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chapter6

# chapter6 - Problem 6.3 This is not a difficult problem but...

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Problem 6.3 This is not a difficult problem but it is a real pain to transfer it from paper into Mathcad. I won't give it to you on the quiz, but know how to do it for the exam. ___________________________________________________________________________ Problem 6.10 See Figure 6.4. The magnitude of L is Ll () l 1 () . hbar . and the z-component is L z m l hbar . The angle between L and the z-axis is found from cos θ L z L Let's do this first for l=1. Notice there is an hbar in both numerator and denominator of the equation for cos( θ ), so the hbar's cancel. For convenience, I will set hbar=1. hbar 1 l1 m l 1 0 , 1 .. L z m l m l hbar . l . hbar . θ m l l , 360 2 π . acos L z m l . θ m l l , 135 90 45 = These are the angles in degrees measured clockwise from the positive z-axis to the vector L. Now let's do it for l=2 l2 m l 2 1 , 2 .. θ m l l , 144.736 114.095 90 65.905 35.264 = ___________________________________________________________________________ 1

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Problem 6.11 The magnetic quantum number, ml, is limited to the values 0, ñ 1, ñ 2, .... ñ l. There are a total of 2l + 1 values. For an orbital quantum number of l = 4, there are nine possible values for m.l: m l 4 3 , 2 , 1 , 0 , 1 , 2 , 3 , 4 , ___________________________________________________________________________ Problem 6.12 The orbital quantum number l may take the values l = 0, 1, 2, ...... n-1, where n is the principal quantum number. Each orbital quantum number has an associated set of magnetic quantum numbers, m.l. The magnetic quantum number m.l is limited to the values 0, ñ 1, ñ 2, .... ñ l. There are a total of 2l + 1 values for each value of the orbital quantum number l. For a principal quantum number n = 4, l takes the values l = 0, 1, 2, 3. l = 0 m.l = 0 l = 1 m.l = -1, 0-, 1 l = 2 m.l = -2, -1, 0, 1, 2 l = 3 m.l = -3, -2, -1, 0, 1, 2, 3 The total number of orbitals with n = 4 is therefore N = 1 + 3 + 5 + 7 = 16. ___________________________________________________________________________ Problem 6.13 s p d f l= 0 1 2 3 Ll l 1 () . hbar . L z m l hbar . because m l ranges from -l to l, the maximum value of m l is l and the maximum value of L z is l*hbar L zmax l hbar . hbar cancels out in the equation for percentage difference, so for convenience, I will just set it equal to 1 hbar 1 percent l l () l 1 . hbar . l hbar . l . hbar . 100 . 2
percent 1 ( ) 29.289 = percent 2 ( ) 18.35 = percent 3 ( ) 13.397 = Beiser's definition of percentage difference is not necessarily what I would call percent difference. Beiser really asked "what percent of L is the maximum value of L.z." Here is the real definition of percentage difference between two quantities A and B: However, on an exam or quiz, do it like Beiser did it. P AB 1 2 () . 100 . ___________________________________________________________________________ Problem 6.14 A spherically symmetric probability-density distribution corresponds to zero orbital angular momentum. This occurs only for l=0; i.e., for s wavefunctions. In table 6.1, when l=0, then m.l=0, and the wave functions φ and θ are independent of angle, which is why the probability-density distribution is independent of angle. ___________________________________________________________________________ Problem 6.15 The radial wave function for the 1s electron in hydrogen is R 1s r 2 a 0 3 2 e r a 0 . From equation 6.24, Pr ()d r . r 2 R 2 . To find the most probable r, we maximize P(r); i.e. take its derivative and set it equal to zero. 0

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chapter6 - Problem 6.3 This is not a difficult problem but...

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