Problem Set 4 Solutions Distributed: February 26, 2016 Due: March 4, 2016

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Eunice Walton
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1 Probl St 4 Solutions Distributd: Fbruary 6, 06 Du: March 4, 06 McQuarri Probls 5-9 Ovrlay th two plots using Excl or Mathatica. S additional conts blow. Th final rsult of Exapl 5-3 dfins th forc constant in trs of th paratrs in th Mors potntial: k = D β. Givn D = J/cul and β = for HCl, k is k = D β = J/cul = 479 N cul = kj n 3 cul N J 0 9 n kj 000 N V x, kj Haronic Mors Vibrational potntials of HCl, sall displacnts V x, kj 5 0 Haronic Mors 5 Th Mors potntial capturs th or ralistic stp ris as x 0 and lvling off to th bond dissociation nrgy as x. For sall displacnts, th haronic potntial and Mors potntial ar vry siilar. CH34: Quantu Chistry

2 Probl St 4 Solutions 5-3 Thr s a strong lin in th IR spctru of H 7 I at 309 c. Givn th intnsity of th lin, this is probably th fundantal vibrational frquncy ω in wavnubrs. Using quation 5.39, ω = k /. πc µ Th rducd ass µ fro th asss of 7 I 7 I = g and H = g fro Wikipdia µ = 7 I H 7 I + H which is.000 g with 4 sig figs. Rarranging quation 5.39, w can solv for k: k = πc ω µ = π s = g c s k = 34. N k = kj n = g 309 c.000 g kg 000 g 3 cul 00 c 0 9 n kj 000 N Th priod τ has units of ti pr cycl, and th frquncy has units of cycl pr ti, so th priod of oscillation will b: τ = ν obs = c s = ωc 00 c 0 τ = fs s s τ = 4.44 fs N kg s 5-4 Givn k = 39 N for 35 Cl 35 Cl, Calculat th fundantal vibrational frquncy and / th zro-point nrgy. Using quation 5.37, ν obs = k π µ with 35 Cl = g CH34: Quantu Chistry

3 Probl St 4 Solutions 3 fro Wikipdia, so µ = 7.48 g ν obs = 39 N / π 7.48 g = N / g g kg kg / s N = s ω = ν obs /c = c = 556 c. Th zro-point nrgy for a haronic oscillator is E 0 = hν E 0 = hν = J s = J s 5-5 Givn that th fundantal fro v = 0 to v = and first ovrton fro v = 0 to v = for C 6 O occur at 43.0 c and c, and using McQuarri s quation 5.43, ω obs = ω v ω x vv + I gt two quations: 43.0 c = ω ω x c = ω 6ω x. Multiplying th first by and adding both, I gt 6.0 c = ω x, and ω x = 3.0 c. Plugging this into th first quation, 43.0 c = ω 3.0 c, and ω = 69.0 c. frquncy. Not that this is about % largr than th fundantal 5- Only do n =. It ll b asir to show this without including any noralization constants in CH34: Quantu Chistry

4 Probl St 4 Solutions 4 front of th wavfunction. Justify why it is okay to ignor th constant if you tak that approach. Fro tabl 5.4, th unnoralizd wavfunction for a haronic oscillator in th first x xcitd stat is ψ x = x. To show that this quation satisfis th Schrödingr quation, w hav to show that it is an ignfunction of th Hailtonian oprator. Ĥψ x = T + V ψ x = ħ d µ dx + kx ψ x = ħ d ψ µ dx + kx ψ x W ll do this in pics. First, th diffrntial part in th first tr, using th product rul: d ψ dx = d dx = d dx x d x dx x αx x x x = x αx αx 3 x x = 3αx + α x 3 x x pulling out ψ x = x d ψ dx = = 3α + α x x x = 3α + α x ψ x using α = kµ 3 ħ + kµ ħ x ψ x kµ, quation 5.45 ħ Plugging this rsult in for th kintic oprator tr in, dx = ħ 3 µ 3ħ = ħ d ψ µ Plugging this rsult into, 3ħ Ĥψ x = = 3ħ k kµ ħ + kµ k µ kx µ kx k µ ψ x. ħ x ψ x ψ x ψ x + kx ψ x CH34: Quantu Chistry

