Chapter 9 Problem Solutions

Size: px

Start display at page:

Download "Chapter 9 Problem Solutions"

Todd Fields
5 years ago
Views:

1 Chapter 9 Problem Solutions. At what temperature would one in a thousand of the atoms in a gas of atomic hydrogen be in the n energy level? g( ε ), g( ε ) Then, where 8 n ( ε ) ( ε )/ kt kt e ε e ε / n( ε) 000 T ( / )( ε ) ( / )(. 6eV) k ln ( ev/k)(ln000) ε ε/, and ε. 6 ev. The P l/ first excited sate in sodium is.09 ev above the S / ground state. Find the ratio between the numbers of atoms in each state in sodium vapor at l00 K. (see Example 7.6.) K multiplicity of P-level : L+, multiplicity of S-level : The ratio of the numbers of atoms in the states is then, ev/k )( 00 K). 09 ev 9 exp (

2 5. The moment of inertia of the H molecule is kg m. (a) Find the relative populations of the J0,,,, and rotational states at 00 K. (b) can the populations of the J and J states ever be equal? If so, at what temperature does this occur? J( J + )h (a) g( J ) J +, εj ε 0 I J 0 N( J) ( ) ( )exp J J + h ( ) exp h J + + ( 0) J N J IkT IkT ( J ( J + ) exp + )[ 0. 79] (. 6 0 J( J + ) 8 ( kg m J s) )(. 8 0 J( J + ) J/K)( 00 K) J( J + ) Applying this expression to J0,,,, and gives, respectively, exactly,.68, 0.880, 0.7, and (b) Introduce the dimensionless parameter. Then, for the populations of the J and J states to be equal, x 7x, x and 6lnx ln 7 7 Using, ln x h / IkT and ln( 5/ 7) - ln( 7/ 5) and solving for T,

3 T 6h Ik ln( 7/ 5) ( ( kg m J s) )(. 8 0 J/K)ln(. ) K 7. Find v and v rms for an assembly of two molecules, one with a speed of.00 m/s and the other with a speed of.00 m/s. v v rms ( ) [ ] (m/s). (m/s) 9. At what temperature will the average molecular kinetic energy in gaseous hydrogen equal the binding energy of a hydrogen atom? For a monatomic hydrogen, the kinetic energy is all translational and solving for T with KE E T 5 E ( / )(. 6 ev) k ( ev/k) K KE kt

4 . Find the width due to the Doppler effect of the 656.-nm spectral line emitted by a gas of atomic hydrogen at 500 K. For nonrelativistic atoms, the shift in wavelength will be between +λ(v/c) and -λ(v/c) and the width of the Doppler-broadened line will be λ(v/c). Using the rms speed from KE(/)kT (/)mv, v kt / m, and λ λ kt / m c ( m) (. 8 0 m 5. pm J/K)( 500K)/( Verify that the average value of /v for an ideal-gas molecule is m / πkt. [ Note : ve av dv /( a )] 0 The average value of /v is n( v) dv v N 0 v m π πkt 8 m/s m πn N πkt / / 0 kt m ve 7 kg) mv / kt m πkt dv < v >

5 7. How many independent standing waves with wavelengths between 95 and 0.5 mm can occur in a cubical cavity m on a side? How many with wavelengths between 99.5 and 00.5 mm? (Hint: First show that g(λ)dλ 8πL dλ/λ.) The number of standing waves in the cavity is 8πL ν g( ν) dν dν c 8πL c c 8πL g( λ) dλ g( ν) dν dλ dλ c λ λ λ Therefore the number of standing waves between 9.5mm and 0.5mm is 8π( m) 6 g( λ) dλ (. 0 mm). 5 0 ( 0 mm) Similarly, the number of waves between99.5mm and 00.5mm is.5x0, lower by a factor of 0. l9. A thermograph measures the rate at which each small portion of a persons skin emits infrared radiation. To verify that a small difference in skin temperature means a significant difference in radiation rate, find the percentage difference between the total radiation from skin at o and at 5 o C.

