CHEMISTRY 373 –TUTORIAL 8 PRACTICE FOR ... - Cobalt

CHEMISTRY 373 –TUTORIAL 8
PRACTICE FOR HYDROGEN ATOM AND ELECTRON SPIN
March 8, 2000
1. Draw contour plots of the 3d orbitals of hydrogen and place the signs on the orbitals. (These are
meant to be sketches not the precise contour plots.)
2. Discuss the differences between radial densities and the contour plots for the hydrogen 3s and
3p orbitals. When you draw contour diagrams for the 3s and 3p orbitals do the radial nodes
affect the signs that you place on the orbital lobes?
3. What is the average distance of the electron from the nucleus in a 2s and a 2p orbital for (a) H,
and (b) Li 2+ ?
(ANS. The normalized wave function for an electron in a 2s orbital for a H-like species is
1
f 2,0,0 = Z 2
Ý2 ? _Þe ?_/2 Y
8
0,0 ÝS,jÞ,
where _ = Zr/a 0 (See Table 13.1 in Atkins). The average distance of the electron in this state
from the nucleus is
a 0
3
K
Ör× 2s = X r 2 2^
0
drX
0^
sinSdSX
D djf2,0,0 rf 2,0,0 ,
0
= 1 8
= 1 8
Z
a 0
3
X
0
K
r 3 drÝ2 ? _Þ 2 e ?_ X
0^
sinSdSX
0
2^
djY0,0 ÝS,jÞ D Y 0,0 ÝS,jÞ,
3
Z a 0
a 0 Z
a 0
4
X
0
K
d__ 3 Ý2 ? _Þ 2 e ?_ ,
= 1 K
8
X d_Ý_
Z 5 ? 4_ 4 + 4_ 3 Þe ?_ ,
0
= 1 a 0
Ý5! ? 4 × 4! + 4 × 3!Þ,
8 Z
= 6 a 0
Z ,
where we have used the fact that the spherical harmonics are normalized and the integral
K
X d__ n e ?_ = n!. For a 2p state, select the 2p z = 2p 0 = f 2,1,0 wave function
0
f 2,1,0 = 1
3
Z 2
2 6 a 0
_e ?_/2 Y 1,0 ÝS,jÞ.
The expectation value or average value of the distance is
K
Ör× 2p = X r 2 2^
0
drX
0^
sinSdSX
D djf2,1,0 rf 2,1,0 ,
0
= 1 24
a 0
Z
K
X _ 5 e ?_ ,
0
= 5! 1 a 0
24 Z
,
= 5 a 0
Z
.
(a) For H, Z = 1, and Ör× 2s = 6a 0 , and Ör× 2p = 5a 0 . (b) For Li 2+ , Z = 3 and Ör× 2s = 2a 0 and
Ör× 2p = 5 a 3 0.)
4. What is the magnitude of the angular momentum for an electron in a 3p orbital in hydrogen?
How many radial and angular nodes does the 3p wave function have?
(ANS. For a 3p state, l = 1 and the wave function can be written f 3,1,0 . Hence,
å2 L f2,1,0 = 2¥ 2 f 3,1,0 . Therefore, the magnitude of the total orbital angular momentum is 2 ¥.
There are n ? 1 nodes in all, l angular nodes and n ? l ? 1 radial nodes. For the 3p state there are

