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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 513
(b) �e mean separation is
⟨r⟩ = ∫ r f (r)dr = ∫
∞
0
r × 4π ( a
π1/2
)
3
r2e−a
2 r2
= 4π ( a
π1/2
)
3
∫
∞
0
r3e−a
2 r2
´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶
Integral G.4 with k = a2
= 4π ( a
π1/2
)
3
× 1
2(a2)2
= 2
π1/2
( 1
a
)
= 2
π1/2
(2Nl
2
3
) = (8N
3π
) l
(c) �e most probable separation is the value of r for which f (r) is a maxi-
mum. Di�erentiating f (r) using the product rule gives
d f (r)
dr
= 4π ( a
π1/2
)
3
× [(2r)(e−a
2 r2) + (r2)(−2a2re−a
2 r2)]
= π ( a
π1/2
)
3
× 2re−a
2 r2 [1 − a2r2]
At the maximum d f (r)/dr = 0, which implies that
r(1 − a2r2) = 0 hence rmp =
1
a
= (2Nl
2
3
) = (2N
3
)
1/2
l
�e solution r = 0 is rejected because this does not correspond to a max-
imum, as implied by the fact that f (0) = 0.
For the case that N = 4000 and l = 154 pm these expressions give
Rrms = N1/2 l = 40001/2 × (154 pm) = 9.73... × 103 pm = 9.74 nm
⟨r⟩ = (8N
3π
)
1/2
l = (8 × 4000
3π
)
1/2
× (154 pm) = 8.97... × 103 pm = 8.97 nm
rmp = (2N
3
)
1/2
l = (2 × 4000
3
)
1/2
× (154 pm) = 7.95... × 103 pm = 7.95 nm
P14D.4 �ere is some lack of clarity in the text over the de�nition of the radius of
gyration, Rg. For a polymer consisting of N identical monomer units, Rg is
de�ned as
R2g = (1/N)
N
∑
i=1
r2i (14.1)
where r i is the distance of monomer unit i from the centre of mass. In other
words, the radius of gyration is the root-mean-square of the distance of the
monomer units from the centre of mass.
A related quantity is the moment of inertia I about an axis, which is de�ned in
the following way
I =
N
∑
i=1
md2i (14.2)