Angular momentum, rigid rotor and the hydrogen atom - Cobalt

CHEMISTRY 373 –TUTORIAL-7
PRACTICE FOR ANGULAR MOMENTUM, RIGID ROTOR, AND HYDROGEN ATOM
March 1, 2000
A. Questions on angular momentum and the rigid rotor.
1. In general, it can be shown that for two operators A and B, the uncertainty product is
ÝAAÞÝABÞ ³ 1 2 X d3 r H D ßA,BàH .
What is the uncertainty product for AL x AL y What is the uncertainty in L z , i.e. AL z
(ANS. I just noticed that this is the same as the last question on the previous assignment.)
2. The 12 C 16 O molecule has an equilibrium bond distance of 112.8 pm. Calculate the following:
(a) the reduced mass.
(b) the moment of inertia.
(c) the frequency, in wave numbers, of the photon emitted when the molecule makes the
rotational transition from l = 1 to l = 0.
Isotopic masses: 12 C, 12.000000 amu; 16 O, 15.994914 amu.
(ANS: The energy spectrum of a rigid rotor can be calculated using the energy expression
or
E l = ¥2 lÝl + 1Þ,
2I
E l = hcBlÝl + 1Þ,
where B = h/8^2cI is called the rotational constant in units of cm 1 . The selection rules are
Al = ±1, where Al = +1 corresponds to absorption and Al = 1 corresponds to emission. For
absortion starting at level l, we have
or
For emission, we find that
l
hcX +1 = E l+1 E l = 2hcBÝl + 1Þ,
l
X +1 = 2BÝl + 1Þ.
l
X 1
= 2Bl.
Note that regardless of whether or not we are dealing with absorption or emission, the
separation between adjacent rotational spectral lines is a constant, 2B.
(a) W = 12.000000×15.994914 × 1
= 1.1385 × 10 23 g= 1.1385 × 10 26 kg
12.000000+15.994914 6.022×10 23
(b) I = Wr 2 = 1.1385 × 10 26 × Ý1.128 × 10 10 Þ 2 = 1.4486 × 10 46 Js 2
l 1
(c) This is an emission line. Therefore, X 1 = 2lB, which for l = 1, becomes X 1 = 2B, where
the rotational constant
B = h
8^2cI = 6.626 × 10 34
= 1.9323
8^2Ý2.998 × 10 10 ÞÝ1.4486 × 10 46 cm1
Þ
1
Therefore, X 1 = 2B = 2 × 1.9323 = 3.8646 cm 1 .)
3. The pure rotational spectrum of 1 H 131 I consists of lines separated by 13.10 cm 1 . Calculate the
bond length of the molecule.
Isotopic masses: 1 H, 1.007825 amu; 131 I, 130.90612 amu.
(ANS. From the answer to the previous question, 2B = 13.10 cm 1 . Thus, B = 6.55 cm 1 . From
the value for B, we can evaluate I, the moment of inertia:

I = h
8^2cB = 6.626 × 10 34
8^2Ý2.998 × 10 10 ÞÝ6.55Þ = 4.2735 × 1047 Js 2
From the value of the reduced mass W = 1.007825×130.90612 × 1
= 1.6608 × 10 24 g
1.007825+130.90612 6.022×10 23
= 1.6608 × 10 27 I 4.2735×10
kg, we obtain r = W = 47
= 1.6041 × 10 10 m = 160.4 pm.)
1.6608×10 27
B. The hydrogen atom.
1. The emission spectrum for the hydrogen atom consists of groups of lines, e.g. the Lyman series,
the Balmer series, etc. Predict the frequencies, in cm 1 , for the first two lines in each of the
series and indicate the part of the electromagnetic spectrum they occur in.
(ANS. For the Lyman spectrum, the quantum number, n f , for the final state is 1. For the Balmer
series, it is n f = 2. For the hydrogen atom, the energy of the n’th level is given by
E n = hcR H
,
n 2
where the Rydberg constant, R H , is in cm 1 and is given by
We
R H =
4
= 1.09737 × 10
32^2P 5 cm 1 ,
2
0 ¥ 2
where W is the reduced mas of the electron and proton. Since the proton is so massive compared
to the electron, we usually take W r m e , the rest mass of the electron. The frequency (in cm 1 )
of emitted or absorbed light is
X = R H
1
n f
2 1 n i
2
.
For the first two lines of the Lyman spectrum, we have n i = 2 to n f = 1 and n i = 3 to n f = 1.
For the first two lines of the Balmer spectrum, n i = 3 to n f = 2 and n i = 4 to n f = 2. Therefore,
we have
Transition Frequency, cm 1 Part of electromagnetic spectrum
n i = 2 to n f = 1 82303 vacuum ultraviolet
n i = 3 to n f = 1 97544 vacuum ultraviolet
n i = 3 to n f = 2 15241 visible
n i = 4 to n f = 2 20576 visible
That completes the answer.)
2. What are the degeneracies of the following orbitals for hydrogen or hydrogen-like ions
(a) n = 1, (b) n = 3, (c) n = 4
(ANS. Neglecting electron spin, the orbital degeneracy of the n’th level in the hydrogen atom is
given by n 2 . Therefore, in (a) the first level is nondegenerate; in (b) the degeneracy is 9-fold,
while in (c), it is 16-fold.)
3. Show that for a 1s orbital of a hydrogen atom or hydrogen-like ion the most probable distance
from the nucleus to the electron is a 0 /Z. Find the numerical values for B 4+ and C 5+ . What is the
average distance of the electron from the nucleus Are average distances and the most probable
distances the same in this state What do you think is the case for excited states, in general
(ANS. The normalized 1s orbital for a hydrogen atom or hydrogen-like ion has the form
f 1s = f 1,0,0 Ýr,S,jÞ = R 1,0 ÝrÞY 0,0 ÝS,jÞ = Z 2 1^ expÝZr/a 0Þ.
where Z is the atomic number and a 0 is the Bohr radius (written as a in Griffiths since he uses
a 0
3

