Nature of Light and
a Modern View of the Atom and QUIZ 

FIRE and LIGHT: Understanding the electronic structure of the atom began with a study of light. This picture with fire is appropriate because visible light is normally created by flames or something that is heated to high temperatures. 

Understanding is Not Easy: Understanding the nature of light and the nature of electrons
is not easy. One reason is we are almost as blind as a nearsighted fetus
with regards to seeing all light that is available. The nature of light
and electron structure can be confusing to the point where we feel we
might be too mentally limited to fully understand it. 

Even Einstein was confused about the nature of light and the nature of electrons. So if you feel confused or if your brain thinks it's going to explode, I will apologize in advance for the following tutorial.  
I say we are almost blind because
we are only seeing about 1.5 percent of the whole electromagnetic spectrum.
Visible light is not different than any of these other forms of light except
for the fact that our eyes evolved to see this narrow slice because that’s
the light region that our sun mostly emitted. 

Let’s say this image is made up of the whole electromagnetic spectrum. Roll the cursor over the image to see how much of the whole spectrum we actually see. Yes, this is how blind we are. We are only picking up
about 1.5 percent of the picture that we could see if we could see the
whole electromagnetic spectrum. 

Now assume this picture is just visible light, which it is. If we could see all the other frequencies of light, we could tell if he had a cellphone in his pocket because it would be flashing (radio frequencies). We could also see communication satellites in space (radio frequency). We could also tell the water had radioactive contamination from sparkles of gamma rays. We could see the train through the trees because the train would absorb radio waves. In the distance we could see that the ground was warm because of a fire put out 2 days ago. Basically, we could see 80 times more information than we do with our eyes. 

Here is 1.5 % of the picture that represents the small amount that we see. Try to guess what it is. Move cursor over the image to see the full image. 

Another problem with our eyes is that it can't tell the difference between a pure color (one frequency) or a combination of color. For example, the white, yellow, and magenta you see on your computer screen here are made up of various colors. The yellow is actually made up of green and red and not yellow. The magenta is blue and red and not pure magenta. Even the white is only 3 colors and not the whole spectrum of the rainbow. So our eyes (and brain) is quite limited in the deciphering of light.  
From an earlier tutorial and how the brain is fooled, it revealed another weakness in our ability to determine brightness. When you look at tiles A and B, it looks like tile A is darker than B; however, they are the same brightness. The brain is fooled by seeing B in the shadow of the green cylinder. (roll cursor over the image to see tiles A and B by themselves.  
So even though we want to trust our eyes, we have to realize that they can fool us and are not giving us a complete image of what's around us. 

We begin our understanding of
light by seeing how it behaves. Here we see it reflects off the lake and
also gets refracted by water drops to make a rainbow. 

We also notice how different materials reflect or absorb
colors. 

FLAME TESTS: Even powders that don’t have a color can produce flames with color. That means that heat is causing some elements to emit certain colors. Different metals emit different frequencies of light. Of course our eyes may be seeing an orange flame but there may be other colors in the flame that we are missing. A prism would reveal that. 

Flame Tests Analyzed: Gustav Kirchoff and Robert Bunsen realized that the flame tests might be a way to discover new elements. 

SPECTROSCOPE: Bunsen and Kirchoff invented a device (a spectroscope) that passed the light from the flame through a prism. That way could see all colors that a specific mineral was emitting. Bunsen invented the "Bunsen Burner" in order to get a clearer flame so it didn't interfere with the colors from the mineral in the flame. Bunsen burners have a opening at the bottom to let mix with the gas before it reaches the flame at the top. This premixing produces a cleaner and hotter flame. 

Using this spectroscope, they discovered two new elements. One they called "rubidium" because it emitted a reddish light. "rubi" means red, which is also why ruby jewels are called rubies. The other element discovered was cesium. Cesium emitted a blue light and was named after the Latin word, Caesius, meaning bluish gray. 

Instead of a flame, one can put an element in a tube and heat it with electricity until it emitted light. If the light goes through a slit and then passes through a prism, the various frequencies of light emitted are separated into a spectrum. Our eyes can see the frequencies that are in the visible light range, but there are other frequencies of light emitted we can't see. 

