Just for fun

LASERS PROVIDE ANTIMATTER BONANZA
A research team used lasers to produce more positrons
(anti-electrons) inside a solid than any previous experiment,
according to the researchers involved. In the 13 March Physical
Review Letters, the team describes firing short pulses from an
intense laser onto thin gold targets and creating a high-density
positron source that could be used to investigate exotic phenomena
near black holes or supernovae.

Hui Chen et al., Phys. Rev. Lett. 102, 105001

COMPLETE Focus story at http://focus.aps.org/story/v23/st8

Studying for the Final + Practice Problems

Several Things I would recommend for studying for the final:

First, go through all posts of the last month on this site, especially things like "Notes on..." or "...what we should know". Try to understand those posts and ask questions, as comments on this web site, regarding anything you do not understand. The more specific the questions are, the better, but just ask about anything that you are confused about. This is very important.

If you ask via a post here, please be clear about what topic and post you are referring too. I will also try to check for questions at the end of all recent posts.

Second, try to understand the last 2 quizzes and most problems from all HW since the midterm, as well as any HW on the relationship of temperature to microscopic motion and on energy and especially potential energy from the first half of the class. Ask questions about that also via comments to this post.

Topics that will be emphasized for the final are:

1) quantum physics: implications of the uncertainty principle; quantum jumps that involve light emission or absorption; and maybe something related to spectroscopy or color.

2) the nature of conducting materials; the difference between a metal and a semiconductor or insulator from an atom/electron counting point-of-view. Bonus if you understand the relevance of the uncertainty principle here.

3) Simple circuits: the relationships between I, V, and R; and also energy disipation and power in circuits P=V I (Watts). Understand that I = amperes = coulombs/second; and P= Watts = Joules/second. Know what that means.

4) the motion of a mass in a potential energy, like the problem (5) on the midterm. If you are truly comfortable with that, it will be valuable.

5) possibly something related to temperature and how, in absolute units, it is fundamentally related to atom motion (in a gas).

Finally, keep checking this site for updated information and, especially, discussion stimulated by student questions.

I'll add practice problems here as I come up with them. These are only a supplement to the above guide and topic outline. (As discussed in class, it is ok to bring a 3x5" card of equations, relationships and units to the final. Please don't get carried away.)

For a circuit with a battery, wire and resistor, describe the what happens in the circuit including the current flow and especially the energy conversion processes.

For a circuit with a battery, wire and a light bulb, describe the what happens in the circuit including the current flow and especially the energy conversion processes.

For a circuit with a 10 Volt battery, wire and 5 Ohm resistor, how much current flows though the resistor? How much heat energy, in Joules, appears in the resistor each second. How much in 5 seconds?

For a circuit with a 10 Volt battery, a resistor and wire (all in series), suppose there is 3 coulombs per second flowing through the wire. How much current flows though the resistor? How much heat energy, in Joules, appears in the resistor each second. How much in 4 seconds?

extreme extra credit: Describe the energy conversion processes associated with an LED (light emitting diode). Is there a quantum jump involved??

For blue light, what is the wavelength, frequency and energy of a typical photon?

For red light, what is the wavelength, frequency and energy of a typical photon?

Discuss the relationship of temperature to microscopic atom motion in a noble gas.

For a composite gas, like air, in thermal equilibrium, do heavier atoms tend to move more slowly than lighter atoms? Explain why. What are your basic assumptions?

Explain the uncertainty principle and its relevance to understanding the origin of the size of atoms.

What is the wavelength, frequency and energy of a green photon?

Present an illustrated discussion of atomic spectra, including what they are, why they were unexpected, and what people infer from atomic spectra regarding the nature of the energy levels of an atom.