6 By the Stefan-Boltzmann law, the total energy density is proportional to the fourth power of the absolute temperature of the cavity walls, as R σt The percentage difference is σt σt T T T 07 K % σt T T 08 K For temperature variations this small, the fractional variation may be approximated by R ( T ) T T T K 0. 0 R T T T 08K. At what rate would solar energy arrive at the earth if the solar surface had a temperature 0 percent lower than it is? Lowering the Kelvin temperature by a given fraction will lower the radiation by a factor equal to the fourth power of the ratio of the temperatures. Using. kw/m as the rate at which the sun s energy arrives at the surface of the earth (. kw/m )( 0. 90) 0. 9 kw/m ( 66%)

7 . An object is at a temperature of 00 o C. At what temperature would it radiate energy twice as fast? To radiate at twice the radiate, the fourth power of the Kelvin temperature would need to double. / Thus, o [( ) K] T T 67 K 800 K( 57 C) 5. At what rate does radiation escape from a hole l0 cm in area in the wall of a furnace whose interior is at 700 o C? The power radiated per unit area with unit emissivity in the wall is PσT. Then the power radiated for the hole in the wall is P' σt A ( W/(m K ))( 97K) ( 0 0 m ) 5 W 7. Find the surface area of a blackbody that radiates 00 kw when its temperature is 500 o C. If the blackbody is a sphere, what is its radius? The radiated power of the blackbody (assuming unit emissivity) is P AeσT A P eσt. 9 0 ( )( m 8 9 cm 00 0 W W/(m K ))(( ) K)

8 The radius of a sphere with this surface area is, then, A πr r A/ π 6. 7 cm. The brightest part of the spectrum of the star Sirius is located at a wavelength of about 90 nm. What is the surface temperature of Sirius? From the Wien s displacement law, the surface temperature of Sirius is m K m K T. 0 0 K λ 9 max 90 0 m. A gas cloud in our galaxy emits radiation at a rate of.0x0 7 W. The radiation has its maximum intensity at a wavelength of 0 µm. If the cloud is spherical and radiates like a blackbody, find its surface temperature and its diameter. From the Wien s displacement law, the surface temperature of cloud is T m K o. 9 0 K 90 K 7 C m Assuming unit emissivity, the radiation rate is R where D is the cloud s diameter. Solving for D, P σ T A P πd / 7 P. 0 0 W D ( W/m K 90 K. πσt π )( ) m

9 5. Find the specific heat at constant volume of.00 cm of radiation in thermal equilibrium at 000 K. The total energy(u) is related to the energy density by UVu, where V is the volume. In terms of temperature, U Vu VaT V The specific heat at constant volume is then U 6σ c V T V T c ( W/m m/s J/K c σ K T ) ( 000K) ( Show that the median energy in a free-electron gas at T0 is equal to ε F / / 0.60ε F. At T0, all states with energy less than the Fermi energy ε F are occupied, and all states with energy above the Fermi energy are empty. For 0 ε ε F, the electron energy distribution is proportional to ε. The median energy is that energy for which there are many occupied states below the median as there are above. The median energy ε M is then the energy such that ε 0 M ε dε εf 0 εdε 6 m )

10 Evaluating the integrals, M εf or ε / / / ( ) ( ), M ( ) εf 0. ε 6ε 9. The Fermi energy in silver is 5.5 ev. (a)what is the average energy of the free electrons in silver at O K? (b)what temperature is necessary for the average molecular energy in an ideal gas to have this value? (c)what is the speed of an electron with this energy? (a) The average energy at T0 K is 5 ε0 εf. ev (b) Setting (/)kt(/5)ε F and solving for T, ε 5. 5eV T F k ev/k (c) The speed in terms of the kinetic energy is v 9 KE 6ε ev 60 0 J/eV 6 F (. )(. ) m/s m 5m 5( 9. 0 kg). Show that, if the average occupancy of a state of energy ε F + ε is f l at any temperature, then the average occupancy of a state of energy ε F - ε is f -f. (This is the reason for the symmetry of the curves in Fig.9.0 about ε F.) K F

