Problem 1. Part b. Part a. Wayne Witzke ProblemSet #1 PHY 361. Calculate x, the expected value of x, defined by

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1 Poblem Pat a The nomal distibution Gaussian distibution o bell cuve has the fom f Ce µ Calculate the nomalization facto C by equiing the distibution to be nomalized f Substituting in f, defined above, this becomes: Ce µ If we set b µ, and thus db, then we can substitute b + µ and db into the equation. It becomes: Ce b db Using I n n e λ with n. Accoding to [Tiple & Llewellyn, p AP 6-7], in this case I π/ λ /. Since e λ is an even function, integating fom to gives I π / λ /. In ou case, λ. So we have: So, C π. C Ce b db Cπ / C π e b db / Pat b Calculate, the epected value of, defined by u uf f What physical intepetation does it have? Since f, as defined in Pat a, and substituting in fo u, we get: Substituting in fo f, we get: Ce f µ Once again, if we set b µ, and thus db, then we can substitute b + µ and db into the equation. It becomes: Cb + µe b db Cb e b db + C be b db + µ Cµe b db Ce b db Fom Pat a, we know that Ce b db. [Tiple & Llewellyn, p AP 6-7] also tell us many wondeful things about how to integate e λ, howeve, we ae integating fom to, and e λ is an odd function, so the fist integal is zeo. So we get: µ This epesents the mean of the distibution. Page

2 Pat c Calculate. Why is this diffeent than? Epess in tems of and. What physical intepetation does it have? Show that Simila to Pat b, the integal in the thid tem is equal to one, leaving the thid tem as µ. The integal in the second tem is fo an odd function fom to, so it is equal to zeo. The integal in the fist tem, howeve, esembles I fom [Tiple & Llewellyn, p AP 6-7], ecept that we ae integating fom to, instead of fom to. This gives usi π/ λ 3/, which gives: We know µ, so we need to calculate. Substituting in u into the equation in Pat b, we get f f Once again, f, so we have: C f Ce e µ µ And yet again, if we set b µ, and thus db, then we can substitute b + µ and db into the equation. It becomes: C b + µ e b db C b 3 + b µ + µ e b db C 3 b e b db +C µ be b db +µ Ce b db C 3 b e b db C3 π 3/ But, C, and substituting this in we get: π C 3 π / So, + µ. π 3 π To find in tems of and, we can substitute µ into + µ, and we get: + This gives us the population standad deviation of the distibution. is also RMS, which is N N., on the othe hand, is N, N which not only has multiple coss tems of i in the numeato, but also has N in the denominato. So, is clealy not equal to. To detemine if we check to see what each side equals. Fom pevious calculations, we have: Page

3 + µ µ and, by the definition of u given above: + f µ f µ + µ f f µ µf µ f Ce µ Ce µ +µ Ce µ Howeve, we have aleady calculated all thee of these integals. Fom ealie in this poblem, we found that Ce µ + µ, fom pat b we know that Ce µ µ, and, by definition, we know that Ce we e left with: µ. So, + µ µµ + µ Since both sides of the equation ae equal to,. Note that it is also possible to use the popeties associated with aveaging to solve this poblem: + + Fom ealie, we know that + µ and that µ, so we have: + µ µ + µ But, µ and µ ae constants, and can be pulled out of the aveages: Pat d + µ µ + µ + µ µ + µ Given µ 3 and, calculate the pobability that <. Note: thee is no analytic fomula fo the esult, so you will have to calculate the integal numeically o look it up in a table. We can calculate z µ 3.5 when. The pobability of getting a value of between µ z 3 3 and µ + z is.86638, accoding to Table C. in [Bevington & Robinson, p5]. Thus, the chance of getting a value of outside this ange is , and the pobability of getting values only below zeo should be half this pobability since it epesents half of the emaining distibution, o.336/.668. Poblem Calculate v RMS v fo H molecules at T 3 K. Using potential enegy, show that Page 3

