Problem 9.1 This is a variation on the example on pages

Transcription

1 Problem 9 his is a variation on the example on pages 00-0 n n g g exp We are given that n()/n()=/000 and we want to solve for Also, we know that g()= and g()=8 We want to solve exp for I'll set this up more generally n n g exp g ln n n g g ln n n g g kln n n g g Now substitute values and get the answer 6 6 n 000 n g g 8 k since we are working with ev 's n g kln n g = 7 0 Problem 9 his is another variation on the example on pages 00-0 n n g g exp We are given =5000 and we want to solve for n()/n(), etc n ( n) n gn ( ) n k since we are working with ev 's

2 ratio( m, n) gm ( ) gn ( ) exp ( ( m) ( n) ) ratio(, ) = (just checking) ratio(, ) = ratio(, ) = ratio(, ) = 5 0 vbar vbar = m/s Problem 97 vrms vrms = 6 m/s 7 0 Problem 98 k Boltzmann's constant in units of ev/k he average energy per molecule is E k E 008 = ev his is much less than the 0 ev needed to raise a hydrogen atom from its ground state to its first excited state: E = Problem 99 k Boltzmann's constant in units of ev/k E 6 Binding energy of hydrogen in ev We need to solve E E k k for = or about 0500 K

3 Problem h 66 0 k 8 0 Boltzmann's constant in mks units m E λ λ k h p 0 0 mass of oxygen average oxygen molecule energy p m E oxygen momentum λ = 6 0 de Broglie wavelength = 0065 so the de Broglie wavelength is only about 65 percent of the molecular diameter Problem 9 he flux is the number of neutrons per square meter per second If we divide the flux by the average neutron velocity, we get the number of neutrons per cubic meter at any time in the beam port: neutrons volume neutrons seconds_for_neutron_to_travel meter m s _meter m n k 8 0 S 0 According to Maxwell-Boltzmann statistics: vbar ρ 8k π m n S vbar ρ = neutrons/m Problem 99 According to the Stefan-Boltzmann law, an object at a temperature radiates an energy R per second per unit time R e σ where e is the object's emissivity and σ is a constant For skin at two temperatures, R R ratio ratio = 0 A percent difference; not huge, but measurable

4 his is "just like" problem ratio Problem 90 ratio = 8 almost a factor of two different R e σ R e σ Problem 9 R R R R 00 7 = 8006 K or 7 = 576 C Problem 9 e σ R e σ R = 9 0 his is in units of watts/square meter o find the rate at which this particular sphere radiates, we need to multiply R by the surface area of the sphere r 5 0 rate R π r rate = 707 watts Problem 95 In the absence of a given emissivity, let's treat the hole as a blackbody, so that Also σ R e σ e

5 R = his is the number of watts/m radiated by the hole o find the rate at which radiation escapes from the hole, multiply R by the area of the hole Don't forget to convert cm to m rate R 0 0 rate = 508 watts Problem 96 R R e σ e σ R R R R = 067 kw Problem 90 According to Wien's displacement law, λ max in MKS units λ max λ max = meters his wavelength of 79 nanometers is lower energy than the lowest energy visible photon, which has a wavelength of about 00 nm hus, these photons are in the infrared See figure on page 5 for a confirmation of this λ max Problem equation 90, page 6 = about 0000 K λ max 5

6 λ max Problem 9 equation 90, page 6 λ max = about 90 nanometers his wavelength is in the infrared; see figure on page 5 Problem 98 he average energy of the free electrons in copper can be found from equation avg 5 avg = ev he average energy of free electrons at room temperature, according to Maxwell-Boltzmann statistics, is bar k Beiser says to use k=005 ev, so bar 005 bar = 008eV he actual free electron energy in copper is bigger by a factor of avg = 6 bar than the average energy of a gas of free electrons at room temperature Problem 99 Part a he average energy is given by equation avg 5 avg = 06 ev Part b he temperature corresponding to this energy is given by equation 9 k Boltzmann's constant in ev energy units avg k = or about 5600 K 6

7 Part c It's safe to do a classical calculation of the velocity of an electron having an energy of 06 ev, because the energy is very much less than the electron rest energy of 5 kev E avg first convert energy to joules From E=mv /, we get v E 9 0 v = m/s a fairly high, but still nonrelativistic velocity Problem 90 Part a Use equation 950 his is relatively easy in Mathcad, not so easy with paper and pencil Let's just pick N= N e k e 00 his is what Beiser uses for a warm "room" temperature n(, ) N 5 F exp Just for kicks, let's plot n() See the end of the document for Beiser's solution = his gives me a reference for scaling 0 9, set a range of energies n, his looks reassuringly like figure 90 he numbers are big because I haven't multiplied by a d Electrons in excited states are those with >F Since we defined N=, we can find the fraction in excited states by integraging n() from F to infinity 7

8 fraction fraction n (, )d n(, )d I did this on purpose Mathcad gripes because some numbers involved in calculating n() are getting too big, so I should integrate not to, but to some "big" number, say times F If I go any bigger than F, I get an overflow fraction = 8 0 about 0% Part b At =08K, 08 7 n, fraction 0 n(, )d Now I can go higher in energy without overflowing fraction = 007 About 7% he above solution is too much for a paper-and-pencil type test problem All of the excited electrons will have energies just slightly greater than F herefore, it is a reasonable approximation to replace in equation 950 by Also, for greater than, the in the denominator of equation 950 is much smaller than the exponential he equation then becomes n( ) N F F exp exp 8

9 he antiderivative of n() is N F F exp exp When you integrate from F to infinity, the value of the antiderivative is 0 at, and evaluated at F it gives integral( ) ( k ) N integral( ) ( k ) N F exp exp Evaluating at the two temperatures: integral( 00) 55 0 = or about 055% integral( 7 08) = 005 or about 5% I would expect that if I gave you the appropriate hints as to what approximations to make, that you could do this calculation for an exam Problem 9 he Fermi energy is calculated from h F m or h m N 8 π V n 8 π where n=n/v is the density of free electrons We can get the number of aluminum atoms per volume by multiplying the density by the atomic mass: n atoms volume mass atom volume mass Don't forget to convert grams to kilograms and centimeters to meters According to able 7, aluminum has s and p electron Beiser tells us that the s and p electrons are close in energy, so we expect all three electrons to become part of the free electron gas hus, we need to multiply the number of atoms by to get the number of free electrons in aluminum h 66 0 m here's where effective mass enters in 9

10 n he /0 converts g-->kg, the (0 ) converts cm -->m, and the/66*0-7 converts amu-->kg h F m n 8 π = Joules = 055 Fermi energy in ev 0