Below are the spectra of 4 different elements. The first one is hydrogen, which emits 4 colors of light that's in the visible light range. Note that other frequencies, such as UV light might be emitted, but we can't see them. The second one is helium. It has 2 electrons and we see 7 colors. Next is mercury whose spectra shows 8 colors. Mercury also produces a lot of UV light which in fluorescent bulbs is normally converted to visible light by the use of certain minerals that capture UV light and emit visible light. The bottom element is the spectrum of uranium. Uranium emits many frequencies of colors. It appears that the elements that have more electrons emit more colors. So there seems to be a connection.  
WAVE VOCABULARY:


WAVE CALCULATIONS: Since light travels at the speed of light (represented by "c") the formula shown connects wavelength, frequency and the speed of light. Dimensional analysis confirms that the formula is correct. Wavelength (in meters per cycle) multiplied by frequency (in cycles per second) ends up as meters per second because cycle cancels out. Meters per second is a measurement of speed. 


Let's do a problem. The laser that is used to attach a detached retina uses a beam with wavelength of 640.0 nm. That's a red color. If asked to calculate frequency, we use the formula (1). To solve for nu (frequency), we divide both sides of the equation by the wavelength lambda (2). We end up with a formula for frequency (3). The speed of light is 300,000,000 meters per second (3.00x108ms^{1}). Note: "s^{1}" means "per second". If you ever see "1" by a unit of measurement, realize it's a short way of saying "per" and it allows you to put the amount in the numerator rather than the denominator. We plug in the speed of light and the wavelength (640 nm). Note: "n" (nano) will not cancel, so we multiply by "n" over 10^{9}. Remember, "n" and 10^{9} are the same, so it's like multiplying by 1. The answer is 4.688x10^{14} cycles per second. Sometimes, books will leave out the word "cycles" and show the number just as "per second" (s^{1}) 

KJZZ (91.5FM) is a service of Rio Salado and Maricopa Colleges. Besides Jazz, it plays programs from National Public Radio. Let's say this is your favorite station and want an antenna with a length that best absorbs this station's radio wave frequency. "91.5" is 91.5 megahertz (91.5 million cycles per second). PROBLEM #1: How many feet should the antenna be, knowing that it needs to be half the wavelength?  
The formula between wavelength and frequency was given above and here again on the left. You can solve for lambda, which means to divide the speed of light (c) by the frequency (v). Sometimes you don't have that formula handy. So you could just use dimensional analysis to make sure the final answer is in meters. Look at the two known measurements. The frequency is cycles per second and speed of light is meters per second. We can't multiply them because we would end up with a strange cyclemeters per seconds squared. We can see that we need to invert the frequency so that the seconds will be in the numerator, so it can cancel the seconds in "meters per second"
PROBLEM #3: What is the length in feet of the antenna best matched to 98.7FM? 

If buying a radio station, you might ask, is there an advantage to having a high or low FM station? In other words, is a higher megahertz signal stronger than a lower one? To answer that, we got to go back into history and find out what Max Planck was doing...  
Despite warnings from his friend, Phillip Von Jolly, Max Planck went into physics. Max started studying the relationship of temperature, color (frequency), and energy coming from material that absorbs & emits all light. They call it a black body since black absorbs light (and emits light). The most practical black body is a chamber that has a small hole in it. The theory at the time predicted that the hot black body would produce a high intensity of light at the higher frequencies of light (such as ultraviolet light) than lower frequencies. However, that wasn’t what the actual data showed. To account for this, Max assumed that higher frequency light needed more energy, therefore reducing its chances of being created. 

The formula for that energy was set to E=hv, where
v is the frequency and h is a conversion constant to turn frequency
in to energy. h is called Planck’s constant. If we want answer in Joules, the constant is He also guessed that the electron oscillations making
light were restricted to whole number increments. For that reason, he
felt energy increased in increments. In other words energy might come
in a packet (a quantum). He didn't think people would believe him, so
he said this was just a mathematical fix and reality really wasn't like
that. 

So Planck's work found out that the color emitted by something hot tells us the temperature (energy output). For example, the yellow parts of the lava are hotter than the red parts.  
From this formula, the distribution of light can be graphed for different temperatures. Notice that that at 3,000 K, which is about the temperature of a standard incandescent light bulb, only about 5% of the light is in the visible range. Most of it is in the infrared region, which we feel as heat. So this kind of light bulb, is barely producing light, but mostly heat. So it's very wasteful. At 6,000 K, which is the temperature of the Sun's surface, the peak color is yellow and there's a good production of visible light. Again, this is why our eyes have evolved to see this range of light. That's what the Sun produces the most of. 