5 Probl St 4 Solutions 5 If w rarrang quation 5.33, k µ h = πν and ħ = π. Plugging this in, w gt, Ĥψ x = 3 hνψ x ψ x is prsrvd, and th ignvalu is 3 hν. W could nglct th noralization constant bcaus for a linar oprator lik th Hailtonian, Ĥ aϕ = aĥϕ for so constant a. 5-6 Suariz how th rsults of ths typs of calculations ight b usd. x, Showing that x = dx ψ xx ψ x = 5 ħ µk, using ψ x = N H αx whr N and H ar th noralization constant and Hrit polynoials for th scond xcitd stat givn on pag 5. To valuat th intgral, I ll start by changing variabls using αx = ξ and αdx = dξ: ξ = dξ N H ξ ξ ξ α α α N H ξ ξ ξ = dξ N α ξ H ξ H ξ ξ I want to look only at th Hrit tr ξ H ξ, and us th rcursion dfinition of th Hrit polynoials, quation 5.55, H n ξ = H n+ ξ + nh n ξ, ξ H ξ = ξ H 3 ξ + H ξ = ξh 3 ξ + ξh ξ = H 4 ξ + 3H ξ = 4 H 4 ξ + 5 H ξ + 4H 0 ξ + ξ H ξ + H 0 ξ If I plug this back in, and rcogniz that th th Hrit polynoials for which n will intgrat to 0, th avrag bcos ξ = dξ N α = 5 α x = 5 α α = 5 α x = 5ħ kµ 5 H ξ H ξ ξ dξ N H ξ ξ dx α N H αx x Going back to x dx N H αx x Th intgral is I usd α = kµ/ħ CH34: Quantu Chistry

6 Probl St 4 Solutions 6 This is th final rsult. It can b usd to calculat th intrinsic rror in a asurnt of position for a quantu haronic oscillator in th scond xcitd stat, x. This liits th prcision with which w can know whr a quantu objct is. 5-7 Suariz how th rsults of ths typs of calculations ight b usd. To do th sa thing as 5-6 for ontu, w want to dfin anothr rcursion rlationship for th Hrit polynoials. W wanna gt d dξ H n ξ starting with th drivativ dfinition of th Hrit polynoials, H n ξ = n ξ d n dξ n ξ, d dξ H n ξ = d n ξ dn dξ dξ n ξ = ξ n ξ dn dξ n ξ + n ξ = ξ n ξ dn dξ n ξ + n+ = ξ n ξ dn = ξh n ξ H n+ ξ dξ n ξ n+ ξ dn+ ξ dξn+ ξ dn+ ξ dξn+ dn+ ξ dξn+ = nh n ξ Fro th rcursion rlationship. Using this rsult and P = ħ d dx, w can siplify th scond drivativ of th wavfunction for a gnral xcitd stat d dx ψ nx = d dx d d = N n dx dx = N n d dx N n H n αx x H n αx x n αh n αx x αxhn αx x = N n 4nn αh n αx x 3 nα xhn αx x αh n αx x 3 nα xhn αx x +α x H n αx x = N n 4nn αh n αx x 3 4nα xhn αx x H n αx x + α x H n αx x CH34: Quantu Chistry

7 Probl St 4 Solutions I nd to work with th scond tr in th final rsult to gt rid of th x 4nα 3 xhn αx x = nα αxhn αx x = nα n H n αx + Hn αx x = 4nn αh n αx x + nαhn αx x Plugging this back in to th scond drivativ, th first tr cancls, and w gt d dx ψ nx = N n nαh n αx x αhn αx x + α x H n αx x = nαψ n x αψ n x + α x ψ n x = nα α + α x ψ n x. Now I can put this into th intgrals for th scond xcitd stat, p = dx ψ x P ψ x, with d dx ψ x = 5α + α x ψ x p = dx ψ x P ψ x = ħ dx ψ x d dx ψ x = ħ dx ψ x 5α + α x ψ x = 5ħ α dx ψ xψ x ħ α dx ψ xx ψ x }{{}}{{} x = 5 α = 5ħ α dx ψ xψ x } {{ } = 5 ħ α = 5 ħ µk 5 α ħ α This is th final rsult, fro which w can calculat th uncrtainty in p. Not that by cobining th rsults fro 5-6 and 5-7, th Uncrtainty principl holds, bcaus if w incras ass or th forc constant, although w can know x with or crtainty, w don t know p. ν obs = c for H 79 Br, and l = 4 p. Th RMSD bond lngth in th ground stat will b givn by x, whr x = l l. Using quation 5.60, x = ħ kµ v +, so in th ground stat, x = ħ. I can also us quation 5.39, kµ CH34: Quantu Chistry

8 Probl St 4 Solutions 8 k = πc ω µ. Cobining ths, x = ħ 4πc ωµ, whr µ = g to 4 sig figs, using 79 Br = g fro Wikipdia. Plugging in valus, x = ħ 4πc ωµ 47 c J s 000 g =.07 0 g kg 0 = p = 64.4 p kg s c J So th RMSD displacnt is lngth. Intrsting. x = 8.03 p, or 5.69% of th ovrall bond 5-4 Using th xprssion at th nd of probl 5-40, th fraction of culs in th ground stat is f 0 = hν/k BT. calculat ν = ω c = s. At 300 K, f 0 = hν/k BT Givn that ω = 650 c for HBr, w can J s s.38 0 = 3 J K K300 = = 00% To th prcision of th asurnt of ω, all th culs ar in th ground stat at 300 K At 000 K, f 0 = 0.49 = 85.% Fwr of th culs ar in th ground stat at 000 K, but still a surprising ajority. CH34: Quantu Chistry

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