CONI'ENTS . . . . . . . . . . . . . . . . . . . Preface •• i . . . . . CRA.PI'ER l: Introduction to Statistical Methods. 1 CRA.PI'ER 2: Statistical Description of Systems of Particles •• ll CRA.PI'ER 3: Statistical Thermodynamics 16 CRA.PI'ER 4: Macroscopic Parameters and their Measurement 18 CRA.PI'ER 5: Simple Applications of Macroscopic Thermodynamics. 20 CRA.PI'ER 6: :Ba.sic Methods and Results of Statistical Mechanics. 32 CRA.PI'ER 7: Silllple Applications of Statistical Mechanics ••• 41 CRA.PI'ER 8: Equilibrium between Phases and Chemical Species •• 54 CHAPrER 9: Quantum Statistics of Ideal Gases •••••• 64 CHAPrER l0: Systems of Interacting Particles ••••• 84 CRAPI'ER 11: Magnetism and Low Temperatures •• 90 CHAPI'ER l2: Elementary Theory of Transport Processes. 93 CRAPI'ER 13: Transport Theory using the Rel.axation-Time Approximation 99 ®.Pl'ER l4: Near-Exact Formulation of Transport Theory ••••• l07 CRAPI'ER 15: Irreversible Processes and Fluctuations ••••••• 117 PREFACE This manual contains solutions to the problems in Fundamentals of Statistical and Thermal Physics, by F. Reif. The problems have been solved using only the ideas explicitly presented in this text and in the way a student encountering this material for the first tillle would probab]_y approach them. Certain topics which have implications far beyond those called for in the statement of the problems are not developed further here. The reader can refer to the numerous treatments of these subjects. Except when new symbols are defined, the notation conforms to that of the text. It is a pleasure to thar,k Dr. Reif for the help and encouragement be freely gave as this work was progressing, but he has not read all of this material and is in no way responsible for its shortcomings. Sincere thanks are also due to Miss Beverly West for patiently typing the entire manuscript. I would great]_y appreciate your calling errors or omissions to my attention. R. F, Knacke CHAP.PER 1 Introduction to Statistical Methods 1.1 There are 6-6-6 = 216 ways to roll three dice. The throws giving a sum less than or equal to 6 are Throw 1,1,1 1,1,2 1,1,3 1,1,4 1,2,2 1,2,3 2,2,2 No. of Permutatiom l 3 3 3 3 6 l -k Since there are a total of 20 permutations, the probability is ~~G = 1.2 (a) Probability of obtaining one ace= (probability of an ace for one of the dice) x (probabil- 5 1 1 6! ity that the other dice do not shov an ace) x (number of permutations)= (b)(l - '6') (5!1!) 5 5 = (5) = .4o2 • (b) The probability of obtaining at least one ace is one minus the probability of obtaining none, or 5 6 1 - (;-) = .667 (c) By the same reasoning as in (a) ve have l 2 2 4 6! (b) ('6') 5!2! = .04o 1.3 The probability of a particular sequence of digits such that five are greater than 5 and five are 1 5 1 5 less than 5 is (2) (2). Then multip:cying by the number of permutations gives the probability irrespective of order. 1.4 (a) To return to the origin, the drunk must take the same number of steps to the left as to the right. Thus the probability is l N N N.I W (-2) = -----<-J-)- - (- 2) C!) ! ! where N is even. (b) The drunk cannot return to the lamp post in an odd number of steps. 1.5 th (b) (Probability of being shot on the N trial)= (probability of surviving N-1 trials) x th 5 N-l l (probability of shooting onesel.f on the N trial)= (b) (b). (c) 6 1.6 m 3 r cl r N = m = O since these are odd moments. Using n = (p clp) (p+q) 2 2 - 2 ~2 and m = (2n - N) = 4 n - 4nN + W-- 4 m = we find m2 = N and ";f+ = 3lf - 2N. 1.7 The probability of n successes out of N trials is given by the sum 2 2 E E w(n) = 'W m i=l j=l with the restriction that the sum is taken only over terms involving w n times. 