (.21)
=.458x10-18J
E = hc/ë; ë=hc/E
=(6.626x10-34J@s x 3x108m/s)/.458x10-18 J
=4.34x10-7m
=434 nm Which checks with one of the line you can see!
Clicker question: Set up for Ephoton type calculation
I mentioned earlier that you can get these line spectra from having an electrical
spark arc through a gas (that is how glow tubes work) or by simply putting the
atoms in a flame. Heat energy kicks electrons to higher states, and the atom
cools light is emitted as electrons drop from highly excited states to lower states
This is the basis for the flame test we use in the lab, where each element gives
off a characteristic color in a flame ( demo ?)
The opposite can also happen. Light of the right wavelength can be absorbed by
an atom to kick an electron from a low orbital to a higher orbital. (Figure 4.30)
We can use the absorption of light in the chemistry lab, not only to identify
compounds, but to measure the amount of a compound in a material.
As good as the Bohr theory was at explaining the hydrogen spectrum. It failed
miserably as explaining the spectrum of any other atom. Many people atried and
tried, but this theory could never be extended to anything else. As a result inthe
next chapter we will have to give up on the Bohr theory with its nice circular
orbital that you see in all the picture books and try something new!