Suppose an electron in an atom has allowed quantum energy levels at exactly 2 eV, 4 eV, 5 eV and 5.5 eV. If 2 eV is the energy of the ground state and all the atoms in a cold gas start out in the ground state, what are the a) energies, b) frequencies and c) wavelengths of possible quantum absorption events in which a single photon is absorbed and the electron jumps from the ground state to another state? (actually, you can start with d) if you like and then do parts a), b) and c)...)
d) How many sharp lines would there be in the absortion spectrum for this atom?
e) Make a graph of an absortion spectrum for this atom.
f) Are these absorptions in the infrared, visible or UV? If visible, what color are they?

For green and orange light, what are the wavelengths, frequencies and energies of typical photons, respectively?

What is the composition of He? What is the composition of an He+ ion? What is the difference between He+ and He? What is the difference between He+ and H?

Balmer Series: emission spectrum

Here is a graphic of an emission spectrum set-up and an emission spectrum. For an emission spectrum, the gas of atoms you wish to study is the source. Photons are emitted from the hot gas of atoms when electrons in the atoms drop down from a higher potential energy state to a lower potential energy state. Energy is conserved, and the electron energy is converted to photon energy in this transition. The photon is created in this process. It does not exist before this transition takes place; it does exist after the transition. The electron falls from a high energy (excited) state to a lower energy state; the energy it loses goes to creating the photon. These photons come out at very specific frequencies.

The role of the prism is to separate the photon beam into its different colors; before the prism all the photons are moving together in a single light beam. After the beam goes through the prism, whatever colors are present are separated, as shown.

The moving aperture allows only one frequency (color) to pass though at a time. As it moves it scans through the spectrum. The graph shows the Intensity measured at the detector as a function of frequency.

An absorption spectrum is the complement of this. The atoms are located in the beam and instead of emitting photons they absorb photons in a process opposite to that of emission. That is, the electron starts out in a low-energy state of the atom and goes to a higher energy state. A photon is anihilated (absorbed) in this process, thus energy is conserved via the equation E1 + hf = E2, where E1 is the energy of the state the electron starts in and E2 the energy of the state it goes to. hf is the energy the photon gives to the electron, after which the photon ceases to exist. (This is a "zero sum game". )

This leads to dark "absorption" lines, in an otherwise broad and featureless reference spectrum. These dark lines are due to the quantum absorption process described in the preceding paragraph. Using the energy it gets from the photon, the electron makes a quantum leap upward to a higher energy state.

Balmer Series Quiz Solution and comments

Let's start by talking about energy units and then doing parts 1) and 4).
We use f=c/wavelength for 1), and then E=hf for 4). Almost everyone got 1), but for many people there was some confusion regarding units of energy for 4). Electron Volts (eV) is a unit of energy that is not the same as Joules. 1 eV = 1.6 x 10^-19 Joules, so an eV is a much smaller unit of energy and one that works well for atoms in the sense that it gets rid of the large negative exponents that occur if one uses Joules.

eV has its origin in the potential energy of an electron, so it makes some sense that it is close to the right size for electrons in atoms; Joules, on the other hand, has its origin in the kinetic energy of a 2 kg mass moving at 1 m/s, and is way off (too big) for atom energies.

Now to the problem:

f=c/lamda = (3 x 10^8 m/s) /656 x 10^-9 m = (3/656) x 10^17 sec-1 = 4.57 x 10^14 Hz .

Getting the exponent right in the 3rd step is critical. Exponents are very important!
For the others you should get: 6.17, 6.91 and 7.31 x 10^14 Hz, respectively. (All are multiplied by 10^14. All are in cycles per second, which is the same as Hz and sec-1.)

4) For 4, you multiply each one by h=4.14 x 10^-15 eV-sec to get the energy in eV.

E= h f = 4.14 x 10^-15 eV-sec 4.57 x 10^14 Hz = 18.9 x 10^-1 eV = 1.89 eV.

[Note that this is a nice, friendly number. No exponents.]

The other ones are: 2.56 eV, 2.86 eV and 3.02 eV.
These photon energies correspond to something like: 1.89 eV (red or maybe orange), 2.56 eV (greenish blue?), 2.86 eV (blue) and 3.02 (violet).

