.;
"-v~
•.• PC1f
U",",J..C
IU
cue ~(a[e
n = 2 in a square-well potential.
~)
An electron is confined to a a square well oflength L. What would be
the length of the box such that the zero-point energy of the electron is equal
to its rest mass energy, m.c2? Express your answer in terms of the parameter
A.c = hlm.c, the 'Compton wavelength' of the electron.
.
8.14(b) t .onfirrn that the wavefuncuon for the first excited state of a
one-dimensional linear harmonic oscillator given in Table 8,1 is a solution
of the Schrodinger equation for the oscillator and that its energy is t1iro.
8.1S(a) Locate the nodes of the harmonic oscillator wave function with
v=4.
8.S(b) Repeat Exercise 8.5a for a general particle of mass m in a cubic box.
8.1S(b) Locate the nodes of the harmonic
v=5.
8.6(a) What are the most likely locations of a particle in a box oflength L in
the state n = 3?
8.16(a) What are the most probable displacements
withv=l?
of a harmonic oscillator
8.6(b) What are the most likely locations of a particle in a box oflength L in
the state n = 5?
8.16(b) What are the most probable displacements
with v=3?
of a harmonic oscillator
8.7(a) Calculate the percentage change in a given energy level of a particle in a
one-dimensional box when the length of the box is increased by 10 per cent.
8.17(a) Assuming that the vibrations
those of a harmonic oscillator with a
the zero-point energy of vibration of
of a homonuclear diatomic molecule
mesCl) = 34.9688711u'
8.7(b) Calculate the percentage change in a given energy level of a particle in a
cubic box when the length of the edge of the cube is decreased by 10 per cent
in each direction.
8.8(a) What is the value of n of a particle in a one-dimensional box such that
the separation between neighbouring levels is equal to the energy of thermal
motion (ikT).
8.8(b) A nitrogen molecule is confined in a cubic box of volume 1.00 m'.
Assuming that the molecule has an energy equal to tkTat T= 300 K, what is
the value of n = (n; + n: + n:)I12 for this molecule? What is the energy
separation between the levels n and n + I? What is its de Broglie wavelength?
8.9{a) Calculate the zero-point energy of a harmonic oscillator consisting
of a particle of mass 2.33 x 10-26 kg and force constant 155 N m-I.
8.9(b) Calculate the zero-point energy of a harmonic oscillator consisting
of a particle of mass 5.16 X 10-26 kg and force constant 285 N m-I.
8.10(a) For a certain harmonic oscillator of effective mass 1.33 X 10-25 kg, the
difference in adjacent energy levels is 4.82 z]. Calculate the force constant of
the oscillator.
8.10(b) For a certain harmonic oscillator of effective mass 2.88 x 10-25 kg,
the difference in adjacent energy levels is 3.17 z], Calculate the force constant
of the oscillator.
8.11 (a) Calculate the wavelength of a photon needed to excite a transition
between neighbouring energy levels of a harmonic oscillator of effective
mass equal to that of a proton (1.0078mu) and force constant 855 N m-I.
8.11(b) Calculate the wavelength of a photon needed to excite a transition
between neighbouring energy levels of a harmonic oscillator of effective mass
equal to that ofan oxygen atom (15.9949mu) and force constant 544 N m-I.
8.12(a) The vibrational frequency of H, is 131.9 THz. What is the vibrational
frequency of O2 (0 = 2H)?
8.12(b) The vibrational frequency of H, is 131.9 THz. What is the vibrational
frequency ofT2 (T = 'H)?
8.13(a) Calculate the minimum excitation energies of (a) a pendulum of
length 1.0 m on the surface of the Earth, (b) the balance-wheel of a clockwork
watch (v= 5 Hz).
8.13(b) Calculate the minimum
excitation energies of (a) the 33 kHz quartz
crystal of a watch, (b) the bond between two
kr= 1177 N m-I.
°
atoms in 02' for which
oscillator wavefunction
of a ,sCl2 molecule are equivalent to
force constant k = 329 N m-I, what is
this molecule? The effective mass
is half its total mass, and
8.17(b) Assuming that the vibrations of a 14N2molecule are equivalent to
those of a harmonic oscillator with a force constant k= 2293.8 N m ",
what is the zero-point energy of vibration of this molecule? The effective
mass of a homonuclear diatomic molecule is half its total mass, and
m(l4N) = 14.0031mu'
8.18(a). The wavefunction, 1If(¢), for the motion of a particle in a ring is of the
form 1If=Neim;. Determine the normalization constant, N.
