# 22 a compact disk has a moment of inertia of 5 kg m 2

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22 · A compact disk has a moment of inertia of about 2.3 × 10 –5 kg m 2 . ( a ) Find its angular momentum L when it is rotating at 500 rev/min. ( b ) Find the approximate value of the quantum number l for this angular momentum. ( a ) L = I ? ( b ) l 2245 L / h _ ? = (2 p × 500/60) rad/s; L = 1.20 × 10 –3 J . s l 2245 1.20 × 10 –3 /1.055 × 10 –34 2245 10 31 23 · If n = 3, ( a ) what are the possible values of l ? ( b ) For each value of l in ( a ), list the possible values of m . ( c ) Using the fact that there are two quantum states for each value of l and m because of electron spin, find the total number of electron states with n = 3. ( a ), ( b ) See Equ. 37-23 l = 0: m = 0; l =1: m = –1, 0, 1; l = 2: m = –2, –1, 0, 1, 2 ( c ) We will first derive the number of electron states for an arbitrary value of n and then substitute the specific value of n . The number of m states for a given n is given by (1). + 2 = 1) + (2 = N 1 n 0 = 1 n 0 = 1 n 0 = m - - - l l l l l The sum of all integers from 0 to p is p ( p + 1)/2, so the first term is 2[( n – 1) n /2] = n 2 n . The second term is just equal to n . So N m = n 2 . Since N , the number of electron states is twice the number of m states, the number of electron states is N = 2 n 2 . For n = 3, N = 18.

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