Problem 1: Proton antiproton atom We can use the expression for hydrogen-like atoms to calculate the energy levels.

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1 Proble 1: Proton antiproton ato We can use the expression or hydrogen-like atos to calculate the energy levels. πz e µ E n = where Z is the total nuber o charges in the nucleus (= 1), e is ( πε ) h n the electron charge (= 1.6 x 1 19 C), μ is the reduced ass o the syste with 1 1 ( ) µ = and 1 = = p = x 1-7 kg, thus μ = ½ p, ε is the perittivity o vacuu (= 8.85 x 1-1 C J -1-1 ) [note: (πε ) -1 is Coulob's constant], h is the Planck constant (= x 1-3 J s) and n is the principal quantu nuber o the syste taking values 1,, e 5 E1 For n = 1, E1 = p =.19x1 J h and or n =, E =, hence ε 3 5 E E E1 = E1 = 1.597x1 J 5 E 1.597x1 J ΔE = h ν,hence ν = = =.65 s. 3 h x1 J s 8 c.99795x1 s c = λ ν, hence λ = c / ν = λ = = = 1.33x1 = Å. ν.65 s The Bohr radius is given by h ( πε ) h ε α = = π µ e π pe = x1 which is 1836/ ties saller than the hydrogen radius due to the dierence in reduced ass o the "ato". Proble : Annulene The nuber o π electrons is 18. Two electrons can occupy each state due to the Pauli exclusion principle. Each state above the lowest is two-old degenerate. Based on these pieces o inoration, we can ill this table: ax nuber o total e - up e - per state to this state States up to = are ully occupied with 18 electrons. The lowest possible transition is ro state = to state = 5. The path oring a circular well is L = 18 x 1. Å. Hence 3 ( ) ( x1 J s) 5 x( 18x1. Å) 9 E = E 5 E = = 3.15x1 J 31 x x1 kg The transition requency is ν = ΔE/h = 5.15 x 1 1 s -1 and the corresponding wavelength is λ = c/ν = n. Proble 3: Cheical bonding: The olecular ion kcal/ol. o + O

2 3. Yes kcal/ol 5. ( ) kcal/ol = 86.3 kcal/ol or (86.3/ A ) kcal/olecule = 1.3 x 1-3 kcal/olecule. 6. ~1.1 Å 7. ~1.6 Å Proble : Electrocheistry: icad batteries io(oh) (s) + H O + e i(oh) (s) + OH E =.9 V T Ec = Ec ln OH F Cd (s) + OH Cd(OH) (s) + e E a =+.89 V T 1 Ea = Ea ln F OH Cd (s) + io(oh) (s) + H O disch arge ch arge i(oh) (s) + Cd(OH) (s) E = E a E c = Ea Ec =.89 V (-.9 V) = 1.99 V. 7 Ah =.7 A x 36 s = 5. C C oles o Cd =.13 oles o Cd.13x11. = 1.7 g o 9685 Cd. Proble 5: Boiler Tank capacity: = V ρ = 3 x.73 g c -3 = 9 kg Heating power P = 116 kw P 116 kj s -3 Consuption rate = = =.7 x 1 kg s ( = 9.73 kg h ) 7 t h.3x1 J kg 9 kg 8 Operation duration t = = = 1.8 x 1 s = 3 h = 1.5 days x 1 kg s t C n H n+ + ½(3n+1) O nco + (n+1)h O CO n = = This ratio does not depend heavily on n; or a 1n 1 CnH + n n representative value o n = 1, the ratio takes the value o 3.1. CnHn + Since t = = 9.73 kg h CnHn + CO -1 CO = 9.73 kg h x3.1 = t CnH n + -1, 3. kg h Proble 6: Aoniu nitrate Mixing is endotheric and the process is adiabatic, thus heat has to be provided by the solution itsel. Since water is at its reezing point, it will tend to reeze, but the solution created will experience a depression o reezing point due to the presence o -1 c 3