Another mystery at the time that seemed
unrelated, was the photoelectric effect.
The device to illustrate the photoelectric effect was a plate and wire
inside a vacuum tube. The battery is arranged so that a negative voltage
is applied to the plate. No current is flowing because the voltage is
too low to cause the electrons to break away from the plate.


When visible light shines on the
plate (like from a flashlight), there is still no current. The classical
view of light was that energy from the light would accumulate
and eventually eject the electrons, but that
never happened. 

When ultraviolet light shines on the plate, electrons are ejected from the surface and current flows. (roll cursor over image to see animation) Electrons go to the other terminal and flow through the circuit. Other electrons leave the negative side of the battery and make it to the plate. If UV light continues, current will continue. At the time this effect could be demonstrated, but no one knew why UV light worked and visible light didn't. This apparatus is often called a photocell or phototube.
It’s used to detect light. Instead of a sheet of metal, semiconductors
can be used. With small semiconductors as light detectors, one can make
video cameras or solar cells. 

Here is a large phototube that is used to detect light. If the detector is metal, then UV light causes current to flow and can be detected. If made from a semiconductor (like computer chip material), then visible light will cause current and be detected. 

The photoelectric effect also happens when UV light from the sun ejects electrons off of the moon dust. With electrons missing they become positively charged and start repelling each other. This causes the moon dust to fluff up and sometimes float a few feet above most of the moon’s surface. 

Returning to the mystery. The mystery was how did light of higher frequency (UV light) eject the electrons? Einstein tackled the mystery by borrowing from Max Planck’s formula, E=hv. 

Einstein proposed that light was not a continuous wave but a packet or particle of energy (later called photons). Energy could only transfer to an electron by being hit by one photon of light similar to the way one particle hits another particle. He knew that higher the frequency, the more energy these photons would have. Only when the photon’s energy was greater than the binding energy of the electron to the metal would an electron be ejected. (move cursor over image to see animation) 

Light had been thought of as a continuous wave with energy
spread over the entire beam. Now with Einstein's explanation of the photoelectric
effect, light needed to be thought of as packets of energy or particles
of light (photons). The view of the universe changed because if energy
was in packets, that meant that time, distance, and about everything else
was broken down into packets. This required a huge rethinking of how we view the universe. 

The world started looking like the way computer modelers see the world. Even though to most people something looks smooth, to the computer modelers, they know it's made up of individual patches (polygons). Smoothness is just an illusion. Smoothness in distance, times, energy, and everything was just an illusion. Like computers are made up of bits, our universe is divided up in small bits of its own.  
Returning to the discussion of the medical laser. Earlier we calculated the frequency of the 640 nanometer light used. (1) Using Planck's formula, we can now calculate the energy of one photon of light . (2) The "h" is a constant that converts frequency to joules. (3) Multiplying the value of "h" by the frequency of the laser light we get the energy of one photon of the laser light (4). It's a very small amount because we are talking about just one photon of light.  
Someone asks how many photons of light are emitted by the laser? That seems like an impossible question to answer. Counting atoms is hard because they are so small. Photons are just as small and also traveling at the speed of light. However, by knowing the energy of each photon and the energy that the laser produces, we can count the photons. Dimensional analysis makes the calculation more certain. 2 watts means 2 joules (J) per second. By multiplying by 0.5 seconds, the seconds cancel and we have energy in joules (J). By taking our energy answer from above (1), we invert that so that joules cancel out. That gives us the final answer of 3.21x10^{18} photons. 

Normal DVD players use a red laser at 640nm. Bluray DVD players use a "blue" laser at 405nm. I was a little disappointed when I learned that the laser light was at 405nm. That's not blue. Blue is between 450nm495nm. Indigo (reddish blue) is 420nm450nm. Violet is 380nm420nm. Some literature says anything shorter than 400nm is ultraviolet. So the Bluray laser light is not even close to blue. The sales people must of pushed for that name because "Bluray" must have sounded better than 'Violetray". The power of the Bluray laser is about 6milliwatts. The blink of an eye is about 350milliseconds. Problem #4: During that time how many photons are emitted from the violet "Bluray" laser? 