11 Using the Fermi-Dirac distribution function f ffd( εf + ε) f ε / kt e + f + f + ε kt ε / kt e + e + e f FD( εf ε) e ε / kt e + / kt + e + / ε / kt ε ε / kt 5. The density of zinc is 7.l g/cm and its atomic mass is 65. u. The electronic structure of zinc is given in Table 7., and the effective mass of an electron in zinc is 0.85 m e. Calculate the Fermi energy in zinc. Zinc in its ground state has two electrons in s subshell and completely filled K, L, and M shells. Thus, there are two free electrons per atom. The number of atoms per unit volume is the ratio of the mass density ρ Zn to the mass per atom m Zn. Then, ε F h m * ( ) ρ 8π m Zn Zn / J s 7 0 kg/m (. ) ( )(. ) - -7 (0.85)(9. 0 kg 8 (65.u)(.66 0 kg/u ) π ) J ev / +

12 7. Find the number of electron states per electronvolt at εε F / in a.00-g sample of copper at O K. Are we justified in considering the electron energy distribution as continuous in a metal? ε N At T0, the electron distribution n(ε) is n( ) ( εf ) / ε At εε F /, ε N n( ) F 8 ε F The number of atoms is the mass divided by the mass per atom, N n ε F ( kg) ( 6. 55u)( eV 7 kg/u) with the atomic mass of copper from the front endpapers and ε F 7.0 ev. The number of states per electronvolt is states/ev and the distribution may certainly be considered to be continuous.

13 9. The Bose-Einstein and Fermi-Dirac distribution functions both reduce to the Maxwell- Boltzmann function when e α e ε/kt >>. For energies in the neighborhood of kt, this approximation holds if e α >>. Helium atoms have spin 0 and so obey Bose-Einstein statistics verify that f(ε)/e α e ε/kt Ae -ε/kt is valid for He at STP (0 o C and atmospheric pressure, when the volume of kmol of any gas is. m ) by showing that of A<< l under these circumstances. To do this, use Eq(9.55) for g(ε)dε with a coefficient of instead of 8 since a He atom does not have the two spin states of an electron, and employing the approximation, find A from the normaization condition n(ε)dεn, where N is the total number of atoms in the sample. (A kilomole of He contains Avogadro s number No atoms, the atomic mass of He is.00 u and 0 xe αx dx π / a / a Using the approximation f(ε)ae -ε/kt, and a factor of instead of 8 in Equation (9.55), Equation (9.57) becomes Vm n( ε) dε g( ε) f ( ε) dε A π h Integrating over all energies, / εe ε/kt dε N 0 n( ε) dε A Vm π h / 0 εe ε/ kt dε

14 . The integral is that given in the problem with x ε and akt, ε/ kt π( kt) εe dε, so that 0 Solving for A, N A N A h ( πmkt ) V / Vm π( kt) π h / Using the given numerical values, V A h ( πmkt ) / kmol A ( J s) [π(.00 u)( kg/kmol which is much less than one., 7 kg/u)( J/K)(9K)] /

15 5. The Fermi-Dirac distribution function for the free electrons in a metal cannot be approximated by the Maxwell-Boltzmann function at STP for energies in the neighborhood of kt. Verify this by using the method of Exercise 9 to show that A> in copper if f(ε) Aexp(ε/kT). As calculated in Sec. 9.9 N/V8.8x0 8 electrons/m for copper. Note that Eq.(9.55) must be used unchanged here. Here, the original factor of 8 must be retained, with the result that N / A h (πm ekt ) V 6 (8.8 0 m )( , J s) [π(9. 0 )(.8 0 J/K)(9K)] Which is much greater than one, and so the Fermi-Dirac distribution cannot be approximated by a Maxwell-Boltzmann distribution. /

Final Exam Practice Solutions

Physics 390 Final Exam Practice Solutions These are a few problems comparable to those you will see on the exam. They were picked from previous exams. I will provide a sheet with useful constants and equations