Problem 24.2 part a
 23
k  1.3806505  10
nm  10
1
 J K
23
N  6.02211415 10
5
M  0.131293 kg
D  0.5 10
2 1
m s
3
T  273  K
1
  3 
 
 8k  T 
 m 
a

2
m
Vaver  209.82
s
D
  7.149  10
Vaver
k T
2
Problem 24.2 part b
mN  2 
1
 N 
 N 
 8k T 
 m 
N

2  P  N
0.0140067
 kg
N
2
 Vaver   DN 

   
VaveN  D 


k T
m
  0.368 nm
2  P 
VaveN 
8
m
VaveN  454.239
s
8
 N  9.907  10
2
 N  0.266 nm
m
m
P  101325 J m
M
ma 
N
Vaver 
9
DN  3  D
Sv  10
Problem 24.6 part a
 23
k  1.3806505  10
D  11.3 10
 11
1
 J K
2 1
m s
S  2.04 Sv
 13
s
nm  10
k T
6     D
23
3
  1.002  10
T  293.15 K
r  1.896 nm
S 6      r
 1  V  
M  m N
M  16.828 kg
1
 kg s
3 1
V  0.740  cm  g
Problem 24.6 part b
m 
m
N  6.02211415 10
Rearranging the Stokes-Einstein equation:
r 
9
  0.998  g  cm
3
1
m
Problem 24.8
Time (h) x (cm)
0
2.5
5.2
12.3
19.1
x/x(0)
4
4.11
4.23
4.57
4.91
Time (s)
1
1.0275
1.0575
1.1425
1.2275
ln(x/x(0))
0
0
9000 0.027129
18720 0.055908
44280 0.133219
68760 0.20498
w=
Slope =
40000 rpm
-1
2.99E-06 s
The slope of this plot (below) gives an expression for S:
S=
1.7041E-13 second
Using the available data and the equation M = S(6r)/(1-V)
r=
1.89051E-09 m
Sedimentaion of Cytochrome c
0.25
y = 3E-06x + 0.0002
R2 = 1
ln(x/x(0))
0.2
0.15
0.1
0.05
0
0
10000
20000
30000
40000
Time (s)
50000
60000
70000
Problem 24.15
First determine the viscometer constant from the water experiment.
 w  1.0015 10
A 
w
 w tw
3
1 1
 kg m
3
s
3
 w  0.998  10  kg m
A  6.69  10
8
tw  15 s
2 2
m s
Second use the viscometer constant and the density of the material to determine the viscosity.
t  37 s
 
mass
V
  A   t
6
V  100  10
  765
3
m
kg
3
m
  0.019 poise
3
mass  76.5 10
 kg
Problem 24.18
T(C)
Viscosity ( cP)
5
0.826
40
0.492
80
0.318
120
0.219
160
0.156
1/T ( 1/K )
0.003595
0.003193
0.002832
0.002544
0.002309
ln()
-0.191161
-0.709277
-1.145704
-1.518684
-1.857899
The slope of this plot gives an expression for E:
10662.705 J/mole
The intecept gives the value of A:
0.008256124 cP
Visocisty of Benzene as a Function of Temperature
0
-0.2
-0.4
y = 1282.5x - 4.7968
R2 = 0.9993
ln(/cP)
-0.6
-0.8
-1
-1.2
-1.4
-1.6
-1.8
-2
0.002
0.0025
0.003
1/T (1/K)
0.0035
Problem 24.24
The value for polyethylene is found by substituting the angle into the equation
given in the problem:
 r 2   nl 2
1  cos 
1  cos 
 nl 2
1  cos 70  32'
1  0.3333
 nl 2
 2nl 2