This is a twopart problem. Like above you need to find the energy of the photons that have wavelength of 405nm. This is done with E=hv. "v" is frequency so we have to convert our 405nm to frequency first. The equation above was λv=c. To solve for frequency (v), you have to divide speed of light (c) by wavelength (λ). Fortunately, the anwswer to E=hv (or hv=E) is in joules which matches with milliwatts because milliwatt is millijoules per second. The blink is 350 milliseconds. So it seems our dimensions are going to be straight forward.
Problem #5: Why were E7 and E8 added to the calculations? Problem #6: What is the formula in A3? Problem #7: What is the formula in L7? Problem #8: A regular DVD uses 640nm laser and its power is 5 milliwatts. What cells in the spreadsheet would be changed in order to calculate the photons it creates in a blink of an eye? 

Mystery of Light 

Let's now return to our mystery of the spectra of elements and how they might explain how the electrons in the atom are structured. 

In an earlier tutorial I showed that the image of the atom showed the electrons floating inside the atom like plums in a pudding with the pudding representing the positive charge. A later experiment showed the positive charge (the protons) concentrated in the tiny nucleus of the atom, and the electrons were thought to be floating randomly in the atom. By understanding the spectra created by heating elements, an understanding of the positions of electrons began to unfold.  
Viktor Rydberg didn’t understand
why hydrogen produced these specific wavelengths of light, but he did come
up with an equation that matched the wavelengths of the lines in the spectrum.
"n_{1}" and "n_{2}" are integers with n_{2} being larger than n_{1}. Let n_{1}=2 and n_{2}=4. The formula says to square n_{1} (2x2=4) and to square n_{2} (4x4=16). These are put in the denominator and subtracted. 1/4  1/16. The answer is 3/16. That is multiplied by the Rydberg constant and then divided by both Planck's constant and the speed of light. The answer will be 1 over the wavelength. To get wavelength just take the answer and divide that into 1. The answer I get is 486 nanometers, which matches the light blue (turquoise) line in the spectrum. So now we have a formula that predicts the colors for hydrogen or any element that has just one electron, such as He^{+} or Li^{2+}. I don't know about you, but this is rather remarkable. This electron and the light it produces are obeying a formula that is just some simple fractions. Like I said, we live in a mathematical universe. 


This formula reveals something very unusual about energy. Remember "n_{1}" and "n_{2}" are integers (whole numbers). That means wavelengths can't be just any length. They are restricted to certain sizes because "n" cannot be fractions. In other words, wavelength can only increase in in increments. If wavelength increases in increments, so does frequency (v). If frequency increases in increments, so does energy (E). Since electrons are the source of the light, they can only have certain levels of energy.  
Neils Bohr theorized
that the quantized energy that Max Planck suggested, Einstein proposed,
and Rydberg calculated could be the reason that certain frequencies of light
were seen coming from hydrogen. He thought hydrogen’s electron could
only occupy certain energy levels within the atom. Light that equaled the
difference in the levels could cause the electron to jump to the higher
level (roll cursor over image to see animation).
When the electron fell back into the first orbit, that same frequency of
light would be emitted. Electrons were now viewed to be in orbit around
the nucleus. As mentioned, the electrons could only orbit at certain distances
which represented distinct energy levels (quantum levels). These energy
levels were labeled n=1, n=2, and so forth. They were called the principle quantum numbers. 

Here we see
a higher frequency of light causing the electron to jump to even a higher
energy orbit. (roll cursor
over image to see animation) 

These are the
three principle light frequencies that are very visible from glowing hydrogen.
Notice that looking at the hydrogen lamp our eyes only see one color (magentalike).
However, if that light is passed through a prism or a diffraction grating
(a flat piece of plastic with grooves), the light will be spread out so
we can see that this magenta light is actually 3 colors. 

This image shows the energy jumps from the electron that produces these 3 colors. The highest frequency (dark blue) is produced when an electron drops from orbit 5 (n=5) down to orbit 2 (n=2). The turquoise blue (middle) is one we calculated earlier using the Rydberg formula. That electron drops from orbit 4 (n=4) down to orbit 2. The lower energy light (red light) happens when the electron jumps from orbit n=3 to orbit n=2. What if an electron dropped from n=5 down to n=1? 

