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.