4 the escape velocity of the eath s gaitational field is v escape GM/R. Compae this vallue with v RMS. The eath s atmosphee contains vey little H. How is it possible that the H the molecules escape with a elatively small v RMS? Why doesn t N escape? [Tiple & Llewellyn, p8] 3kT We know that v RMS m, whee T is the tempeatue, m is the mass of the paticle, and k.38 3 J/K. Substituting, we have: v RMS J/K 3 K kg 99 m/s The escape velocity is the velocity at which some object has sufficient kenetic enegy to ovecome the gavational attaction of some othe object. That is, at infinite distance, the speed of an object initial taveling at escape velocity will be essentially zeo, but the escaping object will be able to each this distance, having ovecome the gavational pull of the efeence object. So, T U. If T mv e, whee v e is the escape velocity and m is the mass of the escaping object, and U GMm m3, whee G is the gavitional constant, M is the mass of the efeence object, kg s m is the mass of the escaping object, and is the initial distance between the cente of the efeence object and the escaping object, then we can easily calculate v e : T U mv e GMm mv e GMm ve GM GM v e If one of these objects is the eath, with mass M kg [Google] and adius km [also Google], then the escape velocity fo the eath is: v e GM m kg s 4 kg, 8 m/s m This escape velocity is much lage than the v RMS of H. Howeve, it is still possible fo the hydogen molecules to escape because the speed of hydogen molecules in the atmosphee follows Mawellian distibution and is not, in fact, a homogenous medium of hydogen molecules all taveling at a speed of v RMS. Some molecules in this distibution will have enough kinetic enegy to ovecome Eath s gavity. Nitogen will also follow a Mawellian distibution, but the v RMS of N molecules at 3 K is: v RMS J/K 3 K kg 55 m/s This is significantly smalle than the v RMS of hydogen molecules. While it is still possible fo individual N molecules to escape Eath s atmosphee, the numbe of nitogen molecules that ae capable of attaining the equisite escape velocity must be significantly lowe than fo hydogen molecules. Accoding to [Tiple & Llewellyn, p33], a gas will escape fom a planet s atmosphee in 8 yeas if the aveage speed of its molecules is one-sith of the escape velocity. The aveage velocity of nitogen molecules is, also accoding to [Tiple & Llewellyn, p3], about 475 m/s. This is oughly /4th the escape velocity fo Eath. I do not know if the 8 figue would scale up naïvely, so that the time equied would be, say, yeas, but if this is the case, then the eath would have to be 3 billion yeas old in ode fo N gas to completely escape Page 4

5 and, accoding to Wikipedia, the age of the eath is only about 4.5 billion yeas old. Even if the 8 figue scaled up as 8+4, this is still billion yeas. Eath still is not nealy old enough fo all the nitogen molecules to have all escaped. Also, it may be that gasses of molecules with an aveage velocity of less than /6th Eath s escape velocity will neve able to entiely escape Eath s gavational field. The book was not entiely clea on this. v RMS O J/K 3 K kg 484 m/s The aveage kinetic enegy, then, of each of these gasses is just E mv m v mv RMS. So we have: Poblem 3 E H kg 99 m/s Compae v RMS and the aveage kinetic enegy pe atom/molecule fo H, He, O, and N gas unde standad tempeatue and pessue. We aleady know fom Poblem that the v RMS of H is about 99 m/s, and fo N is about 55 m/s. Fo O and He, we fist have to calculate the mass of the individual molecules in units that ae actually usable. This is done by dividing the mola/atomic mass by N A 6. 3, Avagado s numbe: m He m O 4.6 g/mole kg 6. 3 paticles/mole g kg 3 g/mole kg 6. 3 paticles/mole g kg Now we can calculate v RMS fo helium atoms and oygen molecules eactly as was done fo nitogen and hydogen molecules: v RMS He J/K 3 K kg 367 m/s 6. J E N kg 55 m/s 6. J EHe kg 367 m/s 6. J E O kg 484 m/s 6. J So, while the vaious v RMS values ae significantly diffeent, the kinetic enegies fo each molecule at standad tempeatue and pessue ae basically identical. This stands to eason, since, accoding to [Tiple & Llewellyn, p34], the aveage enegy of a molecule is independent of mass. This could have been calculated fo each atom/molecule by using the fomula: Poblem 4 E 3 kt J Show that the following blackbody distibutions witten in tems of wavelength o fequency ae equivalent: Page 5

