solution#5 - Chapter 4 Structure of the Atom h Zl =2, Z2 =...

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Unformatted text preview: Chapter 4 Structure of the Atom h Zl =2, Z2 = 79, and 0 = 1° we have 2 a (2) (79) (1.44 x 10-9 eV - m) _12 — __————————— t 0.5° = 1.69 x 10 m 60t( > 2 (7.7 x 106 eV) c° ( ) ' 2 6 (2) (79) (1.44 x 10—9 eV - m) _14 ——— =_______————— t45° =1.48><10 m cot (2) 2 (7.7 x 106 eV) °° ( ) 43 8. From Example 4.2 we know n = 5.90 x 1028 m'3. Thus _ 2122732 2 2 9. f —— 7rnt Got 5 7r (5.90 x 1028 m“3) (10-8 m) 2 (2) (79) (1.44 x 10-9 eV . m) Mme) 2 (5 x 106 eV) 3.49 x 10-4 9. a) With all other parameters equal the number depends only on the scattering angles, so f(90°) _ cot2 (45°) _ m 7 — 0'2” so the number scattered through angles greater than 90° is (10000) (0.217) = 2170. b) Similarly 2 f(70°) _ cot (35°) _ f—(-50—) ————COt2 (25°) _ 0.4435 ) o 2 o f(80 cot (40) =OI3088 f(50°) cot2 (25°) 0 The numbers for the two angles are thus 4435 and 3088 and the number scattered betweel 70° and 80° is 4435 — 3088 = 1347. I . 6 ea . \/144 eV nm C : 1.53c ’U=-——————: — x/47reom'r x/47T60m627‘ — (511000 eV) (1.2 x 10—6 nm) Ch is not an allowed speed. 2 . = e _ 1.44 eV nm 2—600 keV _87reor _ _2 (1.2 X 10—6 nm) Glearly (a) is not allowed and (b) is too much energy. 45 16. For hydrogen: 1.44 eV - nm e _ ec : —__’_”.‘—_""_’____\/C U z ./4' Mom?" — x/47reomcgr ./(511 X 103 eV) (0.0529 nm) 7.30 x 10—30 = 2.19 x 106 m/s H 6 2 2 (2-19 X 10 m/S) ; 9.07 ><1022 m/SQ v '— —— :: f a” ’ 7" 5.29 X10“11 m For the hydrogen-like Li++ Z62 7m;2 2 Z62 = = ———- or v = 4775072 7 47reorm But We also know # 4mm? # (LO — Zme2 Z ZQe"2 32 (1.44 eV - nm) Z 4.79 X 10_4C2 U2 — : / * 47reoaom (5.29 x 10—11 m) (511 x 103 eV/cg) or v z 2.19 X 10’26 = 6.57 X 106 m/s. This is a factor of 3 greater than the speed for hydrogen. 2 1 U2 - = 2.4.5 x 1024 m/s2 “ ” T ‘ (5.29 x10'11m)/3 * 22. From Equation (4.31) on = (1/n)(7‘1/ma0) 1 .055 10-34 J - I_.__l_i————i-——— = 2.19 x 106 m/s 2 0.0073c (9.11 x 10-31 kg) (5.29 x 10—11 m) H n=12 711 1.055 x 10-34 J -s 1 “M’— = 1-0 106 = 00036 2 (9.11 x 10—31L kg) (5.29 x 10—11 m) 9 X m/s c 1 3 3 ll 5‘? e [0 H 1.055 x 10~34 J -s (9.11 x 10~31 kg) (5.29 x 10-11 m) = 7.30 x 105 m/s = 0.0024c n=3z 123 * 23. The photon energy is E___£e:124.0eV-nm=2866V A 434 nm This is the energy difference between the two states in hydrogen. From Figure 4.16, we see E3 = ——1.51 eV so the initial state must be 71 = 2. We notice that this energy difference exist: between n = 2 (with E2 = —3.40 eV) and n = 5 (with E5 = —0.54 eV). 6. As mentioned in the text below Equation (4.38), if each atom can be treated as single-electron atoms (and the problem states we can make this assumption), then 1 R: 1+fi R00 where R00 2 10973731534 x 107 m"1 and m = 00005485799 11. :Using 4He (M = 4.0026 u), R = 0.999863Roo = 1.097223 x 107 m-1 (off by 0.014%) WUsing 39K (M = 38.963708 u), R = 0.9999859Roo = 1.097358 x 107 m-1 (off by 0.001476) 5’Using 238U (M = 238.05078 u), R = 0.9999977Roo = 1.097371 X 107 m-1 (off by 0.00023%) 42. Refer to Figure 4.19 and Equation (4.46). A is inversely proportional to (Z — 1)2 for the K series and to (Z — 7 .4)2 for the L series. a) MU) _ (6— 1)2 _ m _ (92 _ D2 _0.0030 b) MW) g (20 — 7.4)2 _ A(Ca) ’ (74 — 7.4)2 .“ 0'036 48- 3.6 eV, 4.6 eV, 2(3.6 eV) = 7.2 eV, 3.6 eV 4— 4.6 eV = 8.2 eV, etc. with other combinations giving 10.8 eV, 11.8 eV, ligey, 14.4 ev, 15.4 eV, 16.2w, 16.4 eV, 17.2 eV, 18.0 eV. ...
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This note was uploaded on 04/23/2011 for the course PHYS 342 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.

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solution#5 - Chapter 4 Structure of the Atom h Zl =2, Z2 =...

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