This can be solved by using the product rule:(12.22)lettingu=x^2 du= 2 xdx d 9 =e−^2 xdx 9 =−e−^2 x/2 (12.23)and another substitutionu=x du=dx d 9 =e−^2 xdx 9 =−e−^2 x/2 (12.24)(12.25)
(12.26)
So there is only a 32% probability of finding the electron with the radius a 0.
Perhaps more useful is the average (or expectation) value of the radius.
We can calculate it as:(12.27)
Let x=r/a 0 and then use the same approach as above:(12.28)
Since the orbital extends for all values, there is a convention that
orbitals should be represented by the 90% boundary; that is, the
radius at which there is a 90% probability of finding the elec-
tron. You can substitute the value 3a 0 into the integral above and
see that, at this radius, you have a probability also of 3a 0 (actually
you are slightly over). It is this representation that is usually shown
for orbitals (Figure 12.5).= 4 ae 0 −^2 xx
xxx a
−− − −⎛
⎝
⎜⎜
⎞
⎠
⎟⎟ =
32 ∞00
23
4
3
4
3
8
3
2
4
4
03
02
032
00
arera r a xex x∞
− −∞
∫∫=/ ddψψτ
π*drr π d
a= errra/⎛
⎝
⎜⎜
⎞
⎠
⎟⎟ − =
∞ ∞
∫∫1
4
02
0 00 2 4 /
03
0(^20)
a
rera r
∞
−
∫ d
4222 21032
01
22
0
xe−−xxx e x x^1
∫ d(=− + +).3=^3exx
x
eex
x
∫∫−−−−^2 xxx=− + =− e
01
22
01
2
21
22
ddxxx+−e⎛
⎝
⎜⎜
⎞
⎠
− ⎟⎟
1
2
1
2
244
22
(^22)
0
1
22
22
0
1
xe x x
ee
x xx
xx
∫∫
−
−−
=− −
⎡
⎣
⎢
dd⎢⎤⎤
⎦
⎥
⎥=− +
−−(^24) ∫
22
0
1
xe xxe dx
uudd99 9=−∫∫ u
250 PART 2 QUANTUM MECHANICS AND SPECTROSCOPY
Figure 12.5A
boundary-surface
representation of
an s orbital.