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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 531
Solutions to problems
P15B.2 Combining Bragg’s law [15B.1b–648], λ = 2d sin θ, with the expression for the
the separation of planes for a cubic lattice [15A.1a–645], dhk l = a/(h2 + k2 +
l 2)1/2, gives sin θ = (λ/2a) (h2 + k2 + l 2)1/2.
�e di�raction patterns for cubic cells of di�erent types is illustrated in Fig. 15B.10
on page 652. Di�raction is possible for all (hkl) planes for a cubic P lattice, but
re�ections corresponding to (h+k+ l) odd are absent for a cubic I lattice. For a
cubic F lattice, re�ections are absent where two out of h, k and l are odd.�ese
systematic absences in a di�raction pattern allow the lattice type to be assigned.
�e observation of a (100) re�ection, for which (h + k + l) is odd, rules out
cubic I. �e observation of a (110) re�ection, for which h and k are odd, but
l is even, rules out cubic F.�erefore polonium must be cubic P, for which all
re�ections are present.
�e sequence of re�ections expected for cubic P is shown in Fig. 15B.10 on
page 652: the ��h, sixth and seventh re�ections are (210), (211) and (220),
respectively.�e sine of the re�ection angle is proportional to (h2+k2+ l 2)1/2,
which for these three re�ections is 51/2, 61/2, and 81/2: the separation between
the sixth and seventh is indeed larger than that between the ��h and sixth, as
described.
�e cell dimension is calculated from a = λ(h2 + k2 + l 2)1/2/(2 sin θ); each
re�ection allows a separate determination of a. For the (100) re�ection
a = λ(h2 + k2 + l 2)1/2
2 sin θ
= λ(12 + 02 + 02)1/2
2 sin θ
= λ(1)1/2
2 sin θ
= (154 pm) × (1)1/2
2 × 0.225
= 3.42... × 102 pm
Similar calculations for the other re�ections given a = 3.44... × 102 pm and
a = 3.43... × 102 pm.�e average is a = 344 pm .
P15B.4 Combining Bragg’s law [15B.1b–648], λ = 2d sin θ, with the expression for the
the separation of planes for a cubic lattice [15A.1a–645], dhk l = a/(h2 + k2 +
l 2)1/2, gives sin θ = (λ/2a) (h2 + k2 + l 2)1/2. For the same re�ection, a is
proportional to 1/ sin θ, therefore
aKCl = aNaCl
sin θNaCl
sin θKCl
= (564 pm) × (sin 6.00○)
sin 5.38○
= 628.7... pm = 629 pm
�e mass density ρ is ρ = m/V , where m is the mass of the unit cell. If there
are N formula units per cell then ρ = NM/NAV , where M is the molar mass
of a formula unit. Based on the cell parameter a, the ratio of densities of KCl
and NaCl is therefore
ρKCl
ρNaCl
=
MKCla3NaCl
MNaCla3KCl
= (74.55 gmol−1) × (564 pm)3
(58.44 gmol−1) × (628.7... pm)3
= 0.921
�e experimentally determined ratio ofmass densities is 1.99 g cm−3/2.17 g cm−3=
0.917, which is close to the value determined from the di�raction data.�e data
therefore support the X-ray analysis.