1  0.3333
1  cos 70 32'
For a random coil polymer
 r 2   nl 2
Thus, the polyethylene would be expected to have twice the radius of gyration of
a random coil polymer of the same monomer length.
Problem 24.29a
Time (min) x(cm)
0
4.46
80
4.593
160
4.713
240
4.844
Time (s)
0
4800
9600
14400
slope =
intercept =
ln(x/x(0))
0
0.029384639
0.055175882
0.08259206
-1
5.69932E-06 s
0.000753032
Sedimentation of chromatin
0.09
0.08
0.07
ln(x/x0)
0.06
0.05
0.04
0.03
0.02
0.01
0
0
2000
4000
6000
8000
10000
12000
14000
16000
Time (s)
Knowing the slope and that the centrifuge was moving at 18, 100 revolutions per minute,
one can calculate the sedimentation coefficient.
Problem 24.29a
6 1
slope  5.69932  10
s
1
R  8.3144339  joule K
sed 
 12
slope

sed  1.586  10
2
T  293  K
D  4.37 10
V  0.66 10
M 
N0  6.0221367  10
3
3
23
  2   
18100
60 s
s
 11 2  1
m s
3
3
  1.02 10 kg m
1
 m  kg
R T sed
M  270.611 kg
D  1  V  
Problem 24.29b
Assume that the nucleosome is unhydrated and that it is a spherical particle. Use the molar
mass determined above and the molar volume given
1
r 
 3 M V 
 4   N 
0

3
r  4.137  10
9
m
Problem 25.2a
The problem tells you that the reaction is first order in the concentration of
cyclobutane, so the rate law is known. The rate law in terms of the partial
pressure of cyclobutane is
v   k1 Pcyclob
The problem is to express this in terms of the total pressure in a constant-volume
container. To do so, make a table
Time
0 (pressure)
Cyclobutane
P0
Ethylene
0
Total
P0
t (pressure)
P0  x
2x
P0  x
Now, since we know Ptotal at any time, we can get a value for x:
x  Ptotal  P0
Substitution in to the partial pressure of cyclobutane at t gives
Pcyclob  P0  ( Ptotal  P0 )  2 P0  Ptotal
Substitution of this into the rate expression gives an equation expressed in terms
of the total pressure at any time:
v   k1 (2 P0  Ptotal )
Problem 25.2b
For a first-order rate law, there is a relation between the time constant and the
rate constant:
ln 2
k1 
t1/ 2
Rearrangement gives
ln 2
t1/ 2 
 2795 s
2.48  10 4 s 1
Problem 25.2c
This is simply a substitution into the integrated form of the first-order equation:
A(t )
ln
  k1t
A(0)
Then
ln(0.90)   2.48  10 4 s 1t
Solving for t gives the result:
t  424.8 s
Problem 25.3
Since all three of these molecules are in the gas phase
one must determine how the total pressure changes.
P(SO2Cl2)
Time
P (SO2)
0 P0
P0 - x
t
P(Cl2)
P (total)
0 P0
P0 + x
0
x
x
One may show from these values that P(SO2Cl2) = 2P0 - P(total)
which is the function on eshould plot to determine the rate law.
2P0-P(total)
=x
ln(x)
1/x
t (h)
P(total)/kPa
0
11.07
11.07 2.404239 0.090334
3
14.79
7.35
1.9947 0.136054
6
17.26
4.88 1.585145 0.204918
9
18.9
3.24 1.175573 0.308642
12
19.99
2.15 0.765468 0.465116
15
20.71
1.43 0.357674 0.699301
First Order Plot
Second Order Plot
0.8
3
y = -0.1365x + 2.404
R2 = 1
0.7
2.5
0.6
2
1/X
ln (x)
0.5
1.5
0.4
0.3
1
0.2
0.5
0.1
0
0
0
2
4
6
8
Time (h)
10
12
14
16
0
2
4
6
8
10
12
Time (h)
It is obvious from the linearity of the left plot that this decomposition is a first-order reaction.
From the slope of the line one finds that the effective rate constant is 0.1365 h-1 = 3.79x10-5 s-1.
14
16
Problem 25.8
This is a problem to determine the initial order of reaction. This is done by
comparing rates under different conditions. Consider a reaction with two
reactants. The rate may be written as:
R  k[ A] a [ B]b ,
where the superscripts are the orders with respect to A and B. Taking the ratio of
the initial rates for two different conditions
R1
R2
 [ A]1
 
 [ A] 2



a
 [ B]1

 [ B]2



b
If one takes a ratio of two experiments where the concentration of one reactant is
the same, then the ratio allows one to determine the order. Let’s start with the
lactose. The first two lines have a common hydrogen ion concentration.
a
0.00116
 0.010 
 

