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chapter7

# chapter7 - Problem 7.3 For convenience set hbar:= 1 The...

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θ θ 360 2 π := θ 125.264 = _______________________________________________________________________________ Problem 7.9 Remember s p d f l= 0 1 2 3 so an f subshell corresponds to l=3. There are 7 possible m.l values for an electron in this subshell: m l 3 2 , 1 , 0 , 1 , 2 , 3 , := For each m.l, there are two possible values of m.s: m s 1 2 := m s 1 2 + := or That gives a total of 7*2=14 electrons in an f subshell. _______________________________________________________________________________ Problem 7.11 See near the end for a simple solution (one that you would want to use on an exam. To see what's going on, let's make a table of possible electron configurations. The table is too big to fit on this page, so I'll continue on the next. Problem 7.3 For convenience, set hbar 1 := The spin angular momentum has a magnitude S 3 2 hbar := The z-component of the spin angular momentum has a magnitude note this second definition of S.z is "turned off in Mathcad" S z hbar 2 := or S z hbar 2 := Now look at Figure 7.2 in Beiser. The angle between the z-axis and the spin angular momentum vector corresponding to m s =1/2 is θ acos S z S := This is in radians.

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chapter7 - Problem 7.3 For convenience set hbar:= 1 The...

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