9he11.midterm.rev.answers

9he11.midterm.rev.answers - 1 Physics 9HE-Modern Physics...

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1 Physics 9HE-Modern Physics Midterm February 3, 2011 (100 points total) You may tear off this sheet. ---------------------------------------------------------------------------------------------- Miscellaneous data and equations: c = 3.00 x 10 8 m/s e = 1.60 x 10 -19 C 1 eV = 1.60 x 10 -19 J 1 Å = 10 -10 m M Sun = 2 x 10 30 kg M Earth = 5.98 x 10 24 kg r Earth = 6.38 x 10 6 m m e = 9.1094 x 10 -31 kg = 0.5110 MeV/c 2 m p = 1.6726 x 10 -27 kg = 938.27 MeV/c 2 m n = 1.6749 x 10 -27 kg = 939.57 MeV/c 2 1 u = 1.6605 x 10 -27 kg = 931.49 MeV/c 2 m( 1 H) = 1.0078 u G = 6.67 x 10 -11 Nt-m 2 /kg 2 g = 9.81 m/s 2 = 5.67 x 10 -8 W-m -2 -K -4 h = 6.63 x 10 -34 J-s = h/2 = 1.05 x 10 -34 J-s k B = 1.38 x 10 -23 J-K -1 a 0 = 0.529 Å N(t) = N o exp(-t/ O ) = N o exp(-0.693t/ 1/2 ) k C 1/(4  o ) = 8.98 x 10 9 N-m 2 -C -2 R H = 1.09678 x 10 7 m -1 2 2 2 1/2 2 1/2 11 1 0.5 (1 ) ) if v c vc     = v/c = [( 2 -1)/ 2 ] 1/2 T= T o L = L o / 1/ 2 00 1/ 2 ) [1 ] 1 ) for  = c/ (lect.) = f(book) 2 2 2 2 2 2 4 p mu E mc K mc E p c m c  2 2 2 12 1 1 2 ( 1) Sch T GM gH GM E pc h if R T r r c c c           max T = 2.898 x 10 -3 m-K 2 5 21 ( , ) 1 hc kT ch IT e  R(T) =  T 4 h = K max + n = 2dsin 2 coul C radial 2 qq mv F k F r r  2 0 0 2 4 0.529 e a me Å = h/p p = k x p x /2 E t /2 ph v = /k gr v = d /dk ix ix ix ix ix e cos x i sin x cos x e e sin x e e 2 2i sin 2t = 2 sin t cos t cos 2t = cos 2 t – sin 2 t = 2 cos 2 t – 1 = 1 – 2 sin 2 t     0 n n n n 1 n 1 0 n n a 2 n 2 n 2 n ( x ) a cos( x ) b sin( x ), with k , and 2 2 2 n 2 2 n a aver. of over , a
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9he11.midterm.rev.answers - 1 Physics 9HE-Modern Physics...

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