8.18(b) Confirm that wavefunctions for a particle in a ring with different
values of the quantum number ml are mutually orthogonal.
8.19(a) Calculate the minimum excitation energy of a proton constrained
rotate in a circle of radius 100 pm around a fixed point.
inca
-::====::~
to
8.19(b) Calculate the value of 1mII for the system described in the preceding
exercise corresponding to a rotational energy equal to the classical average
energy at 25°C (which is equal to ikT).
8.2O(a) Estimate the rotational quantum number of a bicycle wheel of
diameter 60 cm and mass 1.0 kg when the bicycle is travelling at 20 km h-I.
8.20(b) The mass of a vinyl gramophone record is 130 g and its diameter is
30 ern. Given that the moment of inertia of a solid uniform disc of mass m
and radius r is I =imr2, estimate the rotational quantum number when
the disc is rotating at 33 r.p.m.
8.21(a) The moment of inertia of a CH4 molecule is 5.27 x 10-47 kg m2.
What is the minimum energy needed to start it rotating?
8.21(b) The moment of inertia ofan SF6 molecule is 3.07
What is the minimum energy needed to start it rotating?
X
10-45 kg nr'.
8.22(a) Use the data in Exercise 8.21a to calculate the energy needed to
excite a CH4 molecule from a state with 1= I to a state with 1= 2.
8.22(b) Use the data in Exercise 8.21b to calculate the energy needed to
excite an SF6 molecule from a state with 1= 2 to a state with 1= 3.
8.23(a) What is the magnitude of the angular momentum
when it is rotating with its minimum energy?
of a CH. molecule
8.23(b) What is the magnitude of the angular momentum
when it is rotating with its minimum energy?
of an SF6 molecule
•
8.14(a) Confirm that the wavefunction for the ground state of a onedimensional linear harmonic oscillator given in Table 8.1 is a solution of the
Schrodinger equation for the oscillator and that its energy is
with
I
1
8.24(a) Draw scale vector diagrams to represent the states (a) 5=2' m'=+2'
(b) 1= 1, ml=+I, (c) 1=2, ml=O.
8.24(b) Draw the vector diagram for all the permitted
with 1=6.
states of a particle
--
-----
,-
QUANTUM
THEORY:
TECHNIQUES
AND APPLICATIONS
Hence, we require the values of n I, nz, and n3 that make
2
nl
+ n22 + n32
Therefore,
= 9.
(nl, 112,n3) =
(1, 2, 2), (2, 1,2), and (2, 2, 1) and the degeneracy is
Question. What is the smallest multiple of the lowest energy, EI
E9.7(a)
E = (nT
+ n~ + n5)
h2)
( 8mL2
x
6.E _ (KI(0.9L)2)
KIL2
~
E = (v
+ 2:1)
2
for which E,,, n2n3 does not exist?
II
+ n22 + n32)
or
(~)
8m
x
.
123percentl.
0.81
(k)
1/2
co =;;;
lu»,
K = (nl
= _1_ -1 =10.23~
(KIL2)
IF -
~~'t[~)
K
= L2'
[D
[9.25].
The zero-point energy corresponds to v =·0; hence
11k
1/2
Eo = =tu» = -Ii (-)
22m
1
= (-)
2
x (1.055 x 10-34 J s) x
1/2
155Nm-1
2.33 x 10-
(
26
kg )
= 14.30 x 10-21 J I.
( k)
~
6.E = Ev+1 - E; = tuo [9.26] = Ii;;;
'6 "(OUt-')
6.E)2
Ii
Hence k = m ( -
J)
~~t\t.~
= (1.33 x 10-25 kg) x
1/2
(
[9.25].
21 J)2
4.82 x 1034
l.055 x 10- J s
=
I 278 N m- I
I
The requirement for a transition to occur is that 6.E(system) = E(photon), so
6.E(system)
= tuo [9.26],
he
E(photon)
= h» = -.
A
Therefore, -he = -hco = (h)-
A
2IT
2IT
1/2
= (2n)
(In)
k
A = 2ITe
x ( -k )
m
1/2
x (2.998 x 108 ms-I)
x
(1.673 x 10-27 kg)
I
855Nm-
= 2.63 x 10-6 m = 12.63 Il-m,.
