Homework #6. A circular cylinder of radius R rotates about the long axis with angular velocity

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1 Homwork #6 1. (Kittl 5.1) Cntrifug. A circular cylindr of radius R rotats about th long axis with angular vlocity ω. Th cylindr contains an idal gas of atoms of mass m at tmpratur. Find an xprssion for th dpndnc of th concntration n(r) on th radial distanc r from th axis, in trms of n(0) on th axis. Tak µ as for an idal gas. m 2 n( r) µ ( r) = v + ln 2 v(r) = ωr µ(r) = µ(0) m 2 ω 2 r 2 + ln n(r) ln n(r) n(0) = mω 2 r 2 2 mω 2 r 2 2 n(r) = n(0) = ln n(0) 2. (Kittl 5.2) Molculs in th Earth's atmosphr. f n is th concntration of molculs at th surfac of th Earth, M is th mass of a molcul, and g is th gravitational acclration at th surfac, show that at a constant tmpratur th total numbr of molculs in th atmosphr is N = 4πn(R) MgR R MgR 2 drr 2 r with r masurd from th cntr of th Earth; hr R is th radius of th Earth. Th intgral divrgs in th uppr limit, so that N cannot b boundd and th atmosphr cannot b in quilibrium. Molculs, particularly light molculs, ar always scaping from th atmosphr.

2 Th potntial nrgy of M at r is U = GM M E. f g is th gravitational constant at th r surfac of th Earth g = GM E R 2 U(r) = Mg R2 r µ(r) = Mg R2 r + ln n(r) = µ(r) = mgr + ln n(r) ln n(r) n(r) = MgR 1 R r n(r) = n(r)xp MgR 1 R r N = 4πr 2 n(r)dr = 4πn(R) MgR drr 2 MgR2 r R R 3. (Kittl 5.6) Gibbs sum for a two lvl systm. (a) Considr a systm that may b unoccupid with nrgy zro or occupid by on particl in ithr of two stats, on of nrgy zro and on of nrgy. Show that th Gibbs sum for this systm is =1 + λ + λ Our assumption xcluds th possibility of on particl in ach stat at th sam tim. Notic that w includ in th sum a trm for N = 0 as a particular stat of a systm of a variabl numbr of particls. (b) Show that th avrag thrmal occupancy of th systm is N = λ + λ (c) Show that th thrmal avrag occupancy of th stat at nrgy is N() = λ (d) Find an xprssion for th thrmal avrag nrgy of th systm.

3 () Allow th possibility that th orbital at 0 and at may b occupid ach by on particl at th sam tim; show that =1 + λ + λ + λ 2 = (1+ λ)(1 + λ ) Bcaus can b factord as shown, w hav in ffct two indpndnt systms. (a) = (Nµ ) = λ N whr λ = 0 0 = λ 0 + λ 1 + λ 1 µ = 1 + λ + λ (b) or N = 1 Nλ N = 1 0 λ + ( λ λ λ 1 λ )= 1 + λ + λ N = λ ln λ = λ(1 + ) 1+ λ + λ (c) At E = thr is only on stat. (d) N() = λ λ N E = = 0 λ λ λ 1 λ = 1 + λ + λ 1 + λ + λ or N µ E = 2 ln = µ λ + λ λ = 2 µ λ µ 2 2 λ + λ = µ N λ 2 E = λ ()

4 = λ N = λ λ λ 1 + λ 2 (0+ ) i. =1+ λ + λ + λ 2 = (1+ λ)(1 + λ ) 4. (Kittl 5.7) Stats of positiv and ngativ ionization. Considr a lattic of fixd hydrogn atoms; suppos that ach atom can xist in four stats: Stat Numbr of lctrons Enrgy Ground 1 Positiv on δ Ngativ on δ Excitd Find th condition that th avrag numbr of lctrons pr atom b unity. Th condition will involvδ, λ and. = λ N whr λ = µ N i. = λ 0 δ 2 + λ λ λ 2 δ 2 = δ 2 + λ 2 2 ( + )+ λ 2 δ 2 Lt N = 1. Thn N = λ ln λ = λ λ δ 2 ( )

5 or =λ 2 2 ( + )+ 2λ 2 δ 2 δ 2 + λ 2 2 ( + )+ λ 2 δ 2 = λ 2 2 ( + )+ 2λ 2 δ 2 δ 2 = λ 2 δ 2 δ = λ 2 5. (Kittl 5.10) Concntration Fluctuations. Th numbr of particls is not a constant in a systm in diffusiv contact with a rsrvoir. W hav sn that N = µ from (59). (a) Show that N 2 = 2, V 2 µ 2 Th man-squar dviation ( N) 2 of N from N is dfind by ( N) 2 = (N N ) 2 = N 2 2 N N + N 2 = N 2 N ( N) 2 = µ 2 2 µ (b) Show that this may b writtn as ( N) 2 = N µ n chaptr 6 w apply this rsult to th idal gas to find that ( N) 2 N 2 = 1 N is th man squar fractional fluctuation in th population of an idal gas in diffusiv contact with a rsrvoir. f N is of th ordr of atoms, thn th fractional fluctuation is xcdingly small. n such a systm th numbr of particls is wll dfind vn though it cannot b rigorously constant bcaus diffusiv contact is allowd with th rsrvoir. Whn N is low, this rlation can b usd in th xprimntal dtrmination

6 of th molcular wight of larg molculs such as DNA of MW ; s M. Wissman, H. Schindlr, and G. Fhr, Proc. Nat. Acad. Sci (1976). (a) = Nµ N N = 1 N Nµ N, = 1 µ = µ = ln µ N 2 = 1 N 2 Nµ = 2 2 N, µ 2 (b) ( N) 2 = (N N ) 2 = N 2 2N N + N 2 = N 2 2 N N + N 2 = N 2 N 2 N = µ N µ = 2 µ 2 2 = 1 µ µ N 2 1 N 2 ( N) 2 = N 2 N 2 = N µ

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