Physics Solution Manual for 1100 and 2101

# Also when a nucleus decays the nucleon number is

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Unformatted text preview: e angle θ in Equation (2) is as small as possible. This means making the quantity inside the square root as large as possible. Given that l is an integer, we see that the quantity in the square root of 123 Equation (2) will be of the form , , , K . Therefore, to obtain the smallest value for θ, 234 we must choose the greatest possible value for the orbital quantum number l , which is l = n − 1 = 5 − 1 = 4 . Substituting this value into Equation (2), we obtain l 4 −1 −1 4 o = cos = cos 4 +1 5 = 26.6 l +1 θ = cos−1 31. REASONING a. The maximum number of electrons that the n = 1, l = 0 subshell can contain is 2 (see Figure 30.15). If the atom is in its ground state, the third electron must go into the n = 2, l = 0 subshell. For this subshell, the magnetic quantum number can only be zero ( ml = 0), and the electron can have a spin quantum number of either ms = + 1 or − 1 . 2 2 b. If the atom is in its first excited state, the third electron must go into the n = 2, l = 1 subshell. For l = 1 , there are three possibilities for ml : ml = +1, 0, and –1. For each value of ml , the electron can have a spin quantum number of either ms = + 1 or − 1 . 2 2 SOLUTION a. According to the discussion in the REASONING section, the third electron can have one of two possibilities for the four quantum numbers: n = 2, l = 0, ml = 0, ms = + 1 2 and n = 2, l = 0, ml = 0, ms = − 1 . 2 b. When the third electron is in the first excited state, there are six possibilities for the four quantum numbers: n = 2, l = 1, ml = + 1, ms = + 1 , n = 2, l = 1, ml = + 1, ms = − 1 , 2 2 n = 2, l = 1, ml = 0, ms = + 1 , 2 n = 2, l = 1, ml = 0, ms = − 1 , 2 n = 2, l = 1, ml = − 1, ms = + 1 , n = 2, l = 1, ml = − 1, ms = − 1 . 2 2 Chapter 30 Problems 1565 ______________________________________________________________________________ 32. REASONING For use here, we recall that the l = 0 subshell is denoted by s, the l = 1 subshell is denoted by p, and the l = 2 subshell is denoted by d. In Figure 30.15 we see that the 4s subshell fills before the 3d subshell and the 5s subshell fills before the 4d subshell. We also see in Figure 30.15 that s subshells can contain up to 2 electrons, p subshells can contain up to 6 electrons, and d subshells can contain up to 10 electrons. SOLUTION Following the style in Table 30.3, we find that the ground state electronic configuration for the 48 electrons in cadmium is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 33. REASONING According to the discussion in Section 30.6, the lth subshell can hold at most 2 ( 2l + 1) electrons. For n = 5, we can have the following values for l : l = 0, 1, 2, 3, and 4 . SOLUTION The maximum number N l of electrons that can be put into each subshell is: l = 0 subshell: l = 1 subshell: l = 2 subshell: l = 3 subshell: l = 4 subshell: N 0 = 2 [ 2 ( 0 ) + 1] = 2 N1 = 2 [ 2 (1) + 1] = 6 N 2 = 2 [ 2 ( 2 ) + 1] = 10 N 3 = 2 [ 2 ( 3) + 1] = 14 N 4 = 2 [ 2 ( 4 ) + 1] = 18 The maximum number N of electrons that can be put into the n = 5 shell is: N = 2 + 6 + 10 + 14 + 18 = 50 _____________________________________________________________________________________________ 34. REASONING The orbital quantum number l can have any integer value from 0 up to n – 1. If, for example, n = 4, l can have the values 0, 1, 2, and 3. For a given value of the orbital quantum number l , the magnetic quantum number ml can have any integer value, including 0, from – l to + l . For instance, if l = 2, ml can have the values –2, –1, 0 ,+1, and +2. SOLUTION Of the five subshell configurations, three are not allowed. The ones that are not allowed, and the reasons they are not allowed, are: 1566 THE NATURE OF THE ATOM (b) The principal quantum number is n = 2 and the orbital quantum number is l = 2 (the d subshell). Since l must be less than n, this subshell configuration is not permitted. (c) This subshell has 4 electrons. However, according to the Pauli exclusion principle, only two electrons can be in the s subshell ( l = 0). Therefore, this subshell configuration is not allowed. (d) This subshell has 8 electrons. However, according to the Pauli exclusion principle, only six electrons can be in the p subshell ( l = 1). Therefore, this subshell configuration is not allowed. ______________________________________________________________________________ 35. REASONING AND SOLUTION The table below lists the possible sets of the four quantum numbers that correspond to the electrons in a completely filled 4f subshell: n l ml ms 4 4 4 4 4 4 4 4 4 4 4 4 4 4 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 2 2 1 1 0 0 –1 –1 –2 –2 –3 –3 1/2 –1/2 1/2 –1/2 1/2 –1/2 1/2 –1/2 1/2 –1/2 1/2 –1/2 1/2 –1/2 ______________________________________________________________________________ 36. REASONING The first “noble gas” is the element with an atomic number Z that is equal to the number of electrons that can fit into the first energy level (n = 1), and the atomic number Z of the second “noble gas” is equal to the...
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## This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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