3 dissolved ions. The large aount o heat required or solvation will necessitate soe reezing o water. A Hess cycle o 3 steps will be considered. A. ixing at C with ΔH 1 > B. lowering o the teperature o the ixture to its inal teperature with ΔH < C. reezing o soe water s with ΔH 3 < n The inal teperature is given by θ = K where K is the cryoscopy constant, is the nuber o particles per orula weight or H O 3, n the nuber 8 g o oles o H O 3 n = 1ol 1 8 g ol =, = 1 g. ΔH 1 = Δh s n = kj ol -1 x 1 ol = kj H = c P θ where c P is the olar heat capacity o water and M its olecular M ass (18 g ol -1 ). s H 3 = h where Δh is the olar enthalpy o usion. M ΔH 1 + ΔH + ΔH 3 = ΔH total = because no heat is allowed to be exchanged between the syste and its surroundings. Substituting or θ and solving or s yields the ollowing expression h s nm h s nm nk c P s = + ± + h h h We discount the solution derived ro the + sign as unphysical ( s > ) and arrive at the result s = 8.5 g ice. Hence θ = C. I we ade the sipliication that we expect s <<, then θ is iediately calculated as 3.7 C, which yields a value or s = 9.9g. I this result is used to iprove the θ value using the exact expression, we get θ = C. Then, s can be urther iproved to 8.5 g. The process is spontaneous, irreversible, one where separation o coponents is possible, adiabatic, isobaric, isenthalpic, nearly isoenergetic. The equation ΔG = ΔH - T ΔS can be used here because T varies less than %. ΔG < because the process is spontaneous and ΔH =, hence ΔS >. This is also to be expected ro stability criteria under the constraint ΔH =. Proble 7: Carbon dioxide A rather accurate phase diagra or CO is shown here using data ro Landolt- Börnstein ew Series (P c,t c ) IV/B, p. which 6 has been based on solid liquid Dykyi, J. epas, M.: Saturated Vapor (P 3,T 3 ) Pressure o Organic Copounds, Bratislava, gas Czech.: Slovakian Acadey o Science, Teperature (K) Pressure (kpa) s

4 A qualitative one can be drawn based on the triple and the critical points. Since the roo teperature is well above the triple point, there is no way there can be any solid CO in the ire extinguisher. At 98 K, the CO vapor pressure is 63.1 bar. This value can be estiated by drawing a straight line between the triple and the critical points. It can be calculated also based on the epirical Antoine equation, viz., P B log = A with A = 6.61, B = 78.8 and C = kpa C + T K Proble 8: Iron crystal (a) Let be the atoic radius o iron and a =.87 Å the length o the unit cell edge. Then, as atos touch each other along the body diagonal and ro a Pythagorean theore in the cube: a 3 = = (a 3 )/ = 1. Å The Avogadro nuber ( A ) can be calculated ro the density (ρ) orula. The latter is obtained by inding the nuber o atos per unit cell, ultiplying this nuber by the ass o each ato ole o atos(g / ol) ( ) and, A (atos / ol) eventually, dividing the result by the volue o the unit cell (a 3 ). ote that each bcc unit cell contains two whole spheres, that is Fe atos. ρ(g/c 3 (55.87 / A )g ) = 3 3 a (c ) g A = A g (.87 1 ) c 3 c (b) By applying the Pythagorean theore in the cube, one inds: a + a = () a = = a = 1.7 Å (slightly dierent ro the value ound above or bcc structure, because o the dierent packing, having an inluence on the atoic radius or at least its estiation). As or the density, recalling that each cc unit cell contains our whole spheres, that is Fe atos, once again one has: ρ(g/c 3 (55.87 / A )g ) = = 3 3 a (c ) 5

5 55.87 g = 8. g/c ( ) c The higher value o γ-fe density, as copared with α-fe, points at the act that the cc structure is denser than bcc. cc represents the, so called, cubic close packed structure which, together with the hexagonal close packed, are the ost eicient ways o packing together equal sized spheres in three diensions. (c) and (d) The unit cells below are illustrated by using reduced size spheres. ote that, in hard spheres packing odel the represented atos ust be in contact one to each other. According to the let igure, a perectly itted interstitial ato centered at (½,, ½) in an α-fe cell, would have a radius o: interstitial = ½ a - Fe, where a =.87 A and Fe = 1. Å [(see question (a)]. Thereore: interstitial (α-fe). Å Siilarly, according to the igure in right, a perectly itted interstitial ato centered at (½, ½, ½) in an γ-fe cell, would have a radius o: interstitial = ½ a - Fe, where a =3.59 A and Fe = 1.7 A [(see question (b)]. Thereore: interstitial (γ-fe).53 Å (e) 1 n = 1 Å. Thus: For α-fe: carbon int erstitial.77 A =.A = 3.85 carbon.77 A For γ-fe: = = 1.5 int erstitial.53a Thereore, the carbon ato is roughly our ties too large to it next to the nearest iron atos in α-fe without strain, while it is only 1.5 ties oversize to it in the γ-fe structure. The above estiations explain well the low solubility o carbon in α-fe (<.1 %). () The wavelength (λ) o the X-rays will be calculated ro Bragg's law, assuing irst order diraction: d sinθ = λ, where θ is the angle o diraction equal to 3.6 o and d is the interplanar spacing o the () set o parallel lattice planes, that is, the perpendicular distance between any pair o adjacent planes in the set. The () planes are shown 6

