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.)