The answer is very different in Joules. For example, for the first frequency you would get about 3 x 10^-19 Joules.

We could just forget about eV if people would rather work with Joules. There is sort of a tradeoff between the extra work and confusion of learning a new unit for energy, and the benefit of using a unit (eV) that is more atom friendly. Please let me know your preference via comments here.

Atoms: what we should know, part II

As we discussed today in class, for an electron in an atom there is a sequence of states, or energy levels, in which the electron can reside. (See part I.) These wave-like states have characteristics which are reminiscent of the modes of a string of length L stretched between two posts.

5) The lowest energy state in this sequence is called the ground state. Rather amazingly, the ground state has kinetic energy.

6) This is one of the most surprising results of quantum physics and is intimately related to the uncertainty principle.

7) The uncertainty principle says that the more an electron is confined, the higher its momentum will be. Momentum, speed and kinetic energy (K.E.) all go together, so confining an electron causes it to have kinetic energy, and the more confined it is the more kinetic energy it must have.

8) Just as the fundamental mode of a guitar string has a higher frequency when the string is shortened (f=v/2L), so will an electron have a higher kinetic energy when it is more confined (K.E.~h^2/mL^2). Here L is the length of the string, which is a measure of the confinement of the string wave; and L is the diameter of the electron cloud around the atom, which is a measure of the confinement of the electron. (Smaller L in each case means more confinement.)

9) Note that in each case there is an inverse relationship: smaller L means higher frequency (for the fundamental mode of the string, smaller L means higher kinetic energy for the electron in the atom.

Atoms: what we should know, part I

1) Atoms are composed of a heavy, fixed nucleus which has a positive charge and electrons which are attracted to the positively charged nucleus. Our attention is focused on the electrons, which are light and therefore behave in very interesting and unexpected ways.

2) The electrons in an atom must exist in states which have specific energies. These are called discrete energy states. (Discrete, in this context, means isolated, detached or separate (from other states)-- not part of a continuum.)

3) Transitions in which an electron goes from one discreet energy state to another are the reason behind the sharp-line spectra seen in absorption and emission spectroscopy.

4) The existence of these discrete energy electron states is explained by a "wave theory" of electrons (~1930), which leads to a lowest energy state known as the "ground state" (fundamental), and a sequence of states at higher energies ("harmonics"). This is analogous to the theory of a wave on a string, which leads to a mode of lowest frequency (fundamental) and a sequence of discrete harmonics at higher frequencies.

5) ...to be continued... (see part II)

Quiz due monday --Balmer Series

This is a take-home quiz that is due Monday (at the beginning of class):

Balmer Series Quiz, Due Monday March 9, 2009

In the emission and absorption spectra of a hydrogen atom gas, sharp lines are seen at the following wavelenghts: 656 nm, 486 nm, 434 nm and 410 nm.

1. Determine the light frequency (in Hz) corresponding to each of these wavelengths.

2. a) Sketch a spectroscopy set-up for an absorption spectrum, and then graph:
b) a reference spectrum, and c) an absorption spectrum (intensity vs frequency). Start your frequency axis at zero for both graphs. For the absorption spectrum there is a gas of cold atoms in the beam; for the reference spectrum there are no atoms in the beam and no sharp features.

3. Sketch a spectroscopy set-up for an emission spectrum, and graph an emission spectrum (intensity vs frequency) for the same four line sequence. For an emission spectrum the light source is a hot gas of H-atoms.

4. Using h= 4.14 x 10^-15 eV-sec, calculate the energy of a photon at each of these four frequencies in eV. eV (electron-Volts) is a energy unit which is good for atoms.)

