PC Chapter 10

PC Chapter 10 - 10 Statistical Physics 10-1 Using n j = nj1...

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153 10 Statistical Physics 10-1 Using =++ K 11 22 j jj n np np we obtain: ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29  = + + =++++   +++++++ + 1 11 1 12 2 120 20 6 30 30 60 30 01020 1 287 1 287 1 287 1 287 1 287 120 60 15 120 60 180 30 1301024 1 287 1 287 1 287 1 287 1 287 1 287 1 287 60 0 1 287 n np np np ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29  ++++++   + ++++++++++ ++ = = = == 1 90 180 120 6 15 60 213502 1 287 1 287 1 287 1 287 1 287 1 287 15 4 1 287 30 120 120 180 120 360 120 180 180 360 30 120 60 1 287 1.538 46 1.538 1 0.256. 66 n pE One can find (29 2 through 8 in similar fashion. 10-2 As π π - = 2 B 32 2 2 B 4 2 mv kT N m nv ve V kT , has a maximum for = 0 dnv dv . Therefore, -- - -  = +=    -=  BB 2 B 2 B 2 2 B 2 20 2 2 0. mv dn v mv C v e dv mv Cve
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154 CHAPTER 10 STATISTICAL PHYSICS Thus we have extrema at = 0 v , =∞ v , and  -=   2 B 20 mv kT or = 12 B 2 v m . As (29 → 0 nv at = 0 v and , = 0 v and v are minima and = B 2 v m is a maximum of (29 . Thus = B 2 mp v m . 10-3 A molecule moving with speed v takes d v seconds to cross the cylinder, where d is the cylinder’s diameter. In this time the detector rotates θ radians where ϖ θ ϖ == d t v . This means the molecule strikes the curved glass plate at a distance from A of ϖ θ 2 22 d d s v as - 2 22 Bi 6.94 10 g m and ( 29 ( ( π π π - -  × =  ×  × 23 B 25 BB rms mp 2 rms mp 8 1.38 10 J K 850 K 8 207 ms 6.94 10 kg 32 225 m s 184 m s 6 2 5 02 0.10 m 1.45 cm 60 s 2 225 m s 1.58 cm 1.78 cm v v m vv mm s ss 10-4 [Strength of Absorption Line] = [Population of the initial state] [Transition Probability] Using M-B statistics to calculate the populations of 1 E and 2 E relative to the ground state, 0 E , we find for the strength of absorption lines at 4 K: ( ( ( - - -∆ ×× = = =∆ = = = = = 5 5 0 03 12.41 10 eV 8.62 10 4 eV 13 2 23 1 1.00 at 3 2 2 1.40 at 2 2 0.97 at 1 kT kT S E E eE S E E e S E E At 1 K, ( ( ( ( - - - - × × = = = = = 5 5 5 5 12.41 10 eV 8.62 10 2 12.41 10 8.62 10 1.00 at 3 2 0.47 at 2 2 0.11 at 1 S E EE S E E S E E
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MODERN PHYSICS 155 Energy 2 3 1 K 4 K Absorption Strength 10-5 Fit a curve - BE Ae to Figure 10.2. An ambitious solution would use a least squares fit to determine A and B . The quick fit suggested below uses a match only at 0 and 1 E . (29 - = BE P E Ae thus (29 = 0 PA and ( 29 - = 1 1 BE P E Ae . From Figure 10.2 one finds 0 0.385 P , and this gives = 0.385 A . To determine B use the value -- = == 11 1 0.256 0.385 BE BE P E A ee thus - = 1 0.665 BE e and = -= ln 0.665 0.408 B EE and so ( 29 ( - = 1 0.408 0.385 P Ee . This equation was used to determine the probability as follows 0 0.385 P , ( 29 = 1 1 0.256 PE , 1 2 0.170 , 1 3 0.113 , 1 4 0.075 , 1 5 0.050 , 1 6 0.033 , 1 7 0.022 , 1 8 0.015 . The exact values are 0 0.385 P , 1 1 0.256 , 1 2 0.167 , 1 3 0.078 , 1 4 0.054 , 1 5 0.027 , 1 6 0.012 , 1 7 0.003 9 , 1 8 0.000 717 . These values are plotted below. One sees that this approximation is good for low energy. There is exact agreement for 0 P and 1 and small deviations for the next two values with percent deviations for the higher energy values. 0 2 E 0.10 0.05 0.15 0.20 0.25 0.30 0.35 0.40 4 E 6 E 8 E 0 Probability of finding a particle with a given energy Energy 10-6 (a) +≈ 20 12 10 nn - - -  - ×× =  ×  19 16 2 2 33 1 4.86 eV 1.602 10 J eV exp 4.98 10 1.38 10 J K 1.600 10 K n n .
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PC Chapter 10 - 10 Statistical Physics 10-1 Using n j = nj1...

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