0.00232
 0.020 
Taking the logarithm of both sides gives
 0.693147  a 0.693147 
It should be obvious from this equation that the order with respect to lactose must
be 1.0.
Using the first and third experiments isolates the order with respect to hydrogen
ion.
b
0.00116
 0.001 
 

0.00464
 0.004 
Taking the logarithm of both sides gives the following equation:
 .1386294  b(.1386294)
Once again, it should be obvious that the order with respect to this component is
also 1.0.
So, one may write the rate law in this form:
R  k[ Lactose]1 [ H  ]1
Problem 25.9
The half-life depends on the initial concentration, so one can immediately
conclude that the reaction is NOT first order in acetaldehyde. With two points,
the data are limited, but here is how one would solve the problem. The
integrated general relationship for a reaction gives the following equation (except
for an order of 1):
[ A]  n 1  [ A(0)] n 1  n(n  1)kt
At the half-life, this gives the following result:
 n 1
 A(0) 
 [ A(0)] n 1  n(n  1)kt1 / 2
 2 
This can be transformed to the following equation:
 n 1


 A(0) 
1  2 n 1  n(n  1)kt1 / 2
 2 
Taking the ratio for two different experiments with two different starting
concentrations:
 n 1
t1 / 2,1
 [ A(0)]1 



t1 / 2 , 2
 [ A(0)]2 
From the data given in this problem (only 2 points), one substitutes to give
 n 1
 9.72  10 3 
328



3 
572
 4.56  10 
or
2.1316 n 1  0.57343
Taking the logarithm of both sides, gives an equation for n:
(n  1)0.75687   0.55612
 n  1   0.7348
Or
This gives
n  1.7348
(Your book says this is “2”.)
Once the order is known, one may substitute in either experiment to get k. Using
the first values.
0.7348
 9.72  10 3 mole 
mole 



  9.72  10 3

3 
2
dm 
dm 3 


The rate constant is then:
0.7438
0.7348
 (1.7348)(0.7348)k (328 s )
 dm 3 

k  0.048 
s 1
 mole 
(This is different from your book’s answer. Note how the units depend on the
order. It seems that whoever wrote the answer assumed that, if the reaction were
not first order, it must be second order.)
Problem 25.11
(a) According to the equation for a first-order reaction
CR (t )
 exp k1t 
CR (0)
From the statement of the problem, one knows the fraction remaining at a certain
time:
0.325  exp k1 540 s 
This equation has only one unknown, so one can solve for it. Taking the
logarithm of both sides
ln(0.325)   k1 540 s
 1.12393   k1 540 s
Or
k1  0.00208 s 1
(b) Knowing the rate constant, one may use the equation to calculate the time
when a specific amount is left.
0.100  exp  0.00208 s 1 t
Taking the logarithm
ln(0.100)   0.00208 s 1 t
Solving this equation gives the time for this amount to be left:
t  1107 s


Problem 25.12
This problem requires us to calculate the rate constant for this disintegration from
the half life:
k

ln 2
t1 / 2

ln 2
1.41912  1017 s

ln 2
day
h
min
sec
4.5  109 year  365
 24
 60
 60
year
day
h
min
 4.8843  1018 s 1
The question asks for the number of integrations in one minute for 10 mg of
uranium. We have to calculate how many molecules there are in 10 mg:
10  10 3 g  6.02211415  10 23 molecule mole 1
N 
238.0508 g mole 1
19
 2.52976  10 molecules
The rate of this first-order decomposition is:
dN
  kN
dt
The number of disintegrations can be found, fairly accurately, by multiplying this
by the time, assuming that N does not change greatly over the time (which is true
for such a small rate constant):
dN
N 
t   4.8843  10 18 s 1 (2.52976  1019 molecules)(60 s )
dt
  7414 molecules
The absolute value of this change is the number of disintegrations that occur in
one minute: 7414 disintegrations
Problem 25.18
The concentration of an intermediate in a sequential reaction is given in equation
(25.53):
[ A]0 k A  k A t
[I ] 
e
 e k I t
kI  k A
where I is equivalent to B in the problem. To find the maximum of a function of
time, one takes the derivative with respect to time and finds the value of the time
when the derivative is zero.
[ A]0 k A
d[I ]

