2006年第38届IChO预备试题答案英文

Problem 1: “A brief history” of Hfe in the universe

r=10'^/(1 严=102 K(10 billion degrees)

12 T= 1()2/ (180严= 07x10"^ 10® K(1 billion degrees)

1-3 t= [1(r°/(3x1 03)F S = 10伯S = 3 X 105 yr

1-4, f=(10'^/10Vs = 1O'^s = 3x 10^ yr

1-5, 100 K

1-6- 10 K

a-(f)-(d)-(h)-{i)-(c)-(g)-(j)-{e)-(b)

Problem 2: Hydrogen in outer space

[(8 X 8,3 J K?t moH x 27 K)/(3 J 4)(10'^ kg 010卩)]恢=240 m L

22 volume of cylinder =⑵^^2(3J4)(iO'®cm)^(2.4 x 10° cm s') = 1J x 10小cm's，

2-3- collision/sec = (volume of cylinder) x (atoms/unit volume)

=(1,1 X 10小cmT)(10?6cm?3) = 1.1 x 10" s，

time between collisions = 1/(1J x 10" L) = 9 x 10'^ s = about 3 billion yr

24 (240 m L)(9 X W'6 s) = 2,2 x 10'® m (about 2,000 light yr)

2-5, Speed is proportional to the square root of the temperature. (240 m L)(40/2?7严=920 m s'

2-6- volume of cylinder =⑵恢(3.i4)(io?8cm尸(9.2 x 10° cm s') = 4.1 x 10小cm%' collision/sec = (volume swept per second) x (atoms/unit volume)

=(4.1 X 10小cm? s ')(1 cm?3) = 4.1 x W小s'

time between collisions = 1/(4J x 10小L) = 2,4 x 10'° s = about 800 yr

mean free path = (920 m s')(2?4 x 1s) = 2.2 x 10^ m

人(intergalactic space)M(interstellar space)

=(2,2 X 10'9 m)/(2?2 X 10心m) = about a million

2-7, very small

Problem 3: Spectroscopy of interstellar molecules

3-1- 1002 = 2-9 X 10-^ m K A = 2.9 X 10?5 m

E(photon) = hd入=(6.63 x 10皿J s) (3.0 x 10® m/s) /(2?9 x 10'^ m) = 6?9x10?2i J

32 J: 1

(12x16/28)(1 ?66x 10刃 kg) = 1-14x lO'^^kg

/ = “fl2=(1?14x 10?2«kg)(1?13x 105)2 = 1?45x 104 kg

E(0<^1) = 2 护/8兀引=2(6.63 X E J s)2/[8说 1.45 x kg m^))

= 7-68x10'^ J

£(photon) of Problem 3-1 = 6.9 x 10② J > E(0<->1) = 7,68 x lO^ j

Rotational excitation by the background radiation is feasible.

E(0<->2) = 6 护/8於/ = hcU 2= 8於c//6h

/ = //fl^=({1 ⑵ X 1?66 X 10刃 kg](0.74 x 10虫 m尸=4.55 x W kg m?

2 = 8jt2/c/6h

=[8 於 X 4-55 X 10?48 kgm^xSx 10® m/s]/(6 x 6.63 x 10* J s)

= 271 X 105

r=2.9x 10-^m K/2 = 2-9x 10-^m K/271 x 105 = 107 K

Observation of hydrogen rotational spectra is feasible at 100 K. Problem 4: Ideal gas law at the core of the sun

4-1- protons: (158 g/cnr^x 0.36)/{1.0 g/mole) = 57 mol/cm^ helium nuclei: (158 g/cm^x 0?64)/(4?0 g/mole) = 25 mol/cirP electrons:

57 + (25 x 2) = 107 mol/cm^

Total: 189 mol/cm^

42 volume of a hydrogen molecule = 2 (4/3) 7t 卢

=2x (4/3) n X (0.53 x lO'^cm)^ = 1.2 x 10公 cm?

