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22 ·A compact disk has a moment of inertia of about 2.3×10–5 kg⋅m2. (a) Find its angular momentum Lwhen it is rotating at 500 rev/min. (b) Find the approximate value of the quantum number lfor this angular momentum. (a)L= I?(b)l2245L/h_?= (2p×500/60) rad/s; L= 1.20×10–3 J.s l22451.20×10–3/1.055×10–342245103123 ·If n= 3, (a) what are the possible values of l? (b) For each value of lin (a), list the possible values of m. (c) Using the fact that there are two quantum states for each value of land 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 nand then substitute the specific value of n. The number of mstates for a given nis given by (1).+2=1)+(2=N1n0=1n0=1n0=m∑∑∑---lllllThe sum of all integers from 0 to pis p(p + 1)/2, so the first term is 2[(n– 1)n/2] = n2– n. The second term is just equal to n. So Nm= n2. Since N, the number of electron states is twice the number of mstates, the number of electron states is N= 2n2. For n= 3, N= 18.
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