 ka e  k At  k I e  k I t
dt
kI  k A
Setting this derivative equal to zero gives this equation:
[ A]0 k A
 ka e  k At max  k I e  k I t max  0
kI  k A
Examination shows that there are two factors. The first factor (the collection of
constants) cannot generally be zero. So, the second factor is zero at tmax.
 k Ae  k A t max  k I e  k I t max  0
This equation is rearranged to:
k Ae  k A t max  k I e  k I t max
That can be rewritten as:
kI
e  ( k A  k I )t max 
kA
Taking the logarithm of both sides gives:
 
k I  k A tmax  ln k I 
 kA 
A final rearrangement gives the equation we seek:
k 
1
tmax 
ln I 
k I  k A   k A 
Substitution of the rate constants gives:
 3  106 s 1 
1

tmax 
ln
3  106 s 1  5  106 s 1  5  106 s 1 









1
ln(0.6)  2.55  10 7 s
2  106 s 1
Problem 25.22a
To find the maximum of the intermediate concentration, substitute the values of
the rates constants into the expression:
t max

k
1
ln 1
k1  k 2  k 2



For the bacteriorhodopsin reaction, the rate constants are reported.
t max

 2.0  1012 s 1 
1


ln
2.0  1012 s 1  3.3  1012 s 1  3.3  1012 s 1 
 3.8  10 13 s
Problem 25.23
In parallel reactions, the ratio of the equilibrium concentration of the products is
related to the ratio of the rate constants.
[ B]eq
kC

[C ]eq
kB
The half-life is related to the total rate constant for the disappearance of the
potassium:
ln 2
t1 / 2 
(k B  k C )
These two equations are sufficient to solve simultaneously for the two rate
constants. Using the information from the problem, one has these two equations:
kC
kB

0..893
 8.346
0.107
And
k B  kC

ln 2
1.3  10 9 yr

0.693147
1.3  10 9 yr
Solving these simultaneously gives the final result:
kB
 5.705  10 11 yr 1
k C  4.761  10 10 yr 1
where C represents the first process.

5.3319  10 10 yr 1
Problem 25.29
In a relaxation method such as T-jump relaxation, one follows how a deviation
from equilibrium is ameliorated to return to equilibrium. To do this one, a
mechanism is required, which is given in the problem.
kf
DS 
2 SS
kr
2 SS 
DS
Let us set up a table that indicates the changes that occur:
Condition
Equilibrium
Nonequilibrium
[DS]
[DS]eq
[DS]eq - x
[SS]
[SS]eq
[SS]eq + 2x
In this table, the deviation from equilibrium is measured by x, the deviation of the
double-stranded DNA from its equilibrium value. By the stoichiometry of the
problem, one can then express the concentration of single-stranded DNA. Next,
write out the general formula for the change in concentration of double-stranded
DNA:
d [ DS ]
  k f [ DS ]  k r [ SS ] 2
dt
To obtain an expression for the time-dependent deviation from equilibrium,
substitute into the equation:
d [ DS ]eq  x 
2
  k f [ DS ]eq  x   k r [ SS ]eq  2 x 
dt
The equilibrium concentrations are time-independent quantities. Expansion of
this equation gives:
dx
2

  k f [ DS ]eq  k f x  k r [ SS ]eq
 4k r [ SS ]eq x  4k r x 2
dt
Now, rearrange the terms and neglect any terms in x higher than order one. This
gives the equation that is good for small deviations from equilibrium:
dx
2
 k f [ DS ]eq  k r [ SS ]eq
 k f x  4k r [ SS ]eq x
dt
It can be shown that the first two terms cancel because they arise from the
equilibrium rate expression, so the equation simplifies to:
dx
  k f  4k r [ SS ]eq x
dt
The inverse of the relaxation is the term in parentheses:
1
 
