Chapter 42

# Chapter 42 - 42.1 Solve(a A 4p state corresponds to n = 4...

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42.1. Solve: (a) A 4 p state corresponds to n = 4 and l = 1. From Equation 42.3, the orbital angular momentum is 1(1 1) 2 . L =+ = == (b) In the case of a 5 f state, n = 5 and l = 3. So, () 33 1 12. L =

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42.2. Solve: (a) Excluding spin, a state is described by three quantum numbers: n , l , and m . 3p states correspond to n = 3 and l = 1. The quantum number m takes values from l to l . The quantum numbers of the various 3 p states are displayed in the table below. n 3 3 3 l 1 1 1 m 1 0 + 1 (b) A 3 d state is described by n = 3 and l = 2. Including the quantum number m , the quantum numbers of the various 3 d states are displayed in the table below. n 3 3 3 3 3 l 2 2 2 2 2 m 2 1 0 + 1 + 2
42.3. Solve: (a) The orbital angular momentum is () 1. Ll l =+ = Thus, 2 2 34 34 3.65 10 J s 11 2 1.05 10 J s L ll ⎛⎞ × += = = ⎜⎟ × ⎝⎠ = l = 3 This is an f electron. (b) The l quantum number is required to be less than n . Thus, the minimum possible value of n for an electron in the f state is n min = 4. The corresponding minimum possible energy is min 4 2 13.60 eV 0.85 eV 4 EE == =

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42.4. Solve: From Equation 42.2, the hydrogen atom’s energy is 2 13.60 eV 0.544 eV 5 n En n = −= = The largest l value for an n = 5 state is 4. Thus, the magnitude of the maximum possible angular momentum L is () 14 4 12 0 . Ll l =+ = + = == =
42.5. Solve: A 6 f state for a hydrogen atom corresponds to n = 6 and l = 3. Using Equation 42.2, 6 2 13.6 eV 0.378 eV 6 E == The magnitude of the angular momentum is () 13 3 11 2 . Ll l =+ = + = =

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42.6. Model: No two electrons can have exactly the same set of quantum numbers ( n , l , m , m s ). Solve: For n = 1, there are a total of two states with the quantum numbers given by ( ) 1 2 1, 0, 0, ± . For n = 2, there are a total of eight states: ( ) ( ) ( ) ( ) 11 1 1 22 2 2 2, 0, 0, 2, 1, 1, 2, 1, 0, 2, 1, 1, ±− ± ± ± For n = 3, there are a total of 18 states: () ( ) ( ) ( ) 1 111 2 2 3, 0, 0, 3, 1, 1, 3, 1, 0, 3, 1, 1, ± ± ± ( ) ( ) ( ) 1 1 222 2 2 3, 2, 2, 3, 2, 1, 3, 2, 0, 3, 2, 1, 3, 2, 2, ± ±± ± ±
42.7. Solve: (a) A lithium atom has three electrons, two are in the 1 s shell and one is in the 2 s shell. The electron in the 2 s shell has the following quantum numbers: n = 2, l = 0, m = 0, and m s . m s could be either 1 2 + or 1 2 . Thus, lithium atoms should behave like hydrogen atoms because lithium atoms could exist in the following two states: () 1 2 2, 0, 0, + and 1 2 2, 0, 0, . Thus there are two lines. (b) For a beryllium atom, we have two electrons in the 1 s shell and two electrons in the 2 s shell. The electrons in both the 1 s and 2 s states are filled. Because the two electron magnetic moments point in opposite directions, beryllium has no net magnetic moment and is not deflected in a Stern-Gerlach experiment. Thus there is only one line.

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42.8. Solve: Mg, Sr, and Ba are all in the second column of the periodic table. The electron configuration of Mg ( Z = 12) is 1 s 2 2 s 2 2 p 6 3 s 2 . The electron configuration of Sr ( Z = 38) is 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 10 4 p 6 5 s 2 The electron configuration of Ba ( Z = 56) is 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 10 4 p 6 5 s 2 4 d 10 5 p 6 6 s 2 Assess: It is necessary to recall that the 4 s subshell fills before the 3 d subshell, the 5 s before the 4 d , and the 6 s before the 4 f or 5 d .
42.9.

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## This note was uploaded on 01/30/2011 for the course PHYS 131 taught by Professor E.salik during the Winter '10 term at Cal Poly Pomona.

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Chapter 42 - 42.1 Solve(a A 4p state corresponds to n = 4...

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