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Unformatted text preview: Physics 143 Fall Term 2007 HOMEWORK 12
Due: December 7 Problem 1: (BFG Ch.10 Problem 13)
Calculate the Fermi energy for Cu and Lead. The Fermi energy is: EF where n is the density of free electrons: 2m 3 n
/ M n Am N n n V V Am where n is the number of free electron per atom and m is the mass of one nucleon. 1.67 10 kg m Therefore, Cu: n 1 8.95 10 m 8.44 10 m n 63.5 1.67 10 Am EF C Pb: Notice that: 2m 3 n
/ 1.12 10 J 7.0eV n A Therefore the Fermi energy of Pb is scaled by the factor: EF n
/ / / n n P C P A C C A P EF C / 2 11.4 63.5 1 8.95 207.2 0.96 10 J 0.848 5.9eV EF P 0.848 Problem 2: (BFG Ch.10 Problem 14)
The Fermi Energy. This is actually what we did in the last problem: EF where n is the density of free electrons: n where # of e volumn 2m 3 n
/ mass mass per atom volumn e per atom # of atom volumn e per atom density and mass per atom # of nucleon mass per nucleon . That is: P a g e  13 M Am N n n V Am V where n is the number of free electron per atom and m is the mass of one nucleon. n EF 2m 3 n Am
/ 3 2m m / n A Const. n /A And the prefactor is: Const. 3 2m m 4.12 10 J kg/m 0.26eV kg/m / Problem 3: (BFG Ch.12 Problem 3)
The rms speed of the molecules. The rms speed of monatomic molecules is: v For each inert gas, the rms speed is: Helium (A=4): v Neon (A=20): v Argon (A=40): v Krypton (A=84): v Xenon (A=131): v Radon (A=222): v 3kT m 3kT Am 2.23km/s A 2.23km/s 4 2.23km/s 20 2.23km/s 40 2.23km/s 84 2.23km/s 4 2.23km/s 4 1.16km/s 0.50km/s 0.35km/s 0.24km/s 0.19km/s 0.15km/s P a g e  23 Problem 4: (BFG Ch.12 Problem 20)
The Boltzmann distribution: Due to the Boltzmann Distribution: n ~e In the twostate model: n n e E/ T E E / T E ,E E 0. Therefore, Since E for all T (there is no negative temperature), n e E E / T 1 n That is there is no temperature for which the fraction of needles in State 2 exceeds the fraction in State 1. Problem 5: (BFG Ch.12 Problem 23)
The twostate model: TwoState: State 1: Aligned: E 2eV State 2: Antialigned: E 2eV The fraction of the needles in each state: n e E E / T e V/ n When there are twice as many as needles aligned as antialigned: n e V/ T 2 n 4eV T 6.7 10 K k ln2 T Problem 6: (BFG Ch.12 Problem 31)
What is the probability of finding an electron occupying the ground state in a box at different temperature? (Compare probabilities of being in the first excited state and being in the ground state.) The energy state of 3D box is: E The ground state: 3 1.79 10 J 112eV 2mL Considering the elections with spin, the degeneracy of the ground state is 2. The first excited state: 3 E E E 224eV mL We consider it as a twostate problem, so the fraction of two states is: n 6 E E / T V/ T P e 3e 2 n E
P a g e  33 n 2mL n n T=10K, T=273K, P P 10 9 10 0 0 0 T=24000K, P 9 10 Which means even in very high temperature, comparing with the first excited state, the probability of being in the ground state always goes to 1. November 20, 2007 P a g e  43 ...
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This note was uploaded on 04/27/2008 for the course PHYSICS 143L taught by Professor Scholberg during the Fall '07 term at Duke.
 Fall '07
 Scholberg
 Energy, Work

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