k f  4k r [ SS ]eq
Problem 25.34
For a diffusion controlled reaction, one would expect that the rate constant is
given by
k  4N 0 (rA  rB ) D AB
where DAB is the mutual diffusion coefficient of the two reacting partner, i.e. the
sum of the two independent diffusion coefficients. Comparison of this calculated
diffusion-limited value with the experimental value will allow one to make a
statement about whether the process is diffusion-controlled.
Using the data from the problem gives:
k  4 (6.02211415  10 23 mole 1 )(51.2  10 10 m  2.0  10 10 m)
 (6.0  10 11 m 2 s 1  1.5  10 9 m 2 s 1 )
 6.28  10 7 m 3 mole 1 s 1
 6.28  10 4 dm 3 mole 1 s 1
Comparison of this number to the experimental number (5106dm3 mole-1 s-1)
shows that the reaction occurs much faster (i.e. with a larger rate constant) than
would be expected for a reaction under diffusion control. So, the answer to the
question is that the catalyzed conversion of hydrogen peroxide is NOT diffusion
controlled.
Problem 25.37
(a) To determine the Arrhenius parameters one must make a plot of ln(k) versus
1/T. Taking the data into EXCEL, one gets the following table and plot:
T(K)
270
370
470
570
670
k(M-1s-1)
3.43E+10
3.77E+10
3.99E+10
4.13E+10
4.23E+10
1/T(1/K)
0.0037037
0.0027027
0.0021277
0.0017544
0.0014925
ln(k)
24.25841
24.35293
24.40964
24.44413
24.46805
From the slope and intercept, one obtains the Arrhenius parameters:
E a   R( slope)   8.3144349 joule mole 1 K 1  (95.1627 K )
 791.2 joule mole 1
A  exp(intercept )  exp(24.61085)  4.88  1010 dm 3 mole 1 s 1
(b) Substitution into the definition of the rate constant gives the value at 220K.


791.2 joule

k (200 K )  Ae  Ea / RT  4.88  1010 dm 3 mole 1 s 1 exp 
1
 8.3144349 jouleK 220 K 
 3.17  1010 dm 3 mole 1 s 1
(c) This is a bimolecular reaction. The appropriate equation is (25.133).
Substitution gives
H 
S 
 E a  2 RT  791.2 joule mole 1
  2867 joule mole 1
 2  8.3144349 joule mole 1 K 1 (220 K )

 hC  
 6.6260693  10 34 joule s 4.88  1010 s 1
 R ln 2
A   8.3144349 joule mole 1 K 1 ln 2
 23
1
 e (1.3806505  10 joule K 220 K
 e k bT 
  54.4 joule K 1 mole 1
 


Problem 26.1
The stoichiometric formation reaction is
1
2 NO2 
O3  N 2 O5
3
The problem proposes this two-step mechanism
k1
NO2  O3 
NO3  O2
NO2

NO3
 M
k2
N 2 O5

 M
To start the process, write down the rate of appearance of N2O5 in terms of the
rates of the elementary steps.
d [ N 2 O5 ]
 v2
dt
Now, NO3 is neither a reactant nor a product. Therefore, it must be a reactive
intermediate. That implies the use of the steady-state approximation on it.
d [ NO3 ]
 v1  v 2  0
dt
v1  v 2
This equation says that the rate of the first equation is equal to that of the
second, which gives
d [ N 2 O5 ]
 v1  k1 [ NO2 ][O3 ]
dt
Problem 26.5b
The rate of appearance of HBr at early times is given by the following equation:
1/ 2
k 
d [ HBr ]
 2k 2  1  [ H 2 ][ Br2 ]1 / 2
dt
 k5 
 k eff [ H 2 ][ Br2 ]1 / 2
according to the Chirstiansen-Herzfeld-Polanyi mechanism. The rate constants
and their activation energies have been determined, and are reported in the
problem to be:
Rate constant Ea (kJ/mole)
k1
192
k2
0
k5
74
From the rate equation, one finds the equivalence:
1/ 2
k 
k eff  2k 2  1 
 k5 
Since the various rate constants are thermally activated, one may write:
k eff
k 
 2k 2  1 
 k5 
1/ 2
 2k 2 , 0 e
 k1, 0
 2k 2, 0 
 k 5, 0




