Basic HTML Version
101
n
n
2
i
j
i 1
j 1
r rr cos r r
l
l
=
=
⎛ ⎞
⎛ ⎞
= φ = ⋅ = ∑ ∑ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
Here
φ
is the angle between the vectors.
For a long chain we have n situations where i = j, thus
l l l
i
j
⋅ =
2
(angle is zero
and cos
φ
= 1). The term nl
2
may therefore be separated out:
r
2
=
nl
2
+
l
i
i
=
1
n
∑
⎛
⎝⎜
⎞
⎠⎟
l
j
j
=
1
n
∑
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝
⎜
⎞
⎠
⎟
i
≠
j
=
nl
2
+
l
i
l
j
j
=
1
n
∑
i
=
1
n
∑
⎛
⎝⎜
⎞
⎠⎟
i
≠
j
Then, a very important step follows. Flexible chains do not have a fixed
shape, as we have already seen above. The average for a large number of
molecules (of the same molecular weight (or same n)), or taking the time-
average of a single chain, leads to the same result.
Taking averages over all chains (
n
→ ∞
) we obtain:
r
2
=
nl
2
+
l
⋅
l
j
=
1
n
∑
i
=
1
n
∑
⎛
⎝⎜
⎞
⎠⎟
i
≠
j
=
nl
2
+
l
2
cos
θ
j
=
1
n
∑
i
=
1
n
∑
⎛
⎝⎜
⎞
⎠⎟
i
≠
j
Since the vectors are randomly oriented,
cos
φ
may take any value between -
1 and +1, with an average value of zero. Hence, the terms containing <cos
φ
>
simply cancel. Hence:
cos
θ
i
≠
j
=
0
⇒
r
2
=
nl
2
(n
→ ∞
)
This means that doubling the number of monomers (vectors), corresponding
to a doubling in molecular weight, increases the average squared end-to-end
distance by a factor of 2. The linear dimensions therefore only increase by a
factor of
2
= 1.41.
The end-to-end distance cannot be determined directly from any experiment.