December, 2002 (Mihaly)
1./ The Debye temperature of Argon is 92.0K. What is the speed of sound? (Use Tables in the book to find necessary data.)
Solution: From the definition of the Debye temperature, we get q = hbar v/kB (6p 2N/V)1/3
(Eq. 28 in the book). Look up N/V in Chapter 1, Table 4. The speed of sound is v = hbar / q kB (6p 2N/V)1/3=1060 m/s.
2. There are N electrons on a plane of size LxL. Assuming free electron model, what is the Fermi wave vector? (Hint: work out a two-dimensional formula, based on Eq. 15 in Chapter 6.)
Solution: kF=(2pN)1/2/L
3./ What is the London penetration depth of the superconducting indium? (Table 1 of Chapter 6 may be helpful.)
Solution: lL= (mc2/4pnq2)1/2. The Table of values at the end of the book has the “electron radius” e2/mc2 = 2.8x10-13cm. Table 1 has n=11.49x1022cm-3. We get lL=1.57x10-6cm = 1.57x10-8m = 15.7nm.
4./ Cesium Chloride has a simple cubic lattice, with lattice spacing a=4.11Angstrom. A powder sample of CsCl is placed in an X-ray beam of wavelength 1.2Angstrom. What are the three lowest values of 2Q (the angle between the incident and scattered X-ray)?
Solution: The diffraction condition in a powder sample is |G|=|DK|; the absolute value sign indicates the fact that there is always a crystal in the powder that has the right orientation relative to the incident beam. Since |DK|=2k sin Q, with k=2p/l and in a cubic lattice |G|=(2p/a)(h2+k2+l2)1/2, we get 2/l sin Q = (1/a)(h2+k2+l2)1/2. This yields
sin Q = (l/2a)(h2+k2+l2)1/2, where h, k, l, are integers. The lowest value is obtained when two of them are zero, and the third one is 1, yielding sin Q = (l/2a). The next one corresponds to h=1, k=1, l=0, (or similar) sin Q = 1.41 (l/2a). The third one is h=1, k=1, l=1 leading to sin Q = 1.63 (l/2a). The angles are: 2Q = 16.8o, 23.8o, 29.3o.
Taking the Bragg condition, 2d sin Q =nl, and using d=a, and n = 1,2,3 yields the first number right, but the other two are incorrect.
5./ With the incident light perpendicular to the surface, the optical reflectance of a metal is well approximated by the Hagen Rubens formula: R=1-(2w/ps)1/2, where w is the angular frequency of the light, and s is the conductivity (CGS units are used). This works when the frequency is low, w<<s, and also w<<1/t, where 1/t is the relaxation rate.
a./ Express n as a function of w and s
b./ Prove that R=1-(2w/ps)1/2
You may use e(w) = 1+4pis/w, and R=[(n-1)2+K2]/ [(n+1)2+K2], where n+iK = e1/2.
Solution.
Since w<<s and e(w) = 1+4pis/w ~ 4pis/w and e1/2= (1+i) (2ps/w)1/2. (Here we used |1+i|2=2). This means n=K=(2ps/w)1/2.
Of course, since w<<s, n and K are much larger than 1. This helps in the calculation of the reflectivity. We also use n=K:
R=[n2-2n+1+K2]/[n2+2n+1+K2] ~ [2n2-2n]/[2n2+2n] = [n-1]/[n+1] = [1-1/n]/[1+1/n] ~
1-2/n.
Since n =(2ps/w)1/2, and 2/n=(2w/ps)1/2, we proved the HR formula.