1 2 2 Then W(n) = E w. E v. If ve sum over all i ••• m, each sum contributes . i=l i m=l m (vi-1~ ) and 2 l W' (n) = (w + w 1 2 N N! n N-n W' (n) = E .....,...,,.._.__..,.. v w by the binomial theorem • n=O n!(N-n)! 1 2 Applying the restriction that w must occur n times we have 1 N! n N-n W(n) = n!(N-n)! "W'l w2 'We consider the relative motion of the two drunks. With each sillr..i.ltaneous step, they have a probability of 1/4 of decreasing their separation, 1/4 of increasing it,and 1/2 of maintaining it O'J taking steps in the same direction. Let the number of times each case occurs be n , n , 1 2 and n , respectively. steps is 3 2 The drunks meet if n = n • Then the probability that they meet after N steps irrespective of 1 2 the number of steps vhere we have inserted a parameter x which cancels if ~=n • We then perfonn the unrestricted 2 sum over n n n and choose the term in which x cancels. By the binomial expansion, 1 2 3 2 2 1 1 1 N 1 N 1/2 -1/2 N (4 4X 2) (2) pr= x + + = (x + x ) l 2N 2N 2N ! 1/2 n -1/2 2U-n Expansion yields P' = (2) ~ n! 2N- ! (x ) (x ) Since x must cancel we choose the terms where n = 2N-n or n = N Thus 1.9 (a) 1n (1-p)N-n ~ -p (N-n) ~ -Np n << N thus ( 1-p )N-n ~ e -Np N! (b) (N-n)! = N(N-1) ••• (N - n+l) ~ 1f if n << N r N! n ( )N-n n -Np "'n -A (c) W( n) = n!(N-n)! P 1-p ~ n! p e = n! e 1.10 0) "'n (a) E n! = e -t. e "' = 1 n=O = n ->.. - " n >.. e (b) ll = L-.," n.I n=v (c) n2 = n~ n~n e ->-. = e ->-. (~) 2 L. ~~ = >-. 2 + >-. -- - 2 2 (.6n) = n - .n..!-;) = >-. 1.11 (a) The mean number of misprints per page is l. -l Thus W(n) = -n>-..,n e ->-. = e n! -1 and W(O) = e = .37 3 2 -1 {b) P = 1 - 2: ~ = .08 n:::() n. 1.12 (a) Dividing the time interval t into small intervals ~t ve have again the binomial distribution for n successes in N = t/~t trials N! n ( )N-n W( n ) = , (N ) , P 1-p n. -n • -,.._n "- In the lilllit ~t ~ o, W(n} ~ n! e as in problem 1.9 vhere )... = ~ , the mean number of disintegra- tions in the interval of time. =4-n, -4 n 0 1 2 3 4 5 6 7 8 (b) W(n) e n. W(n) .019 .(176 .148 .203 .203 .158 .105 .()51 .003 1.13 2 2 We divide the plate into areas of size b. Since b is much less than the area of the plate, the probability of an atom hitting a particular element is much less than one. Clearzy n << N so we may use the Poisson distribution =6,n -6 W(n) e n. 1..1.4 We use the Gaussian approDJDation to the binomial distribution. (21.5-2COr W (n) = _ _;l.;;;..__ e - 2(400/4 = .013 J;;¥f J..1.5 The probability that a line is in use at e:n:y instant is 1/30. We vant N lines such that the probability that N+l or more lines are occupied is less than .01 vhen there are 2000 trials during the hour. rN. ___1,;; ;..__ • 01 = l. - e n=O~ cr vhere -n = 320C0O = 66.67, 4 The sum may be approxilllated by an integral JN¥z° l - 2 2 .99 = -- exp [-(n-n) /2a ] dn 1 ,Ji;' a (n-n+½;& J _ 2 .99 = -1 e y 12 dy afier a change of' integration variable. -1 ~ (-n- )/cr 2 The lower limit may be replaced by -eo with negligible error and the integral f'ound f'rcan the tables of' error functions. N--n~l We f'ind = 2. 33 a N = 66.67 + (2.33)(8.02) - .5 ::: 85 1.16 {a) The probability f'or N molecules in Vis given by the binomial distribution N-N N ! V N V o W(N) = N!(No-N)! (-) (1 - -) Vo Vo 0 V Thus N=NP=No v- 0 0 V V ) N - (1 - - oV V (N-i)2 0 0 (b) = Ff- i l (c) If V << V, :::- 0 N ~o (d) 1.17 Since O <<; < O, the integral. is eval.uated on the path -kb residue (ib) = ~ i :Fork< O kb kb -residue (-ib) = ~ Q(k) = e 8

Solutions Manual for Fundamentals of Statistical and Thermal Physics by Frederick Reif

Similar Solutions Manual for Fundamentals of Statistical and Thermal Physics by Frederick Reif