a 0 for other purposes). The radial probability density (the probability of finding the electron
between r and r + dr) for this state is
r 2 D
f 1,0,0 f 1,0,0 = 1^
3
Z
a 0 r 2 expÝ2Zr/a 0Þ.
To find the extremal points, take the derivative with respect to r and set the derivative to zero.
Thus,
/
/r r2 D
f 1,0,0 f 1,0,0 = / /r
= 1^
1
^
Z
a 0
3
Z
a 0
3
r 2 expÝ2Zr/a 0Þ ,
2r 2Z a 0
expÝ2Zr/a 0Þ,
= 0.
Therefore, r = a 0 /Z. The Bohr radius, a 0 = 0.529 × 10 10 m. For B 4+ , Z = 5. Therefore, the
most probable distance is r = 1.058 × 10 11 m = 10.58 pm. For C 5+ , Z = 6. The most probable
radius is 8.8167 × 10 12 m or 8.8167 pm. Notice that as the nuclear charge increases, the most
probable radius decreases which is consistent with increased electrostatic attraction between the
electron and the more highly charged nucleus of the heavier elements. The average distance is
^ 2^
dS sinS X dj
D f1,0,0 Ýr,S,jÞrf 1,0,0 Ýr,S,jÞ,
0
Ör× = X
0
K
dr r
2 X 0
3 K
= 4 Z
a0
X dr r 3 2^
expÝ2Zr/a
0
0ÞX dS sinS
0^
X D djY0,0 ÝS,jÞY 0,0 ÝS,jÞ,
0
3
= 4 Z a 0
a0 2Z
= 1 4
a 0
Z
a 0
3!,
4
X
0
K
ds s 3 expÝsÞ,
= 3 ,
2 Z
which is not the same expression as for the most probable distance for this state. For excited
states, they will not be the same either.)
4. Prove that if f 1 and f 2 are two eigenfunctions of a Hamiltonian, H, with the same eigenvalue,
E, then any linear combination of the two wave functions also has the same energy eigenvalue,
E.
(ANS. We have
Hf 1 = Ef 1 , Hf 2 = Ef 2 .
Let Vf 1 + Wf 2 be some linear combination of f 1 and f 2 , where V and W are possibly complex
constants. Then
HÝVf 1 + Wf 2 Þ = VHf 1 + WHf 2 ,
= VEf 1 + WEf 2 ,
= EÝVf 1 + Wf 2 Þ,
and the Hamiltonian, H, has the same eigenvalue, E.)
5. The following expressions give the 2p orbitals of hydrogen and hydrogen-like ions (b and c are
normalization constants):

Show that
f 2p+1
f 2p0
f 2p1
f 2p+1
= brexp Zr
2a 0
sinS e ij ,
= cr exp Zr
2a 0
cosS,
= brexp Zr
2a 0
sinS e ij .
= brexp Zr
2a 0
x + iy
r ,
f 2p0 = cr exp
2a Zr z
0
r ,
f 2p1 = brexp
2a Zr x iy
0
r .
Find the linear combinations of the wave functions for the 2p +1 and 2p 1 orbitals that give the
real atomic orbitals
f 2px = cr exp
2a Zr x
0
r ,
Note that
f 2py
= cr exp Zr
2a 0
y
r .
f 2pz = f 2p0 = cr exp Zr
2a 0
z r .
The subscript on the real orbital designation is actually related to the angular function!
(ANS. The equations for polar coordinates are
x = rsinS cosj, y = rsinS sinj, z = rcosS.
We can write the complex wave functions as
f 2p+1 = brexp
2a Zr sinS e ij ,
0
f 2p0
f 2p1
= brexp
2a Zr
0
= brexp
2a Zr
0
= cr exp
2a Zr
0
ÝsinS cosj + sinS sinjÞ,
cosS,
= cr exp Zr
2a 0
z r ,
x + iy
r ,
= brexp Zr
2a 0
sinS e ij ,
= brexp
2a Zr
0
= brexp
2a Zr
0
ÝsinS cosj i sinS sinjÞ,
x iy
r .
To find the real functions, we subtract f 2p+1 from f 2p1 to get
Therefore, f 2px is given by
f 2p1 f 2p+1
= 2brexp Zr
2a 0
x r .

1
f 2px =
2 Ýf 2p 1
f 2p+1 Þ = 2 brexp
2a Zr x
0
r ,
where c = 2 b. Check that f 2px is normalized. Now, add the two functions
f 2p1 + f 2p+1 = 2ibrexp
2a Zr y
0
r .
Therefore, we set
1
f 2py = i
2 Ýf 2p 1
+ f 2p+1 Þ = 2 brexp
2a Zr y
0
r .
The relationship for f 2pz is obvious.)