6 ũf df 8πf c 3 e hf kt hf df uλ dλ 8πhc λ 5 λkt dλ e hc Since c λf, we can substitute in fo fcλ in the fist equation to deive the second equation. We also need to substitute df cλ dλ. ũf df 8πf c 3 e hf kt 8π cλ 8πc λ c 3 c 3 hf df hcλ cλ dλ e hcλ kt hc c kt λ dλ λe hcλ 8π h λ 5 e hc λkt c dλ 8πhc λ 5 e hc λkt dλ Since this is the same as the ight hand side of the second equation, we can conclude that ũf df uλ dλ. Poblem 5 Integate Plank s law ove all wavelengths to deive Steffan s law: that the total powe adiated by a black body is whee using R T 4 π5 k 4 5h 3 c 3 e π4 5 Plank s law can be witten in two ways. The fist is: uλ 8πhcλ 5 e hc/λkt The above fom is a spectal enegy density, which has units of enegy pe unit volume pe unit fequency. Anothe way to wite Plank s law is: Iλ πhc λ 5 e hc/λkt This has units of powe pe unit aea pe solid angle whateve that means pe unit fequency. Since R is a powe pe unit aea, the Iλ is the fom of Plank s law that we want to use. If we use uλ, the esulting equation will be diffeent by a facto of 4 c. We can integate this, using hc λkt λ hc kt, and dλ λ uλ dλ kt hc : ˆ πk4 T 4 h 3 c πhc λ 5 e hc/λkt dλ πhc λ 3 dλ e hc/λkt λ πhc hc kt e, hc λ kt dλ, 3 πhc k 3 T 3 3 h 3 c 3 e πk 4 T 4 3 h 3 c e πk4 T 4 π 4 h 3 c 5 π5 k 4 T 4 5h 3 c π 5 k 4 5h 3 c T 4 T 4 3 e kt hc kt hc Page 6

7 Poblem 6 Set the deivative of Plank s law equal to zeo to show Wien s displacement law: that λ m T const m K whee λ m is the wavelength of maimum intensity adiation fom a blackbody. The Sun s suface tempeatue is 58 K. Calculate λ m. What colo is the sun? Plank s law is: 5λ 6 e hc/λkt λ 6 e hc/λkt hce hc/λkt λ 7 kt e hc/λkt 5 + hce hc/λkt λkt e hc/λkt Now we can substitute get: hc λkt hce hc/λkt λkt e hc/λkt into the equation, to e e 5 This Wikipedia cannot be solved in tems of elementay functions. It can be solved in tems of Lambet s Poduct Log function but an eact solution is not impotant in this deivation. Howeve, we can estimate faily easily. Fist, let s eaange the function so that it eads: 5 e e 5 e uλ 8πhcλ 5 Now we can let e, and estimate 5. Substituting that back into the ight side of the equa- e hc/λkt tion, we have 5 e Substituting this value back into the equation, we get Diffeentiating, we get: 5 e Substituting that back into the equation one last time, we get du dλ 8πhc 5λ 6 5 e Between the e hc/λkt last two iteations, the value of changed by less than λ 5 +8πhc e hc/λkt ehc/λkt hc.5. This is a easonable appoimation. So, we can say that λ kt Now if we substitute this back into hc λkt, we get, 8πhc 5λ 6 e hc/λkt + hce hc/λkt with a little shuffling aound: λ 7 kt e hc/λkt λt hc Setting this equal to zeo, we get: k m kg 3. s 8 m s 8πhc 5λ 6 e hc/λkt + hce hc/λkt λ 7 kt e hc/λkt.38 3 kg m s K.9 3 m K Appoimation eos aside, this agees well with the the poblem statement. The colo of the sun, if it is at 58 K, is λ nm This makes the sun s colo of highest contibution geen, close to blue. Howeve, the sun s colo is still white. Fo details, please see Page 7