Physics 206, Modern Physics II Mid-Term Examination Solution. 1

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1 Physics 06, Modern Physics II Mid-Term Examination Solution.. Consider a molecule with a permanent electric dipole moment p placed in an electric field E, with p aligned either parallel or anti-parallel to E. (a) Recall that the energy of a dipole in an electric field is given by E p E, and show that this system has two allowed energy states separated by pe. (b) For simplicity assume a ground state energy of 0 and an excited state energy of pe and degeneracies in the ratio g(pe)/g(0) /. For a collection of N molecules obeying Maxwell-Boltzmann statistics, calculate the ratio of the number of molecules in the excited state to the number in the ground state at temperature T. (c) For high T such that k B T pe, the ratio of the number of molecules in the upper state to the number in the lower is to. Taking reasonable estimates of p C m and E V/m, find the temperature at which the ratio has fallen by a measurable 0% to.9 to. (d) Calculate the average energy E at T and show that E 0 as T 0 and E 4pE/3 as T (e) Find E totatl from E, and show that the heat capacity for this two-level system is ( NkB C ) ( ) ( pe k B T e pe/k BT ( + epe/k BT ) (f) Sketch C as a function of pe/k B T. Find the value of pe/k B T at which C is a maximum, and explain, in physical terms, the dependence of C on T. ) Solution (a) Let the energy of the parallel case be E U and the energy of the anti-parallel case be E D, we then have: E U p E pe cos(0) pe E D p E pe cos(80) pe E E D E U pe

2 Physics 06, Modern Physics II Mid-Term Examination Solution. (b) Let n(pe) be the number of molecules in the excited state with energy pe. Then the ration of the number of molecules in the excited state to that in the ground state is: (c) Using Equation () we get; n(pe) n(0) g(pe)ae pe/k BT g(0)ae 0 e pe/k BT ().9 e pe/k BT ln(.9/) pe k B T pe T k B ln(0.95) (d) the average energy E is given by: K E n(pe) (pe) + n(0) 0 n(pe) + n(0) n(pe) n(0) pe n(pe) n(0) + using Equation (), we get: E e pe/kbt pe e pe/k B + pe + epe/k BT As T 0, e pe/kbt and E 0, and as T, e pe/kbt and E 4pE/3. (e) The total energy is E total the average energy E times the total number of molecules N, i.e. E total NE NpE + epe/k BT

3 Physics 06, Modern Physics II Mid-Term Examination Solution. 3 The heat capacity C is given by: C de total dt ( NkB ) ( ) pe k B T e pe/k B T ( ) + epe/k BT (f) Let x pe/k B T, then the heat capacity C becomes: C Nk Bx e x ( + ex ) () When T, x 0, and we find from Equation () that C 0. A plot of Equation () as C/(Nk B /) vs x is shown in Figure ()..0.5 C/(NK B /) x Figure :

4 Physics 06, Modern Physics II Mid-Term Examination Solution. 4 Equation () can be written as: C Nk Bx e x ( + ex ) ( ) NkB x e x + e x + 4 ( ) ex NkB x e x ( ) ex NkB x + e x + 4 ( ) ex NkB ( (3) x + e x + ex) 4 C ( (4) x + e x + ex) 4 Nk B Now, using e x + x + x /! + x 3 /3! + and e x x + x /! x 3 /3! + in Equation (4) we get: C Nk B ( x + e x + ex) 4 [ x + ( x + x /! x 3 /3! + ) + ( + x + 4 x /! + x 3 /3! + ) ] ( + x + ) ( x + ) (5) x x 3! 4 x x 3! As T 0, x and the right hand side of Equation (5) 0, i.e. C 0 as T 0 and as T. To find the value of pe/k B T at which C is maximum, we set the derivative of Equation () to zero, i.e. dc dx 0 Nk B Nk B [ ] xe x ( + ex ) + x e x ( + ex ) x e x ( + ex ) 3 ] x e [ x ( + + x xex ex ) ( + ex ) (6)

5 Physics 06, Modern Physics II Mid-Term Examination Solution. 5 The condition for maximum C is obtained from Equation (6) as: 0 + x xex ( + ex ) + e x + x + xex xe x + x x ex (7) we can find the solution of Equation (7) by plotting the two functions: y x+ x and y ex on the same graph and locate the point of intersection. The plot of the two functions is shown in Figure (). The two curves intersect at x.65, so Figure : this is the value of pe/k B T at which C is maximum. The temperature at which