1/ 2
 k1, 0
 2k 2, 0 
 k 5, 0




1/ 2
e
e
1/ 2
 E a , 2 / RT
 E a , 2 / RT
 Ea , 2 / RT
e
e
 k1, 0 e  Ea ,1 / RT

 k e  Ea , 5 / RT
 5, 0




1/ 2

 ( Ea ,1  Ea , 5 ) / RT 1 / 2
 ( E a ,1  E a , 5 ) / 2 RT
E a ,1  E a , 5
 k1, 0 
) / RT
 ( Ea , 2 
2
 e
 2k 2, 0 

 k 5, 0 
This is of the form of an activated rate constant, with the activation energy being
E a ,1  E a ,5
E a ,eff  E a , 2 
2
Using the data in the table, one can estimate the effective activation energy for
this process at early times.
192kJ mole 1  74kJ mole 1
E a ,eff  0kJ mole 1 
2
 59 kJ mole 1
Problem 26.7
The mechanism has three elementary steps:
k1
2A
A2 
k 1
2 A 
A2
k2
A  B 
Product
By the assumption of pre-equilibrium, the two first steps occur sufficiently fast
that an equilibrium concentration of A is maintained. The equilibrium expression
is
k1
[ A] 2
K 

k 1
[ A2 ]
The formation of product only occurs in the second (actually third) step:
d [ Product ]
 k 2 [ A][ B]
dt
Rearrangement of the equilibrium expression allows one to express the
concentration of A in terms of the concentration of [A2]:
[ A]  K 1 / 2 [ A2 ]1 / 2
Substitution gives a final expression for the rate of creation of product:
1/ 2
k 
d [ Product ]
 k 2 K 1 / 2 [ A2 ]1 / 2 [ B]  k 2  1  [ A 2 ]1 / 2 [ B]
dt
 k 1 
Note that this derivation predicts the order with respect to A2 to be 1/2 and the
order with respect to B to be 1. These predictions can be tested by experiment.
Problem 26.10
This is an initial-rate analysis of enzyme catalysis, of the type that is carried out
quite often in biochemical kinetics, in this case the decomposition of hydrogen
peroxide by catalase. The reaction presumably obeys the Michaelis-Menten
mechanism. The particular data here are given in the table.
[H2O2] (mole dm-3)
0.001
0.002
0.005
Rate (mole dm-3 s-1)
1.38E-03
2.67E-03
6.00E-03
1/[H2O2] (dm3 mole-1)
1000
500
200
1/Rate (s dm3 mole-1)
7.25E+02
3.75E+02
1.67E+02
The data can be reduced using a Lineweaver-Burk plot. In this method, a plot of
the inverse of the rate versus the inverse of the substrate concentration is
expected to be linear:
KM
1
1
1


v
v max
v max [ H 2 O2 ]
The data are shown in the table, and the plot is given in the figure below:
800
y = 0.6977x + 26.557
R2 = 1
700
1/Rate (dm 3 mole -1 s)
600
500
400
300
200
100
0
0
200
400
600
800
1000
1200
1/C (dm 3 m ole -1)
From the slope of the line and the intercept, one may determine the MichaelisMenten parameters.
v max  1 / intercept  1 / 26.557 dm 3 mole 1 s
 3.77  10  2 mole dm 3 s 1
K M  slope  v max  0.6977 s  3.77  10 2 mole dm 3 s 1
 0.0263 mole dm 3
The remaining rate constant is found from the known catalase concentration:
v max
3.77  10 2 mole dm 3 s 1
k2 

[Catalase]
3.5  10 9 mole dm 3
 1.08  10 7 s 1
Problem 26.11
v0 [I]=200 mole dm -3 1/[S]
v00; [I]=0
[S]
1/v0
1/v00
(mole dm ) (mole dm s )
(mole dm s )
dm mole s dm mole
s dm3 mole-1
0.299
0.071
0.018
3.344
14.085
55.556
0.500
0.100
0.030
2.000
10.000
33.333
0.820
0.143
0.042
1.220
6.993
23.810
1.22
0.250
0.070
0.820
4.000
14.286
1.75
0.286
0.105
0.571
3.497
9.524
2.85
0.333
0.159
0.351
3.003
6.289
5.00
0.400
0.200
0.200
2.500
5.000
5.88
0.500
0.250
0.170
2.000
4.000
-3
-3
-1
-3
Initial rate (s dm3 mole-1)
60.000
-1
-3
3
y = 16.314x + 1.2817
R2 = 0.9959
50.000
40.000
30.000
y = 3.8838x + 1.5477
R2 = 0.9856
20.000
10.000
0.000
0.000
0.500
1.000
1.500
2.000
2.500
3.000
3.500
4.000
1/[S] (dm mole )
3
Without inhibitor
-1
With inhibitor
a. Determining the Michaelis constants from the slope and intercept
vmax =
Km =
-3 -1
0.646 mole dm s
-3
2.509 mole dm
b. The fact that both the slope depends on the concentration of
inhibitor, but not particularly the intercept, shows that this is competitive inhibition.
Ki =
-3
62.5 mole dm
-1
Problem 26.19
a. Start with the equation for the Michaelis-Menten mechanism:
[ S ]0
v0  v max
[ S ]0  K m
Multiply through to give
v0 [ S ]0  K m   v max [ S ]0