6 shaded in the igure. Let a be the length o the cubic unit cell edge. Then, ro previous data or α-fe: a =.87 A, so the distance between adjacent () planes is: d = a = 1. Å Thereore, ro Bragg's law: λ = d sinθ = 1. sin(3.6 o ) λ 1.55 Å This value corresponds to the K α1 transition o iron. Proble 9: Cyclodextrine a) V = a x c. b = a b c sin(β) = 77 Å 3, V = 77 Å 3 / = 1868 Å 3 b) FW = g ol -1, ρ = FW /(V A ), hence ρ = g c -3 Proble 1: Inrared spectroscopy 1. The nuber o vibrational odes is given by 3-6 or non-linear and 3-5 or linear olecules, where is the nuber o atos in the olecule. Hence CO: 1, H O: 3, C 6 H 6 : 3, C 6 : 17. The act that the olecules have a vibrational inra red absorption iplies that the olecules have a peranent dipole oent, hence these diatoic olecules ust be heteronuclear. For a siple haronic oscillator-type diatoic olecule, the eigen requency is given 1 k by the equation ν =, where k is the orce constant and μ the reduced ass o π µ the olecule. In the absence o any urther inoration, nothing can be said about the reduce asses or the orce constants (though bond strength is not related to k, but ore likely to bond dissociation energy). The eigen requency is equal to the requency o the absorbed photons because the vibrational energy is given by E v = (v+½) h ν and the energy or the resonance transition is ΔE = E v=1 E v= = h ν. Proble 11: adioactivity and cheical reactivity 1. γ-rays are a or o high energy electroagnetic radiation.. There are no stable (non-radioactive) isotopes beyond Bi. 3. There are nuerous light isotopes that are radioactive.. Xe copounds such as XeF are coercially available. 5. Cs is the eleent with the lowest ionization potential (3.89 ev). Proble 1: Carbon dating a) Let be the 1 C/ 1 C ratio in living systes and the sae ratio ound in a saple coing ro a syste that died t years ago. Then, the ollowing relation λt between the is true: = e, where λ (= ln / t ½ ) the disintegration constant or 1 C. The above equation becoes ln t = λ t ½ = ln ln ln.5 = 57y = 11y ln b) The β decay schee is based on the nuclear reaction n p + β + ν e, where p is a proton and ν e an electron antineutrino. In the case o 1 C we have 1 C 1 + β - + ν hence C becoes a (coon) 1 ato. c) I an organic olecule contains 1 C, the consequence o its disintegration can be grave or the structure o the olecule, causing great daage to the olecule in 7

7 the vicinity o the 1 C ato. At least the cheical bond is raptured since 1 is a cheically dierent ato than 1 C. Free radicals ay also be created. d) The total carbon inside a huan body o 75 Kg is 75 kg x.185 = 13.9 kg. The total radioactivity () is =.77 Bq / g x13.9 kg = 385 Bq The aount o 1 C present is estiated ro the total radioactivity as ollow: d = λ dt Then A t ½ 57 y 6x6xx365.5 s 15 = = A = 385s = 1.x1 atos = 1.66 nol λ ln.693 y Proble 13: Uraniu a) alpha decay: X(A,Z) X(A-, Z-) + He + (p+n) (ΔΑ = -, ΔΖ= -) beta decay: X(A,Z) X(A, Z+1) + β + ν e (ΔΑ =, ΔΖ = +1) Since changes in the ass nuber (ΔA) are due to the eission o alpha particles only, in each series we have: total alpha particles eitted = ΔA total / Alpha eission also changes the atoic nuber (Z) (ΔZ = -), so the total decrease in Z due to the total alpha particles eitted would be twice their total nuber. But Z o the inal (stable) eleent o the radioactive series is higher than the expected Z based on alpha eission. This dierence in Z is due to the nuber o beta particle eitted. Thus 38 U 6 Pb, α = ΔA/ = (38-6)/ = 3/ = 8, β = α - ΔΖ = 18 -(9-8) = 6 35 U 7 Pb, α = ΔA/ = (35-7)/ = 8/ = 7, β = α - ΔΖ = 1-(9-8) = b) This occurs when an alpha decay (ΔZ = -) is ollowed by two successive beta decays (ΔZ = +). c) For each radioisotope o uraniu we can write 35 Ν = 35 Ν exp(-λ 35 t) and 38 Ν = 38 Ν exp(-λ 38 t) where the nuber o nuclei at tie t, at tie t = and λ = ln / t ½ =.693 / t ½ the disintegration constant. At t =, 35 = 38, then 38 exp( λ 38t) 99.3 = = = 1 35 exp( λ 35t).7 Thus λ 35 t - λ 38 t = ln 1 =.95 λ 38 =.693/t ½ =.693/.5x1 9 = 1.5x1-1 y -1 λ 35 =.693/t ½ =.693/7.1x1 8 = 9.76x1-1 y x1 y 9 t = = = 6.x1 y 1 ( ) x1 y 8. d) The energy released by the coplete ission o 1g 35 U is E = (1/35) x 6. x 1 3 x MeV = 5.13 x1 3 MeV and the energy released upon cobustion o 1 g C is E = (1/1) x 6. x 1 3 x.1 ev =.6 x 1 3 ev =.6x1 17 MeV Hence the aount o carbon that would release the sae aount o energy as the ission o 1 g 35 U is = (5.13x1 3 )/(.6x1 17 ) =.9 x 1 3 kg C 8