6 Physics 06, Modern Physics II Mid-Term Examination Solution. 6 C is maximum is then given by:.65 pe k B T T pe.65k B K

7 Physics 06, Modern Physics II Mid-Term Examination Solution. 7. The rotational motion of molecules has an effect on the equilibrium separation of the nuclei, a phenomenon known as bond stretching. To model this effect, consider a diatomic molecule with reduced mass µ, oscillator frequency ω, and internuclear separation R when the angular momentum is zero. The effective potential energy for nonzero values of l is then U eff (r) µω (r R ) + l(l + ) µr (8) (a) Minimize the effective potential U eff (r) to find an equation for the equilibrium separation of the nuclei, R l, when the angular momentum is l. Solve this equation approximately, assuming l µω R /. (b) Near the corrected equilibrium point, R l, the effective potential again is nearly harmonic and can be written approximately as U eff (r) µω (r R l ) + U (9) Find expressions for the corrected oscillator frequency ω and the energy offset U by matching U eff and its first two derivatives at the equilibrium point R l. Show that the fractional change in frequency is given by ω ω 3l(l + ) µ ω R 4 Solution (a) At the equilibrium separation R l, the effective potential energy U eff is minimum, 0 du eff dr Rl µω (R l(l + ) l R ) µrl 3 [ ] ( ) l(l + ) R l R + µ ω when l µω R /, the second tern on the right hand side of Equation (0) is small and is only a small correction to R, which is the equilibrium separation at R 3 l (0)

8 Physics 06, Modern Physics II Mid-Term Examination Solution. 8 zero angular momentum. For a small angular momentum, l, then R l is very close to R. We then can replace R l in the second term in Equation (0) by R. [ ] ( ) l(l + ) R l R + () µ ω R 3 (b) From Equation (8) we see that the energy offset U, is the effective potential energy at R l, U U eff (R l ) l(l + ) µω [R + µ ω [ l(l + ) l(l + ) µrl µω Rl 4 R 3 l ] R + ] + l(l + ) µr l The first tern inside the square brackets on the right of Equation () is small and can be ignored, we then have: () U l(l + ) µr l The force constant K is related to the frequency ω of molecular vibration and the effective potential energy U eff by: K µω d U eff dr So, it is posible to relate the frequency at angular momentum state ω to the frequency at a state of zero angular momentum ω, by taking the second derivative of Equation (8) and equate it to µω l, d U eff µω 3l(l + ) dr + µrl 4 µωl ωl ω 3l(l + ) + µrl 4 Since R l R and for l µω R /, we than can write: ω l ω + 3l(l + ) µr 4 The second term on the right is small and ω l ω, so ω l ω (ω l ω )(ω l + ω ) ω ω

9 Physics 06, Modern Physics II Mid-Term Examination Solution. 9 the fractional change is then; ω ω 3l(l + ) µ ω R 4

10 Physics 06, Modern Physics II Mid-Term Examination Solution To obtain a more clearly defined picture of the Fermi-Dirac distribution, consider a system of 0 Fermi-Dirac particles sharing 94 units of energy. By drawing diagrams like Figure (3), show that there are nine different microstates. Using n j n j p + n j p + calculate and plot the average number of particles in each energy level from 0 to 4E. Locate the Fermi energy at 0 K on your plot from the fact that electrons at 0 K fill all the levels consecutively up to the Fermi energy. (At 0 K the system no longer has 94 units of energy, but has the minimum amount of 90E.) 5E 0E 5E 0 Figure 3: One of the nine equally probable micro-states for 0 Fermi-Dirac particles with a total energy of 94E Solution Figure (4) shows all possible arrangements. It should be noted that filling the lowest 0 level takes only 90 units of energy. In addition, keeping the total energy limited to 94 units it is not possible to promote any particle from the first six levels, because any of these particles can only go levels that are already occupied. So we restricted

11 Physics 06, Modern Physics II Mid-Term Examination Solution. 5E 0E 5E Figure 4: All possible equally probable micro-states for 0 Fermi-Dirac particles with a total energy of 94E. Open circles represent the original locations of the grey ones. to promoting particles from levels with energies 6-9 units. The probability of each micro-state is then /9. The first six states (0 5E) have particles each, so, n 0E through n 5E ( ) ( ) n 6E ( ) ( ) ( ) n 7E ( ) ( ) ( ) n 8E ( ) n 9E (. ) n 0E ( ) ( ) ( ) ( ) ( ) ( ) ( )

12 Physics 06, Modern Physics II Mid-Term Examination Solution. ( ) n E 9 + ( ) n E 9 + ( 0. ) n 3E 9 0. n 4E 0.00 ( ) ( ) ( ) 9 The plot of n ave n vs energy states is shown in Figure (5). At 0 K all low energy n ave.0 E F Energy States Figure 5: levels are full, that is all levels from 0E up to 9E, so the Fermi energy is then 9E as shown in Figure (5).