Spectroscopy

Optical spectroscopy

Electrons exist in energy levels within an atom. These levels have well defined energies and electrons moving between them must absorb or emit an energy equal to the difference between them. (Energy is conserved.) In optical spectroscopy, the energy absorbed to move an electron to a more energetic level and/or the energy emitted as the electron moves to a less energetic energy level is in the form of a photon (a particle of light). Because this energy is well-defined, an atom's identity can be found by the energy of this transition. The wavelength of light can be related to its energy. It is usually easier to measure the wavelength of light than to directly measure its energy.

Optical spectroscopy can be further divided into absorption, emission, and fluorescence.

In atomic absorption spectroscopy, light is passed through a collection of atoms. If the wavelength of the light has energy corresponding to the energy difference between two energy levels in the atoms, a portion of the light will be absorbed. The relationship between the concentration of atoms, the distance the light travels through the collection of atoms, and the portion of the light absorbed is given by the Beer-Lambert law.

Waves on strings: websites

Here are some links to web sites where you can explore the nature of waves on strings. Understanding waves (on strings) is relevant and helpful for understanding the quantum physics issues we will be covering for the next few days.

http://www.ngsir.netfirms.com/englishhtm/StatWave.htm

http://hyperphysics.phy-astr.gsu.edu/hbase/Waves/string.html

http://id.mind.net/~zona/mstm/physics/waves/standingWaves/standingWaves1/StandingWaves1.html

Homework help section Thursday. 6 PM

There is a HW help section as usual in Nat Sci 2 annex room 101 at 6 PM.
HW for this week is posted tow posts below this.

Homework on Conducting Materials and Circuits: Solutions and Discussion

This includes solutions to the HW problems from last week. and some additional discussion contextualizing and nuancing the solutions.

Problem 1. a) Describe the nature of copper (Cu) and aluminum (Al) paying particular attention to how many electrons remain localized (with an atom core) and how many electrons go into non-localized states and become part of the electron sea.

a) Cu has 29 electrons. 28 of those stay with their atom core, which then becomes a Cu+ ion (since it has 29 protons and only 28 electrons). One electron per Cu atom goes into an electron sea. The electrons of the electron sea are not associated with any particular atom core (Cu+ ion) and, indeed, their wave-functions extend through the entire solid. These non-local electrons in the electron sea are the electrons that enable the conduction of electricity.

b) They are there waiting to conduct electricity all the time. When a voltage is applied, they move freely, but they are always there waiting to flow. The Cu+ cores form a rigid crystal lattice. The non-local electrons are like a liquid waiting to flow when a voltage is applied, which is like tilting a pipe with water in it.

c) It difficult to explain in simple terms why some electrons leave their atoms, and go into the non-local states of the electron sea, because it involves the wave nature of electrons and the uncertainty principle. Basically, the benefit, and motivation for some electrons to do that, is that by going into non-local states the electrons can lower their kinetic energy. This is because non-local states have very uncertain position and thus can have very low momentum and kinetic energy (which is proportional to momentum squared).

Al is similar in all respects except that the details of the electron counting are different. For Al, 3 electrons per Al atom leave “home” to go into the non-local states of the conduction electron sea. This leaves behind a simple 10 atom Ne-like core which is the ion Al+3, since there are 13 protons and only 10 electrons.

Cu has a less simple core. In the Cu core there are 18 electrons in a argon-like “noble gas” configuration, and 10 more electrons in a filled 3d mini-shell. Don’t worry if you don’t know what that means, and if you do, great! These d mini-shells come up once you go beyond the lightest elements, and it is just a complexity we have to live with if we want to relate our basic knowledge to real materials.

2. a) Describe the nature of pure silicon (Si). How many electrons does it have? What are their roles?

2) Si has 14 electrons. It forms a crystal lattice in which each atom has 4 nearby neighbors. Si atoms are very social and like to share an electron with each of their 4 neighbors. These shared electrons form strong bonds between Si atoms which are called covalent bonds. (This is the strongest form of chemical bond.) Anyway, as far as counting goes, each Si puts a total of 4 electrons into covalent bonds (one for each of its closest neighbors) and the remaining 10 electrons form a neon-like core. There are no conducting electrons in Si.
b) What happens when you mix in some phosphorous (P) atoms in silicon? Suppose the phosphorous atoms substitute for some of the Si atoms. How and why does that change the nature of the material and how does it affect its conductivity?