When the electron drops from n=5 down to n=1, more energy must be released. It's just like a ball being dropped from a higher height. When light has more energy, its frequency goes up and wavelength gets shorter. Violet is the color with the highest frequency that we can see. If the frequency goes higher than violet, we get ultraviolet, which we can't see. (safety tip: The hydrogen lamp above is putting out UV light as well as the visible light we see. UV light can harm eyes. (My graph is a bit out of scale. The jump from n=2 to n=1 is actually bigger than the jump for n=5 to n=2, which is why it produces UV light.)  
The only jumps we haven't mentioned are the little jumps from the higher "n" orbits. The n=5 to n=4 jump produces light at a lower frequency than red. It's infrared light. The n=4 to n=3 jump also releases infrared light. This is one reason the hydrogen lamp shown above will feel warm. Note: There are higher orbits than n=5; however, the orbits get closer and closer. If an electron is pushed to higher orbits, it usually just escapes from the atom. 

If any two "n" orbits are plugged into the Rydberg equation, all of these frequencies (wavelengths) can be calculated. As you should easily guess, I would use a spreadsheet to do the calcuations. Again, I think it amazing that we humans figured this out. Atoms are extremely small, but we are making progress in understanding what electrons are doing inside an atom by looking at its light and using mathematics. This is like understanding what is happening inside of car radio by just listening to the music. On top of that, doing that with a car radio that is so small that it's invisible. 

The below spreadsheet will calculate the nanometer wavelength of light jumping from upper level 3 down to level 2. The answer is 656 nm which matches the wavelength of red light in the sprectrum shown earlier. By changing E2 and G3, it would calculate the wavelength light emitted from any jump from upper level to any lower level. (Later I will allow for input so you can try it).
Problem 9: The above calcuation was for an electron jumping from n=3 to n=2. What would be the wavelength of light emitted if an electron jumped from n=4 to n=2? Problem 10: Regarding the light emitted in problem 9, what is its color? Problem 11: What would be the wavelength of light emitted if an electron jumped from n=4 to n=1? (This will be in the UV range). 

Realize that hydrogen has just one electron. These are the different orbits the electron can jump to if subjected to heat or light. In the early 1900's this was the picture of the atom and the electrons. 

Remember, I did warn you about being confused and feeling mentally limited. Unfortunately, what's been covered so far isn't the worse part. So take a break and then continue at your own risk.  
The spectra of elements showed that light waves also behaved like a particles (photons). Who was brave enough to ask, "If waves could behave like particles, can particles behave like waves?"  
Louis De Broglie was brave enough to ask that question and to come up with an equation that related the mass of the particle to its wavelength. This is very bizarre. Particles are supposed to move in straight lines and behave like particles not waves. Electrons were thought of as particles, but now de Broglie was about to change that.  
To prove that electrons acted like waves, they shot electrons at a salt crystal. As the electrons passed through the crystal, they interfered with each other the same way light waves (and other waves) interfere with each other. Notice that some points are brighter than others. That's because the electrons are interacting with each other to enhance their effect in some regions and decrease their effect in others. It's like waves that add to each other if the crests of the waves cross each other or cancel themselves if a crest of one wave meets the trough (low) part of another wave.  
Remember when I first introduced a joule, I said it was the energy used to push 1 liter of water (1 kilogram) across 1 meter distance so its speed would get up to 1 meter/sec speed in 1 second time. All those words become 1 kg·m^{2}/s^{2} 
Visible light has wavelengths that are too large to detect very small objects. The waves simply go around it. According the de Broglie's formula, the electron will have a much smaller wavelength than visible light, so that we can see much smaller objects . Let's calculate how much smaller. From his equation, we put Planck's constant on top and divide by the mass of the electron times the speed of the electron. Let's say the electron microscope creates a beam of electrons traveling at 2/3 the speed of light. If we do just these calculations, we don't end up with nanometers. If we used the equivalent to 1 joule shown, all units cancel except meters. To get nanometers we multiply by nano over 10^{9}. Now we get the wavelength of these electrons as 0.00364 nanometers. The average wavelength of visible light wavelengths is 550 nanometers. So by dividing the electron's wavelength into the visible light wavelength, we find that electron waves are about 151,000 times smaller. So images will be 151,000 times sharper.  
With 151,000 times smaller wavelengths, the images can be a lot sharper with more detail being shown. Here is an image of a cell taken with an electron microscope.  
How can something
be a particle and a wave? Here’s an animation I made that addresses
this strange concept. This wave motion would be n=1. (roll
cursor over image to see animation). I started off with the electron
as a particle orbiting but also defining a wavelike oscillation. Next I
show just the oscillation without the electron particle. Finally I changed
the wave into an electron cloud representing the probability of the electron’s
position. In textbooks you might see the electron represented by a particle
in orbit or as a cloud. That's because of electron's dual nature of being
both wave and particle. 