13 Physics 06, Modern Physics II Mid-Term Examination Solution The Fermi energy of aluminum is.63 ev. (a) Assuming that the free electron model applies to aluminum, calculate the number of free clcctrons per unit volume at low temperatures. (b) Determine the valence of aluminum by dividing the answer found in part (a) by the number of aluminum atoms per unit volume as calculated from the density and the atomic weight. Note that aluminum has a density of.70 g/cm 3. Solution (a) The free electron density n is related to the Fermi energy E F through: ( ) ( ) h /3 3n E F m e 8π ( ) ( ) 3/ 8π me E F n 3 h ( ) ( 8π ( ) free electrons/m 3 (b) The number of atoms/m 3 n can be calculated using the density ρ, Avogadro s number N A, and the molecular mass M: The valence of aluminum is then: n ρn A M atoms/cm atoms/m 3 n n ) 3/

14 Physics 06, Modern Physics II Mid-Term Examination Solution A one-dimensional model for the electronic energy of a diatomic homonuclear molecule is described by the potential well and barrier shown in Figure (6). In the simplest case, the barrier width is shrunk to zero (w 0) while the barrier height increases without limit (U ) in such a way that the product Uw approaches a finite value S called the barrier strength. Such a barrier known as a delta function barrier can be shown to produce a discontinuous slope in the wavefunction at the barrier site L/ given by dψ dx L/ + dψ dx L/ ms ψ(l/) (3) Solve the wave equation in the well subject to this condition to obtain expressions for the energies of the ground state and first excited state as functions of the barrier strength S. Examine carefully the limits S 0 and S and comment on your findings. U w 0 L L/ Figure 6: Solution To the left and to the right of the barrier location the wave function psi is that of a free particle with a wave number k given by: k me (4)

15 Physics 06, Modern Physics II Mid-Term Examination Solution. 5 so the wave functions are: ψ (x) A sin(kx) + B cos(kx) 0 x L (5) ψ + (x) F sin(kx) + G cos(kx) L x L (6) The infinite walls at x 0 and x L requires that ψ (0) ψ + (L) 0. using these boundary conditions in Equations (5) and (6) we get: 0 A 0 + B B 0 (7) 0 F sin (kl) + G cos (kl) G F tan (kl) (8) Using Equations (7) and (8) in Equations (5) and (6) we get: ψ (x) A sin(kx) 0 x L ψ + (x) F {sin(kx) tan(kl) cos(kx)} { ( ) } sin(kl) F sin(kx) cos(kx) cos(kl) ( F cos(kl) ) {sin(kx) cos(kl) cos(kx) sin(kl)} L C sin(kx kl) x L (0) The waves can be either symmetric or antisymmetric at the middle point. Let us examine each case separately. (A) Antisymmetric Case: If the waves are antisymmetric at the middle point then the wave functions on either side of the mid-point must vanish at the mid-point, i.e. ψ ± (L/) 0 () (9) This also means that the barrier has no effec on the wave functions and consequently the slope at the mid-point is continuous. dψ dx dψ + dx Ak cos(kx) Ck cos(kx kl) at the mid point the two slopes are equal, so we get: Ak cos(kl/) Ck cos(kl/ kl) Ak cos(kl/) Ck cos( kl/) Ak cos(kl/) Ck cos(kl/) A C

16 Physics 06, Modern Physics II Mid-Term Examination Solution. 6 To satisfy the condition in Equation (), we must put kl/ nπ. Using this in Equation (4) we can get the energy E for the antisymmetric case as: E n ( k) m π m(l/) n n,, () (B) Symmetric Case Sine the waves are symmetric around the mid-point, the two wave function must be equal at that point, i.e. A sin(kl/) C sin(kl/ kl) C sin( kl/) C sin(kl/) A C (3) It is given that the slopes of the two functions at the mid-point are different and the difference is given by Equation (3) from which and Equation (3) we get: Ck cos(kl/ kl) Ak cos(kl/) ms A sin(l/) Ck cos( kl/) Ak cos(kl/) ms A sin(l/) Ak cos(kl/) Ak cos(kl/) ms A sin(l/) k cos(kl/) ms sin(l/) ( ) tan(kl/) kl msl tan(x) αx (4) where x kl/ and α /msl. The roots of Equation (4) can be found graphically by plotting the functions y tan(x) and y αx and determining the values of x at the points of intersections of the two functions as shown in Figure (7). Let the points of intersection be x n, this gives us: k n x n L E n k n m (5) Only values of x n (and k n ) larger than zero should be considered, since the wave function is symmetric and k 0 makes the wave functions vanish everywhere.