P has 15 electrons. It wants to fit in. It substitutes for a Si atom. P wants to fit in so it too provides 4 electrons to covalent bonds with neighbors and has a 10 atom neon core. But there is one electron left over. That one goes into a non-local state joining other electrons form other P atoms in an electron sea that can conduct electricity. This is where the conductivity of the semi-conductor Si comes from. It is from the extra electron in the phosphorous “doping” atom.

3. (extra credit) What are unusual macroscopic and microscopic characteristics associated electrical conduction of a superconductor?

On a macroscopic level, superconductors are unusual in that they conduct electricity with ZERO resistance. Other metals have resistance. Superconductors have no resistance at all. They are “frictionless” even if there is some disorder in their structure. This was very surprising. It took more than 50 years for scientists to solve the mystery of superconductors. Part of the solution leads to the surprising result that electrons in superconductors pair with each other! On a microscopic level the electrons in a superconductor are paired. For reasons that are difficult to explain and related to a quantum property known as “phase” pairs of electrons move with zero resistance.

4. Gold is difficult since it has so many electrons to wonder about. Wow, it has 79 electrons and the nearest noble gas is Xe, which has 54 electrons. The bottom line is this : 1 electron per Au atom goes into the electron sea. Those are the ones that are responsible for the conductivity. Of the ones left behind, 10 are in a filled d-level “mini-shell”, 14 are in a filled f-level “mini-shell”. I am amazed that the counting then seems to work: 54 + 14 + 10 +1 = 79. The last 1 is the one that counts for conductivity. Sorry. Didn’t realize this would be quite so hard. My mistake. I thought it would be interesting because gold is a familiar material and a good conductor. Copper, silver and gold are all in the same column of the periodic table and all similar metals.

5. Ge is like Si except the noble gas core is the next one down and there is a filled d-level mini-shell in addition to the 4 electrons in covalent bonds (as in Si). So it is a semiconductor with a phenotype very similar to Si.

6. This is very hard problem, but important since GaAs is a very common and useful semiconductor (lasers, LED’s…). Ga and As are on either side of Ge, so if each As gives each Ga one electron then they can formed a semiconductor with 4 covalent bonds to nearby neighbors just like Si and Ge. In the since that each Ga receives one electron from each As, it is sort of like a NaCl, where each Na loooses an electron to each Cl forming a lattice of Na+ and Cl-. Through this ionic electron exchange, however, it becomes a semiconductor, like Ge and Si. This raises the interesting question "Why isn't NaCl a semiconductor?" which we won't answer here.

Chapter 23, problems:

2) I=V/R can be rearranged as R=V/I.

R=120 Volts/20 amps = 6 Ohms.

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3) P= I V. 1200 Watts = I x 120 Volts. I = 1200 Watts/120 Volts =10 amps.

R=V/I = 120 Volts/10 amps = 12 Ohms

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4) Capacity is 60 ampere-hours. Headlights draw 6 amps.

6 amps x 10 hours = 60 ampere-hours, so they can do this (draw 6 amps) for 10 hours.

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6) P=I V . In this case 4 Watts = I x 120 Volts, so I =4/120 = (1/30) amps

b) R=V/I = 120 V/(1/30 amps)=30x120 Ohms = 3600 Ohms.

c) 4 Watts = 4 Joules/sec. There are about 365 x 24 x 60 x 60 seconds in a year. So the total power utilized in one year is 4 x (365 x 24 x 60 x 60) (Joules).

d) to get watt hours in one year we multiply 4 watts times the number of hour in a year, which is 365 x 24. Then divide by 1000 to convert from watt-hours to kilowatt-hours.

Total cost would then be: $0.15 x [4 x (365 x 24)].

