Chapter41 - 41.1. Solve: (a) A 4p state corresponds to n =...

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41.1. Solve: (a) A 4 p state corresponds to n = 4 and l = 1. From Equation 41.3, the orbital angular momentum is L =+ = 11 1 2 () hh . (b) In the case of a 5 f state, n = 5 and l = 3. So, L = 33 1 12 .
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41.2. 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 33 3 l 11 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 l 22 2 m 2 1 0 + 1 + 2
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41.3. Solve: (a) The orbital angular momentum is Ll l =+ () 1 h . Thus, ll L + = = × × = 1 365 10 105 10 12 2 34 34 2 h . . J s J s 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 EE min == = 4 13.60 eV 4 0.85 eV 2
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41.4. Solve: From Equation 41.2, the hydrogen atom’s energy is E n n n =− ⇒ = 13 60 5 2 . e V 0.544 eV The largest l value for an n = 5 state is 4. Thus, the magnitude of the maximum possible angular momentum L is Ll l =+ () = 14 4 12 0 hh h .
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41.5. Solve: A 6 f state for a hydrogen atom corresponds to n = 6 and l = 3. Using Equation 41.2, E 6 2 6 = =− 13.6 eV 0.378 eV The magnitude of the angular momentum is Ll l =+ () = 13 3 11 2 hh h .
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41.6. Visualize: l is the orbital quantum number. It is an integer: l = 0, 1, 2, … L is the actual numerical value of the orbital angular momentum, with units of J s. It is related to l by Ll l =+ () 1 h .
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41.7. Solve: (a) A lithium atom has 3 electrons, 2 are in the 1 s shell and 1 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: 2, 0, 0, 1 2 + () and 1 2 . Thus there are 2 lines. (b) For a beryllium atom, we have 2 electrons in the 1 s shell and 2 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 1 line.
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41.8. 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 2 states with the quantum numbers given by 1, 0, 0, 1 2 ± () . For n = 2, there are a total of 8 states: 2, 0, 0, 2, 1, 1, 2, 1, 0, 2, 1, 1, 1 2 1 2 1 2 1 2 ± −± ± ± For n = 3, there are a total of 18 states: 3, 0, 0, 3, 1, 3, 1, 0, 3, 1, 1, 1 2 1 2 1 2 1 2 ± ± ± 3, 2, 2, 3, 2, 1, 3, 2, 0, 3, 2, 2, 1 2 1 2 1 2 1 2 1 2 ± ± ± 1,
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41.9. Visualize: The quantum number s = 1 2 is the spin of an electron. It plays a role analogous to l , although it is not an integer. S is the actual numerical value of the spin angular momentum. It is related to s by Ss s =+ () 1 h .
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41.10.
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This note was uploaded on 02/27/2010 for the course PHYS 161,260,27 taught by Professor Staff during the Spring '08 term at Maryland.

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Chapter41 - 41.1. Solve: (a) A 4p state corresponds to n =...

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