17 Physics 06, Modern Physics II Mid-Term Examination Solution y(x) x/ p Figure 7: Now if S, the the slope α of the straight line y αx becomes zero, the line lies on the x-axis, and will intersect with tan(x) at x n nπ with n,, 3,. The energy levels E n corresponding to various waves are then given by: E n π m(l/) n n,, 3, (6) Equations () and (6) are identical, that is the energy levels of the asymmetric case and the symmetric case with S are degenerate. If S 0, then α and the line becomes vertical and can intercept the tan(x) curve at x n (n + )π with n 0,,, 3,. the energy levels E n are then given by: E n π ml (n + ) n 0,,, 3 (7) The ground state energy is then given by: and the first state is; E π 4mL E 9π 4mL

18 Physics 06, Modern Physics II Mid-Term Examination Solution As a model for a diatomic molecule, consider two identical point masses m connected by a rigid massless rod of length R. Suppose that this molecule rotates about an axis perpendicular to the rod through its midpoint. Use the Bohr quantization rule for angular momentum to obtain the allowed rotational energies in this approximation. Compare your result with the correct quantum mechanical treatment of this model. Solution The angular momentum L of this system is: L mv R + mv R mvr According Bohr s theory the angular momentum must be multiples of, i.e. L mvr n v n mr n,, 3, The energy of rotation E B according to Bohr s theory is then: E B mv + mv ( ) n m mr mr n n,, (8) Energy of rotation E Q according to a correct quantum mechanical treatment is given by: ( ) E Q {l(l + )} I cm Where I cm is the moment od inertia about the center of mass, ( ) R ( ) I cm m R + m mr ( ) E Q {l(l + )} l 0,,, (9) mr

19 Physics 06, Modern Physics II Mid-Term Examination Solution. 9 Comparing Equations (8) and (9) we see that l(l + ) replaces n in the Bohr result. However, the two expression are indistinguishable for large l and n, but they disagree markedly for small values of l or n. In particular, E Q can be zero while E B has a minimum value of /mr, when n.

20 Physics 06, Modern Physics II Mid-Term Examination Solution Silver has a density of kg/m 3 and a resistivity of Ω m at room temperature. (a) On the basis of the classical free-electron gas model, and assuming that each silver atom contributes one electron to the electron gas, calculate the average time between collisions of the electrons. (b) Calculate the mean free path from τ and the electron s thermal velocity. (c) How does the mean free path compare to the lattice spacing? Solution (a) The average time τ between collisions in an electron gas is: τ m e ne ρ where m e is the mass of the electron ( kg) n is number of electrons in the gas per m 3, ρ is the resistivity ( Ω m), and e ( C) is the electronic charge. n can be calculated from Avogadro s number ( ), the density of silver (.05 g/cm 3 ), the atomic number of silver (08), and the number of electrons each silver atom contributes the electron gas (), The mean time τ is then: (b) the mean free path L is given by: n electrons/cm electrons/cm electrons/m 3 τ ( ) s L v rms τ

21 Physics 06, Modern Physics II Mid-Term Examination Solution. The root mean square speed of electrons at a temperature T K can be calculated from: The mean free path is then: m evrms 3 k BT 3kB T v rms m e m/s L v rms τ m

22 Physics 06, Modern Physics II Mid-Term Examination Solution. 8. Assuming that conduction electrons in silver are described by the Fermi free-electron gas model with E F 5.48eV, repeat the calculations of Problem 7. Solution It is possible to replace v rms by v F : (a) The average time between collision is: τ m e ne ρ σm e ne (30) where σ /ρ is the conductivity and ρ is the resistivity, m e is the mass of the electron, e is the electron s charge, and n is the number of electrons per unit volume. (b) (c) σ ρ (Ω m) (3) Let A v be Avogadro s number (particles/mol), d the density (g/cm 3 ) and M the molar mass. Assuming that there is one free electron per atom, then the free electron density is: n A vd M e/cm 3 Using Equations (3) and (3) in Equation (30) we get e/m 3 (3)

23 Physics 06, Modern Physics II Mid-Term Examination Solution The natural line width emitted by a collection of Cr 3+ ions in ruby in thermal equilibrium at 90 K is about 0.4 nm for the nm line. This is the thermal line width. Show that the emission from a ruby laser is much more monochromatic than ± 0. nm if laser emission is confined to a single cavity mode. Take the length of the ruby between polished faces to be 0 cm, n.8, and dn/dλ 0. Solution The laser operation is confined to a single mode, the laser line width will be equal to the change in wavelength between adjacent lines. So: λ λ L(n λ dn/dλ) (694.4) nm λ laser ± 0.00 nm So the condition: L m λ n restricts the laser wavelength to about /00 of the thermal line width.

eigenvalues eigenfunctions

eigenvalues eigenfunctions Born-Oppenheimer Approximation Atoms and molecules consist of heavy nuclei and light electrons. Consider (for simplicity) a diatomic molecule (e.g. HCl). Clamp/freeze the nuclei in space, a distance r