,/2.0....
Since A <X ml/2,
$S'·C3a...
J
)
(a) co=
(T)
Anew = 21/2Aold = (21/2)
x (2.63Il-m)
= 13.nll-ill I.
1/2
[elementary physics].
6.E = Iua [harmonic oscillator level separations, 9.26]
= (l.055 x 10-34 J s) x
981
. ms
l.Om
(
-2) 1/2= 13.3 x 10-
34
J
I.
1/2
[11
=
1 N m].
183
STUDENT'S SOLUTIONS MANUAL
184
=
(b) !)'E =; h»
~)
(6.626
X
~)':i<:~?:'~)=
10-34 JI1z
13.3 x 10-
33
J I·
The Schrodinger equation for the linear harmonic oscillator is
glttt~)
!i2 d21/f
- 2m d.x2
1
+ 2kx21/f
= E1/f [9.24].
The ground-state wavefunction is
1/fo
= Noe-x2/2a2
with a =
(If)
mk
[9.29aJ
1/4 = (~)
m2{j)2
1/4
k=£
ma4
(a)
.
Performing the operations
d!O = ( _a~x)
d
2
--d.x1/fo
= (
2
1/fo,
1) x ( 1 ) x 1/fo + (1)
--x
a2
--1/fo
a2
--x
a2
=
1
x2
-1/fo
a4
--~"'f{io
a2
= (X~
-a4
- :2)
1/fo,
and substituting into the Schrodinger equation,
--
n2
2m
(X2-
a4
1 2 1/fo
- - 12 ) 1/fo + =kx
a
2
= E01/l0
which implies
Eo
2
= _n
2m
(X2
+~
_~)
a4
a2
(b)
-
But Eo is a constant, independent of x; therefore the terms that contain
only if
x
must drop out, which is possible
n2
1
---+-k=O
2ma4
2
which is consistent with k =
n2
Eo = -- 2
2ma
1
n2/ma4
[
=-nw
2
using
as in (a). What is left in (b) is
(j)
( k)
= -;
1/2
and k
n2
= ma
4
]
.
Therefore, 1/fo is a solution of the Schrodinger equation with energy
1
2nw.
gllS-(~)
~)
The harmonic oscillator wavefunctions have the form
1/fv(x)
= NvHv(y)
exp
(
-21i )
with y = ~ and
a =
(n2)mk 1/4 [9.28J.
The exponential function approaches zero only as x approaches ±oo, so the nodes of the wavefunction
are the nodes of the Hermite polynomials.
H4(y)
=
16i - 48i
+ 12 =
0 [Table 9.1J.
QUANTUM
= i, we have
Dividing through by 4 and letting z
4z2
12z
-
so z =
THEORY:
TECHNIQUES
AND APPLICATIONS
185
a quadratic equation
+3=0
-b
±.Jb2 - 4ac
12 ± Jl22 - 4 x 4 x 3
3 ±.J6
2x4
2a
2
Evaluating the result numerically yields z = 0.275 or 2.72, so y = ±0.525 or ±1.65. Therefore
x =1 ± 0.525a or ± 1.65a I·
Numerical values could also be obtained graphically by plotting H4(y).
COMMENT.
~'I1-COv)
~)
For a two-particle oscillator
I
I
1
meff
m\
rn
-=-+-.
meff
replaces
m
in the expression for
co,
where
(See Chapter 13 for a more complete discussion of the vibration of a diatomic molecule.) Here ml
so meff = m12. Thus
Eo = ~1Uu =
2
mcJ5Cl)
~n(~)1/2
2
=
~n(2k)1/2
2
meff
m
= 34.96888 u = (34.96888 u) x (1.66054 x 10-27 kg/u) = 5.807 x 10-26 kg.
34
_ (1.05457 x 1O2
JS)
Eo -
x
») _I
1
((2) x (329Nm26
5.807 x 10- kg
~.t<6~
)
E9::t6(a)
We require
f
1{1*1{1 d. = 1,
that is,
r:
10
N2e-im'¢eim{¢d¢
r:
= 10
2
N 2_~
-
2n
,
N-_~1
)\/
2n
.
N2d¢ = 2nN2 = I.
\/2
-
6
-21
5. 1 x 10
J.
I
= mo ;