2 nodes in total, 1 angular node and 1 radial node.)
5. Show that the real 2p x and 2p y orbitals are not eigenfunctions of the å L z operator.
(ANS.The real 2p x and 2p y functions can be expressed in terms of the complex functions as
follows:
Therefore, we find that
Similarly, we see that
f 2px =
1
2 Ýf 2,1,?1 ? f 2,1,1Þ, f 2py = i
1
2 Ýf 2,1,?1 + f 2,1,1Þ.
å
Lz f 2px =
1
2
å
Lz f 2,1,?1 ? å L z f 2,1,1 ,
= ?¥ 1 2 Ýf 2,1,?1 + f 2,1,1Þ,
= i¥f 2py .
å
Lz f 2py = i
1
2
å
Lz f 2,1,?1 + å L z f 2,1,1 ,
= i¥ 1 2 Ý?f 2,1,?1 + f 2,1,1Þ,
= ?i¥f 2px .
From these equations, we see that the real 2p orbitals, 2p x and 2p y , are not eigenfunctions of the
å
Lz operator.)
6. (a) If the spin component S z of an electron is measured, what possible values can result?
(ANS. + 1 2 ¥ and ? 1 2 ¥.)
(b) The set of common eigenfunctions of å S 2 and å S z are denoted by J = f 1
2 , 1 2
Show that
å S x J = 1 ¥K, å
2 S x K = 1 ¥J, 2
and K = f 1
2 ,? 1 .
2
å S y J = 1 2 i¥K,
å S y K = ? 1 2 i¥J.
(ANS: Write å S x = 1 2 Ýå S ? + å S + Þ and å S y = i 2 Ýå S ? ? å S + Þ. The rest follows from å S + J = 0,
å S + K = ¥J, å S ? J = ¥K, å S ? K = 0.)
7. (c) If the spin component S x of an electron is measured, what possible values can result?
(ANS. According to Postulate 3, what will be measured are the eigenvalues of the operator å S x .
Let e be an eigenfuction of å S x with eigenvalue V. Thus,
å S x e = Ve.
Since the eigenfunctions J and K of the spin operator å S z form a complete set (Postulate 4), we
can express e as a linear combination of J and K:
e = aJ + bK.
Using the results from part (b), we have
å S x e = 1 Ýå S
2 ? + å S + ÞÝaJ + bKÞ,
This implies that
= ¥ 2 bJ + ¥ 2 aK,
= VaJ + VbK.
Va = ¥ 2 b, Vb = ¥ 2 a.

Using the second to substitute for b in the first equation, we obtain
or
V 2 = ¥2
4 ,
V = ± ¥ 2 .
Therefore, we would measure ±¥/2 for S x .)
(d) According to Postulate 4, the spin functions, J and K, form a complete set. This means that
any one-electron spin function can be written as a linear combination of J and K. Construct
eigenfunctions of å S x having the eigenvalues obtained in (c).
(ANS. From Part (c), suppose that V = +¥/2. Then, a = b and we write
e+ ÝxÞ = aÝJ + KÞ.
We fix a by normalization, i.e., a = 1/ 2 . If V = ?¥/2, then b = ?a and we have
e ? ÝxÞ = aÝJ ? KÞ.
Again, fix a by normalization. The final result is
Confirm that
e+ ÝxÞ =
1
2 ÝJ + KÞ, e ? ÝxÞ =
1
2
S x e ± ÝxÞ = ± ¥ 2 e ± ÝxÞ .
ÝJ ? KÞ.
That completes the analysis.)
(e) Suppose a measurement of S z is carried out and a value of + 1 ¥is obtained. If a measurement
2
of S x is then carried out, give the probabilities for each outcome.
(ANS. If S z is measured and a value of + 1 ¥ is measured, then the particle will be in the state J.
2
If the property S x is measured then either the eigenvalue + 1 ¥ or ? 1 ¥ will be obtained with
2 2
probability X db J D e+ ÝxÞ 2
=
1
or X db 2 JD ÝxÞ
2
e ? =
1
, respectively.)
2
8. Show that J and K are each eigenfunctions of å S x
2
but not
å S x .
(ANS. From Part (b), we have already seen that J and K are not eigenfunctions of å S x .
Furthermore, we have
å S x
2
J =
1
2 ¥å S x K = 1 4 ¥2 J,
å S x
2
K =
1
2 ¥S xJ = 1 4 ¥2 K,
where we have used the identities in Part (b).)