Here I am showing
wave and particle nature of a p orbital in a similar manner as before(roll
cursor over image). Similar to what n=2 in the below image. The electron
wave is trapped by the atom like waves bouncing back in forth in a box.
For a wave to be stable, its frequency much match the speed of its reflection. When that happens, the wave appears to be standing still. They call this a "standing wave." 

Bohr’s atom showed electrons as particles similar to planets going around the sun. de Broglie’s atom treats electrons more as waves with wave patterns that fit symmetrically within the atom. In both cases, the energy levels of the electrons must go up by an incremental quantity (quantum). Both of these played a role in the modern theory of the atom and the structure of the electrons. As the progression of elements were built by adding one proton and one electron at a time, the position of the protons was always in the center of the atom in the nucleus. However, electrons repelled each other, so as each electron got added for each new element, they would find a position and shape that maximized their distance from each other. Amazingly, the way they positioned themselves followed a fairly basic pattern. 

To
understand the pattern of how electrons positioned themselves, let's first
discuss pattern recognition. Pattern recognition means making observations
and seeing an underlying scheme. It also requires a person to recognize
similarities and differences. Problem 12: In the sheet music shown, the F chord on the guitar is playing 4 notes. Are any of these notes being played with the G chord? (Hint: look at the fingering chords at the top. See if they have any common dots.) Problem 13: Are any of F chord notes played with the Bb (B flat) chord? 

Problem 14:
What is the "?" in the following progressions? 14a) 1, 4, 9, 16, 25, 36, ? 14b) 1, 3, 6, 10, 15, 21, ? 14c) kilo (10^{3}), mega (10^{6}), giga(10^{9}), tera(10^{12}), peta(10^{?}) 

The above exercise involves simple numbers that progress in a definite pattern. Amazingly, that pattern is the same that determines the size, shape, and orientation of electrons that make up the elements. I find this astonishing that all the elements of the universe are built from this rather simple pattern. There's a huge diversity of elements (gases & metals, various colors, soft & hard, safe & toxic, light & heavy, inert & reactive, and many more differences. However, underneath there's just 3 particles (protons, neutrons, and electrons) and a simple progression of how the atoms of the elements are built. Add one proton with one or two neutrons and place it in the center. Add one electron and place according the pattern shown above in your problem. In the problem the first left value was for "n". This sets the size of the electron, which can also be thought of the size of its orbit.  
n=2, l=0, m=0, s=+1/2 n=2, l=0, m=0, s=1/2 n=2, l=1, m=1, s=+1/2 n=2, l=1, m=1, s=1/2 n=2, l=1, m=0, s=+1/2 n=2, l=1, m=0, s=1/2 n=2, l=1, m=1, s=+1/2 n=2, l=1, m=1, s=1/2 
The "n" sets the size of the electron. This also sets what period (row) the atom will belong in. In your exercise you saw the box with "n=2". I didn't show the +1/2 and 1/2 because I wanted to simplify the pattern slightly. The +1/2 and 1/2 are the two "spins" that electrons can have. With n=2, l can go from 0 to 1. m can go from minus l to plus l. Each one can have a +1/2 and 1/2 spin (s). That makes 8 combinations. That's why there are 8 elements in period (row) 2 on the Periodic Table.  
Again, "n" is the size of the electron and determines what period the element belongs in. "l" determines the shape of the electron. For electrons that have l=0, it means their shape is spherical. If l=1, then the shape is dumbbell shape. If l=2, the shape has four lobes. If l=3, the shape has eight lobes. See below image. The "m" numbers represent orientation. These sets of numbers are called the quantum numbers for that electron. "Quantum" implies a whole number progression. Notice there are no fractions. So the change of electron size, shape, and orientation does not happen gradually but in jumps, which is the essence of quantum chemistry (quantum physics). 

As mentioned, the "l" value determines the shape of the region (or orbital) that the electron will most likely be found. These shapes (orbitals) were given letter names before the idea of quantum numbers were proposed. Below is how the orbitals got their names. Before moving on, realize that electrons are not confined to these regions drawn. The shapes are the boundary where the electrons spends 90% of its time. 10% of its times its farther away. 