7) Power is I V which is 9 amps x 110 V = 990 Watts. Watts is just an word that means Joules per second. To find the number of Joules generated in one minute we multiple Joules/second by 60 seconds to get 990 Joules/sec x 60 seconds = 60 x 990 Joules.

8) 9 amps is just another word for 9 coulombs per second. Total number of coulombs in one minute (60 seconds) is: 9 coul/sec x 60 = 540 coulombs.

Error checking is welcome. Have I made any mistakes on these? Please feel free to comment.

Homework on Light; newest version with solutions added.

This is the final up-to-date version of this assignment with solutions added. This assignment was due on Friday (March 6).

Ch 26. Problems: 3 (1st part only), 5, 6, 7
[hint for 5: radio waves travel at the speed of light=3 x 10^8 m/s. time is distance/speed.
For 7, someone in the class got a wavelength to atom size ratio of about 10,000. Does that seem about right? Feel free to post your result and comment.]

3. time =distance/speed of light = 500 sec

5. time =distance/speed of light = 1.4 x 10^8 sec (a much longer time than 500 sec)

6. f=c/lamda = 5 x 10^14 sec-1 . It is important to get the exponent right!

7. lamda =c/f = 0.5 x 10^6 meters = 500 nm. In these problems (6 and 7) it is important to get the exponents right and to be mindful of any nm to meter conversions. Atom size is typically 0.1 to about 0.3 nm. So that is about 1000 times (or a little more) smaller than the wavelength of visible light.
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Ch 30: Exercises: 3, 5, 15, 48
Problem "1" (there is only one problem, it has no number)

3. blue light emitted corresponds to a greater change in energy of the electron in the atom.

5. If you double with wavelength, the frequency decrease by a factor of 2 (f=c/lamda), so the energy of a photon would be half as much. (Not twice.)

15. infrared is lowest frequncy range, UV is highest frequency and therefore has highest photon energies. Visible is in between IR (lowest photon energies) and UV (highest phton energies).

48. see attached jpg image. After you do this you can sort of see the solution visually from the picture!

Problem: One key thing for this problem is to realize that wavelength is inversely related to frequency, so if the frequency for a transition is higher, the wavelength must be shorter!

Balmer Series Problem: [Because this problem was posted late, it will count as extra credit and, because it is long, triple credit. I suggest you do it because it is relevant to things we will emphasize regarding both the quantum nature of the atom and spectroscopy.]

In the emission and absorption spectra of a hydrogen atom gas, sharp lines are seen at the following wavelenghts: 656 nm, 486 nm, 434 nm and 410 nm.

a) Determine the light frequency (in Hz) corresponding to each of these wave-lengths.

b) Sketch a spectroscopy set-up for an absorption spectrum, and then graph an absorption spectrum of intensity vs frequency. (This means intensity as a function of frequency, with intensity on the vertical axis and frequency on the horizontal axis. It is highly preferable to start the frequency axis at zero frequency; that way it is easier to see patterns of frequencies.) For your sketch of the spectroscopy set-up, you can limit it to a source, a source aperture, a prism, a moving aperture and a detector. (The book shows mirrors and that is more correct, but for your sketches you can skip the mirrors for simplicity.)
Remember that for an absorption spectrum there is a cold gas of atoms in the beam. Actually, there should be two graphs: one of the intensity vs frequency when there is no H-atom gas in the beam, which is called a reference spectrum; the other the actual absorption spectrum that you get when the H-atom gas in in the beam. (For the reference spectrum assume a broad, featureless spectrum spanning the entire visible range as well as some IR and UV.)

c) Sketch a spectroscopy set-up for an emission spectrum, and then graph an emission spectrum of intensity vs frequency for the same four line sequence. For an emission spectrum the source is a hot gas of H-atoms.

d) Using h= 4.14 x 10^-15 eV-sec, calculate the energy of a photon at each of these four photon frequencies in eV.

