P hysics 318 Solution f or P roblem Set 8 1. rate=number of simulated emissions/per second. rate = N2 B21 u(f ) where the B21 is the coecient of stimulated emission, N2 is the number of atoms in the upper lever, u is the electromagnetic energy density. an
P hysics 318 Solution f or P roblem Set 4 1. (a)an atom with three p-shell valence electrons l1 = l2 = l3 = 1, and s1 = s2 = s3 = 1 2 assume l = l1 + l2 |l1 l2 | l |l1 + l2 | 0 l 2 and l3 = 1,0 l + l3 3 0 l 3 so l can be 0, 1, 2, 3 Notation: a lot of stud
P hysics 318 Solution f or P roblem Set1 Each subsection counts for 10 points, if you have questions, please come to my oce crow 220, or send me email dhu@physics.wustl.edu, I think my oce hour is Monday 12-1pm, if this time is not good for you ,you are a
P hysics 318 Solution f or P roblem Set 6
1. See above. 2. (a) from the problem that the Fermi energy of aluminum is 11.63 ev at low temperature, low temperature means that T 0 1
and for free electrons, they are fermions, so the mean number of free electr
P hysics 318 Solution f or P roblem Set 10 1. From the problem, the 0 has a mass m of approximately 135M ev/c2 So, m0 c2 = 135M ev mc = c 197.3M ev 1015 m = 1.463 1015 m = 2 m0 c 135M ev
While the fermi is dened as 1015 meters. So the range is the order o
P hysics 318 Solution f or P roblem Set 9 1 When k near the origin, that means k 0 2 (k a)2 using Taylor expansion, cos(kx a) 1 (kx2a) + . . ., cos(ky a) 1 y2 + . . ., cos(kz a) 1
(kz a)2 2
+ .
l
(kx a)2 (ky a)2 (kz a)2 +1 +1 ] 2 2 2 2 2 2 kx + ky + kz 2
P hysics 318 Solution f or P roblem Set7 1 The maximum value of f (x) = and f (x) = x3 df (x) is given by = 0. ex 1 dx
xex = 0 x 2.82 that means h = 2.82kB T 2.82kB T At T = 2.7K , = = 1.59 1011 s1 h
(ex 1) 3x2 ex x3 = 0 3x2 (ex 1) x3 ex = 0 3ex 3 (ex 1)2