TEST BANK FOR Materials Science and Engineering an intro 6th Edition By Bill Callister

2.1 (a) When two or more atoms of an element have different atomic masses, each is termed an isotope.
(b) The atomic weights of the elements ordinarily are not integers because: (1) the atomic masses of the atoms generally are not integers (except for 12C), and (2) the atomic weight is taken as the weighted average of the atomic masses of an atom's naturally occurring isotopes.
2.2 Atomic mass is the mass of an individual atom, whereas atomic weight is the average (weighted) of the atomic masses of an atom's naturally occurring isotopes.
2.3 (a) In order to determine the number of grams in one amu of material, appropriate manipulation of the amu/atom, g/mol, and atom/mol relationships is all that is necessary, as
#g/amu = 1 mol6.023 x 1023 atoms⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 g/mol1 amu/atom⎛ ⎝ ⎜ ⎞ ⎠ ⎟
= 1.66 x 10-24 g/amu
(b) Since there are 453.6 g/lbm,
1 lb-mol = 453.6 g/lbm()6.023 x 1023 atoms/g-mol()
= 2.73 x 1026 atoms/lb-mol
2.4 (a) Two important quantum-mechanical concepts associated with the Bohr model of the atom are that electrons are particles moving in discrete orbitals, and electron energy is quantized into shells.
(b) Two important refinements resulting from the wave-mechanical atomic model are that electron position is described in terms of a probability distribution, and electron energy is quantized into both shells and subshells--each electron is characterized by four quantum numbers.
2.5 The n quantum number designates the electron shell.
2
The l quantum number designates the electron subshell.
The ml quantum number designates the number of electron states in each electron subshell.
The ms quantum number designates the spin moment on each electron.
2.6 For the L state, n = 2, and eight electron states are possible. Possible l values are 0 and 1, while possible ml values are 0 and ±1. Therefore, for the s states, the quantum numbers are 200(12) and 200(−12). For the p states, the quantum numbers are 210(12), 210(−12), 211(12), 211(−12), 21(-1)(12), and 21(-1)(−12).
For the M state, n = 3, and 18 states are possible. Possible l values are 0, 1, and 2; possible ml values are 0, ±1, and ±2; and possible ms values are ±12. Therefore, for the s states, the quantum numbers are 300(12), 300(−12), for the p states they are 310(12), 310(−12), 311(12), 311(−12), 31(-1)(12), and 31(-1)(−12); for the d states they are 320(12), 320(−12), 321(12), 321(−12), 32(-1)(12), 32(-1)(−12), 322(12), 322(−12), 32(-2)(12), and 32(-2)(−12).
2.7 The electron configurations of the ions are determined using Table 2.2.
Fe2+ - 1s22s22p63s23p63d6
Fe3+ - 1s22s22p63s23p63d5
Cu+ - 1s22s22p63s23p63d10
Ba2+ - 1s22s22p63s23p63d104s24p64d105s25p6
Br- - 1s22s22p63s23p63d104s24p6
S2- - 1s22s22p63s23p6
2.8 The Na+ ion is just a sodium atom that has lost one electron; therefore, it has an electron configuration the same as neon (Figure 2.6).
The Cl- ion is a chlorine atom that has acquired one extra electron; therefore, it has an electron configuration the same as argon.
2.9 Each of the elements in Group IIA has two s electrons.
2.10 (a) The 1s22s22p63s23p63d74s2 electron configuration is that of a transition metal because of an incomplete d subshell.
3
(b) The 1s22s22p63s23p6 electron configuration is that of an inert gas because of filled 3s and 3p subshells.
(c) The 1s22s22p5 electron configuration is that of a halogen because it is one electron deficient from having a filled L shell.
(d) The 1s22s22p63s2 electron configuration is that of an alkaline earth metal because of two s electrons.
(e) The 1s22s22p63s23p63d24s2 electron configuration is that of a transition metal because of an incomplete d subshell.
(f) The 1s22s22p63s23p64s1 electron configuration is that of an alkali metal because of a single s electron.
2.11 (a) The 4f subshell is being filled for the rare earth series of elements.
(b) The 5f subshell is being filled for the actinide series of elements.
2.12 The attractive force between two ions FA is just the derivative with respect to the interatomic separation of the attractive energy expression, Equation (2.8), which is just
FA = dEAdr = d−Ar⎛ ⎝ ⎜ ⎞ ⎠ ⎟ dr = Ar2
The constant A in this expression is defined in footnote 3. Since the valences of the Ca2+ and O2- ions (Z1 and Z2) are both 2, then
FA = Z1e()Z2e()4πεor2
= (2)(2)1.6 x 10−19 C()2(4)(π)8.85 x 10−12 F/m)()1.25 x 10−9 m()2
= 5.89 x 10-10 N
2.13 (a) Differentiation of Equation (2.11) yields
dENdr = Ar(1 + 1) − nBr(n + 1) = 0
4
(b) Now, solving for r (= ro)
Aro2 = nBro(n + 1)
or
ro = AnB⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1/(1 -n)
(c) Substitution for ro into Equation (2.11) and solving for E (= Eo)
Eo = −Aro + Bron
= −AAnB⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1/(1 - n) + BAnB⎛ ⎝ ⎜ ⎞ ⎠ ⎟ n/(1 - n)
2.14 (a) Curves of EA, ER, and EN are shown on the plot below.
5
(b) From this plot
ro = 0.24 nm
Eo = -5.3 eV
(c) From Equation (2.11) for EN
A = 1.436
B = 7.32 x 10-6
n = 8
Thus,
ro