Reading on light, spectroscopy and atoms

Please read the posts below on this web site on: Notes on light and photons, and Notes on atoms: part1, atomic spectra.

In addition, there is related material in our conceptual physics book is in chapters 30, 31 and 32. Chapter 30 has a discussion of emission and absorption spectra. Chapter 32 (page 623) has a brief discussion of a pattern observed by a Swiss schoolteacher, J. Balmer, which we will cover in class (Atomic Spectra: Clues to Atomic Structure). You can find out more about that by searching on the phrase "Balmer Series", which is very famous and probably the most important experimental result which lead to the discovery of quantum theory.

This is a difficult and confusing subject and you are very much encouraged to post any questions you have either here, or at the end of the posts: Notes on Light..., or Notes on Atoms...

Be sure to come to Wednesday's class.

Wednesday's Class ---Don't miss it!

We will have a special visitor on Wednesday giving a presentation designed to elucidate the nature of atoms and atomic spectra to an audience of non-scientists. Please don't miss this class. I think it will be very interesting and very relevant to what we are learning for the rest of the quarter.

Notes on Atoms: part 1, atomic spectra

Today (Monday) we began our treatment of the quantum atom. I am hopeful that this will be the most interesting and important part of this conceptual physics course. Understanding the quantum nature of matter is a big thing.

This has two aspects. One is bizarre experimental results which were discovered in the 1900's. In particular spectroscopic measurement showed the presence or absence of sharp lines at particular frequencies (colors) in emission and absorption spectra respectively. This were quite a surprise and mystery, and they were completely unexplained until the wave concept for electrons was imagined and the wave equation for electrons* was discovered. *Also known as the "Schrodinger equation".

The other aspect focuses on the theoretical nature of atoms when the wave-like character of the electron is appropriately incorporated. In this class we will not do the wave calculations of the Schrodinger equation --that is done in physics 101 and 139-- but what we can and will do is look at the consequences of the wave theory: the phenomenology of atoms which emerges after the grungy math is done, which is really the most interesting part!

On Monday we discussed the nature of a spectrometer --how it involves a source, an aperture, a prism, a second aperture, and a detector. The 1st aperture "colimates" the beam, making it so that all the light the gets through it is going in the same direction. The prism splits the source beam into colors (frequencies). The second aperture slowly moves across the beam allowing only one color, or frequency range, to get through at a time. Then the detector measures the intensity of what gets through. By measuring what gets through (intensity) as a function of the position on the 2nd aperture one obtains what is called a "spectrum": a measured graph of Intensity as a function of frequency.

There are two types of spectra that are important to us. One is an emission spectrum, in which the light source is a heated atomic gas. In this case one will see zero intensity at many frequencies (or a very low non-zero background) punctuated by dramatic narrow spikes (peaks) at which there is a lot of intensity at a particular frequency. These are called "bright line spectra". "Line" refers to the narrowness of the peaks; they are so narrow they look like vertical lines in the graph. "Bright" means high intensity.

The other type of spectrum is an absorption spectrum. For these the source is essentially an ordinary light bulb, i.e., a hot piece of metal. This will emit a continuous spectrum with no sharp lines or anything dramatic. The drama begins when you put a cold gas of atoms in the beam. One then finds that the gas of atoms has absorbed almost all of the light at very specific frequencies, thereby creating lines in the spectrum where the intensity that reaches the detector is close to zero. These are called "dark line spectra".

We looked at both emission and absorption spectra for the example of a hydrogen atom where there were sharp lines at: 4.6 x 10^14 Hz, 6.2 x 10^14 Hz, 6.9 x 10^14 Hz and 7.3 x 10^14 Hz.
These correspond to red, green, blue and violet light respectively. On Wednesday we will begin the task on unravelling the mystery of where these sharp lines come from and why.

On Wednesday we will have a special visitor, Nina X. McCurdy, who has been working on explaining the origin of quantum spectra to a general audience. This will be a very important class. Please do not miss it!