hydrogen gas: V/n = RTJp = (0,082 atm L K，moP) x 300 K /1 atm

=24.6 L/mole = 4,1 x Umolecule = 4J x 10'^ cm^/molecule

** X 10-24 cm3/4J x 10-20 cm3 = 3 x 10?5 = 0.003 % liquid hydrogen: (2 g/mole) / (0.09 g/cm?) / (6 x 10羽 molecule /mole)

=37 X 10?23cm3

(1.2 X W 公 cm3)/(3?7 x cm^) = 0.03 = 3% solar plasma: neglect volume of electrons (4/3)(71)(14 X 10山 cm)3 (1 X 57 mol/crrP + 4 x 25 mol/cm^)(6 x 10^^ moM)

=1J X 10? J 1 X 10」。％

Volume occupied is extremely small and ideal gas law ts applicable.

4-3, From 4-1, we know there are 189 moles of particles/cnr?.

r = pV/n/?=(2.5x 10")(1 X 10'^)/(189)(0-082) = 1.6 x 10^ K

Problem 5: Atmosphere of the pla nets

After almost one half-life, the molar ratio between Pb-206 and U-238 is 1.

Mass ratio: Pb-206/U-238 = 206/238 = 0.87

(1/2)mvfe2= GMm/R

v^ = (2GM/R} = [(2)(6.67x10 " N m2 kg ?2)(5.98 x 10^^ kg)/(6.37 x 10® m)]

% =1.12x104 ms'

hydrogen atom: {SRT/TI M)虫

=[(8)(8.3145 kg moN K ')(298 K)/(3.14)(1.008 x IO? kg moN)]”2

=2500 m s' (22% of the escape velocity)

nitrogen molecule:

2500 m s' X (1/28)恢=470 m L (4% of the escape velocity)

The fraction with speed exceeding the escape velocity is much greater for hydrogen atoms than for nitrogen molecules. 5-1,

莖U T 豐 Pb + 8；He + 6j ；e

5-3, 5-4,

a. JupIter: large mass, low temperature, H/He retained at high pressure

b? Venus: lost light elements, rich in carbon dioxide, high pressure

c.Mars: small mass, rich in carbon dioxide, low pressure

d.Earth: lost light elements, carb on dioxide converted to oxyge n

through photosynthesis

e.Pluto: very small mass, lost light elements, very low atm os pheric p ressure 5-5,

5-6-

H-H He : :o=c=o: ：N=N：：O=O :

HICIH

5-7, He (4K) V H2 (20K) < N2 (77K) < O2 (90K) vChUH 12K)

Dispersion force is greater for larger molecules.

Nitrogen with the triple bond has a smaller bond length than oxygen. Nitroge n

also has less lone pair electro ns to be involved in disp ersion.

Problem 6: Discovery of the noble gases

6-1- In 1816 Prout published a hypo thesis that all matter is com posed ultimately of hydrogen.

(Later, Harlow Shapley, an eminent astronomer, said that if God did create the world by a word, the word would have been hydrogen.) Prout cited as evidenee the fact that the specific gravities of gaseous elements appeared to be whole-number multiples of the value for hydrogen.

62 28 NHa + 21 O2 + 78 N2 + Ar t 92 N2 + 42 hbO + Ar

6-3- [(92)⑵(14-0067) + 39-948)/93 = 28.142

64 78 Nz + 21 O2 +Ar + 42 Cu -> 78 N2 + 42 CuO + Ar

6-5, [(78)⑵(14-0067) + 39-9481/79 = 28.164

72 4 NH3 + 3 O2 T 2 N2 + 6 H2O

Molecular weight of pure nitrogen = (2)(14,0067) = 28.013

**/28.013 = r0054

The discrepancy would increase about 7-folcl (0.0054/0.0008),

volume of air = 1000 = 10^ liter

(106)/22.4 = 4?5 X 1()4 mol of atr

weight of argon = (4.5 x 10^)(0,01)(40) = 1,8 x 10^g = 18 kg

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