The
quality of the spectroscopic lines were labeled sharp,
principle, diffuse,
and fundamental. It was believed that
the different orbitals were responsible for the quality of lines; for example,
orbitals that created “sharp” lines were given the name “s”.
“p” orbitals made the principle lines, etc. This turned out not
to be true, but the names stuck. So these names exist along with the quantum
names of "l=0", "l=1",
"l=2", "l=3"
which are s, p, d, & f. 

The third letter was "m". That tells us the orientation of the electron. Here we have p orbital electrons meaning their "l" equaled one. I'm showing them as double lobes (dumbbells) and as double lobe electron clouds. If m=1, then the electron is oriented along, let's say, the x axis. If m=0, then the electron is oriented along the y axis. If m=1, then the electron is oriented along the z axis. These x, y, z orientations keep the electrons as far away from each other as possible. Now the +1/2 and 1/2 means there can be two electrons in each of these orientations if one has a +1/2 spin and the other has 1/2 spin. So in this picture, there could be a maximum of 6 electrons in these three p orbitals.  
Below are images of p orbitals. Here was see the p orbitals (l=1) for different "n" levels. The "l" level only goes as high as one less than the "n" level. So n=2 is the first chance that l=1 (and p orbitals) is possible. "m" can only go from the negative "l" number up to the positive "l" number. So if l=1, then m is 1, 0, to 1. So there's only 3 orientations. This is handy, since x, y, z are 3 orientations.  
Below are the "n=6 and n=7" levels of p orbitals. Realize that these images are the regions that just one electron occupies. Again, the edges of these shapes are placed to show were the electron spends 90% of its time. It actually goes farther out meaning this boundary is more "fuzzy" than it is being drawn.. Actually, two electrons can occupy this orbital if they have opposite spins (+1/2 and 1/2). I want to credit Mark Winter and Sheffield University and its Orbitron website for creating such great 3d renderings of these orbitals. Visit http://winter.group.shef.ac.uk/orbitron/ to see more.  
Here are the d orbitals, which is when the count of "n" gets to 3 because "l" can then count up to 2 (n1). When "l" gets to 2, then "m" can count from 2 up to +2. That's five orientations. So row (period) 3 of the Periodic Table have elements that can possess d orbitals.  
Below is a comparison of p, d, and f orbitals. f orbitals are "l=3". At l=3 then "m" can go from 3 to +3, which is 7 levels. In the picture below only 6 of those are being shown.  
I had never seen a book or website show an atom with multiple orbitals being represented. I spent several days using Google's 3d Sketchup program to create s, p, and d orbitals. I then put them all together (see below). So what you see is a krypton atom with all the orbitals being represented. The s orbitals are spherical. The p orbitals go off in the x, y, and z direction. The p orbitals for "n=2" fit inside the p orbitals for "n=3". The d orbitals go in 4 directions that fit in between the p orbitals. In summary, the electrons find spaces that keep them as far apart as possible (electrons repel). Someday I will show the animation that I created that shows the orbitals appearing one by one.  
Below is a picture I did that shows all five d orbitals for n=3 together. There's a picture earlier of all five of them separately, but here I put them all together. Again, I've never seen them superimposed, so I was interested in how the looked. It's quite artistic actually.  
The above image is only showing 5 electrons. If each one is holding a pair of electrons with opposite spins, then the image could contain 10 electrons. Imagine gold with 79 electrons. Many of those are f orbital electrons, which are even more complex. So there's a whole whirlwind of activity in just one atom. I'd love to see an animation of that. 

Below it all comes together to show how these quantum numbers actually define the Periodic Table of the Elements (see image below). The first two columns on the left are when l=0, which are s orbitals. "m" is only zero, so there is only one orientation (spherical). The +1/2 and 1/2 spins gives us two elements in this block. The right green block represents "l=1" which is the p orbitals. Since "m" has 3 orientations and there is a + and  spin for each orientation, there is a progression of 6 electrons that cover these combinations. That's why this block is six elements wide. The middle yellow block is for "l=2", which is the d orbitals. As mentioned, m goes from 2 to +2 giving us 5 orientations. With the two spins, it takes 10 steps to fill up the d orbitals, which is why there's a run of 10 elements that go across this middle section. Finally, the Periodic Table always shows a separate block of elements. This is for "l=3", which is the f orbitals. Since m can go from 3 to +3, that's 7 steps. With both spins, that means there it takes 14 electrons to fill up the f orbital and why there are 14 elements in that row.  
I
know this table is too small to read, but if you click on it, the full size
image will load (Remember, you might have to click it to make it full size.
Also use back arrow to come back to this page). This table shows the quantum
number of the last electron added for that element. This table also has
the order of filling for electrons, which is always the lower energy levels
first. This explains why the 3d orbitals are not in Period 3. The
energy level for the 4s orbitals is less than 3d orbitals.
So the 4s orbitals (on the left) come first followed by the 3d
orbitals. So these 3d orbitals show up on Period 4. In the filling order at the top, the 4f orbital energy is just after 6s orbitals. That's why there's an indication on Periodic 6 column 3 "*", that the 4f orbital group comes next. 

The road to get an understanding of the atom was not easy. It required a close look at the light that comes from flames and a brave rejection of the accepted ideas about nature. We've learned that light is not continuous, that energy comes in packets and electrons behave as particles and waves. There's a simple view of electrons orbiting the nucleus and a wave view of them oscillating back and forth in certain regions according to a progression of values of n, l, and m. The motto I get from all this is that "Nothing is as simple as it looks and nothing is as complicated as it looks." 

If you go more into chemistry or physics, the information covered here is valuable. Otherwise, it may be interesting but not that valuable; however, we did touch upon how flame tests reveal the identify of elements. That has some practical uses.  
Here is a product that you add to the
fire in a fireplace, woodstove, or campfire. Notice that it has a warning
about being harmful or fatal if swallowed. It also probably isn't that
healthy to breathe either. A different ad for this product says it produces
blue and green colors (red and yellow are already present in wood fires).
For more information and pictures of a fireplace with these colorglo crystals check out this webpage: 

Problem 18: Water softeners use either sodium chloride (NaCl) or potassium chloride (KCl). The potassium chloride is more expensive but healthier if you are drinking the softened water. In a flame test the chlorine (Cl) does produce light, but it's UV light so you don't see it. The black container shown is filled with salt. Let's say you took a sample and placed it in a flame and you got the purple flame. Is this KCl or NaCl salt? (use the two links above for hints).  
Problem 19: At the front of your textbook, you have
the Periodic Table that is laid out with these colored blocks. What is
the orbital letter name and the "l" number for: Bonus question: What is the quantum number for the last electron of gold (Au)? List n, l, m, and spin. (Hint: Go up 5 panels to refer to the Periodic Table that has all of that information.) 

Ultraviolet light is listed as light with wavelengths between 400nm and 10nm. It is also listed has having energy between 3 electron volts (3eV) and 124 electron Volts (124eV). 3 electron volts is the kinetic energy an electron will acheive when accelerated between 2 plates with a 3 volt difference. In other words, 2 doubleA batteries (1.5 V each) would be able to produce electrons having 3 electron volts of energy. When those electrons slam into a target, they give off light of equal energy. In this case light with 3 electron volts of energy. That energy can be converted to joules, and then we an use the E=hv formula to find frequency, and then use λv=c to find λ (wavelength). However let's use a 9 volt battery which can make UV light of shorter wavelength (more energy). Problem 20: What wavelength (in nm) could be produced with 9 volts? 

Earlier you learned how to turn wavelength into energy in Joules using E=hv. So that gives a hint to first turn 9 electron volts (9 eV) into joules. That conversion is looked up on the Internet. I like http://www.onlineconversion.com/ for conversions. After it is in joules, then you have E, which is equivalent to hv. Let's start with the two pertinent formulas: E=hv and λv=c. This could be an occassion where algebra would be useful. We want to solve for λ (wavelength). Divide both sides by v: λv/v=c/v gives us λ=c/v. We aren't given frequency, just energy (9eV). So let's replace the frequency (v). In E=hv, we can solve for v by dividing both sides by h: E/h=hv/h, giving us E/h=v. So we can replace the v in λ=c/v with E/h. That gives us λ=c/(E/h). This format is a little awkward because we are dividing by a fraction. If this were 5 ÷(3/4), were would invert the 3/4 and multiply (5 x 4/3).We can do the same here so c ÷E/h becomes c x h/E. After that we get λ=hc/E. So in a spreadsheet the speed of light (c) and Planck's constant goes in the numerator and the Energy goes in the denominator. We add a few more cells to do needed conversions.
Problem 22: What formula is placed in A3? 

Congratulations on getting through this tutorial and quiz. If in my CHM151 class, send answers to chm151@chemistryland.com. Use subject line of "Chapter 7Modern Atom"  
Number of different visitors (not just hits) since Nov 9, 2009