K 
v0 1  m   v max
 [ S ]0 
Rearrangement gives the equation sought:
 v 
v0 
v max  K m  0 
 [S ]0 
 v0
Division by [S]0 gives:
 Km
b. Make a plot of v0 versus v0/[S]0 according to this equation
0.4
0.35
v0 (mole dm -3 s -1)
0.3
y = -0.0446x + 0.4639
R2 = 0.9953
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
6
7
-1
v0/[Sucrose] (s )
[Sucrose]
(mol dm-3)
0.029
0.059
0.088
0.117
0.175
v0
(mole dm-3 s-1)
0.182
0.266
0.31
0.33
0.372
1/[Sucrose]
(mole-1 dm3 s)
5.494505495
3.759398496
3.225806452
3.03030303
2.688172043
The intercept is vmax.
vmax =
0.4639
mole dm-3 s-1
0.0446
mole dm-3
The slope is -Km.
Km
v0/[Sucrose]
(s-1)
6.27586207
4.50847458
3.52272727
2.82051282
2.12571429
v0
[ S ]0
Problem 26.20
This is a calculation of the energy contained in a photon of light. The appropriate
equation for this is Planck’s equation:
hc
.
E p  h 

Let us start with the calculation of the energy of a single 290-nm photon.
6.6260693  10 34 J  s 2.99792458  10 8 m s 1
hc
Ep 


290  10 9 m
19
1
 6.8498124  10 J photon

a. To find the number of photons, one must divide the MED by the energy per
photon.
50  10 3 J cm 2
photons
MED
 7.2994700  1016
N 290 

19
1
Ep
6.8498124  10 J photon
cm 2
Since the energy of the 320-nm photon is different, the number at that
wavelength will be different.
photons
N 320  8.0545975  1016
cm 2
b. This requires one to calculate the time it takes for the solar to deposit the
equivalent of the MED:
50  10 3 J cm 2
MED
t 

 34.5 s
solar flux
1.45  10 3 W cm  2
That is not a long time!

Problem 26.21
This problem is a mechanism, for which we are supposed to define the rate law.
The mechanism is
CH 3 CHO  h
 CH 3   CHO 
CH 3   CH 3 CHO
k1

CH 4
 CH 3 CO 
k2
CH 3 CO  
CO  CH 3 
k3
2 CH 3  
C2 H 6
First, one writes the rate sought:
d [CO ]
R 
 v2
dt
There are several reactive intermediates in this problem on which one applies the
steady-state approximation:
d [CH 3 ]
 0  v0  v1  v 2  2v3
dt
And
d [CH 3 CO ]
 0  v1  v 2
dt
This latter equation shows that v1
 v2 .
Combining the two equations gives v0  2v3 . From this equation one gets a
relationship that allows the steady-state concentration of the methyl radical:
I a  k 3 [CH 3 ]2
Or
I a
[CH 3 ] 
k3
Since v1 and v2 are the same, according to the first equation, one may write
R  v 2  v1  k1 [CH 3 ][CH 3CHO ]
Substitution to eliminate the concentration of methyl radical gives the expression
one seeks.
k  1/ 2
R  k1 [CH 3 ][CH 3 CHO]  1 1 / 2 I a1 / 2 [CH 3 CHO]
k3
It is predicted to be first order in the acetaldehyde and only the ½ order in the
absorbed light intensity.
Problem 26.22
The problem involves manipulation of equation 26.97:
kf
k f f 
 
k f  k ic  k isc  k Q [Q]
One is given kic and f and told that kisc amd kQ[Q] are negligible. This gives the
equation:
kf
 .
k f f 
k f  k ic
This still requires, apparently, one to determine kf. Multiplying through gives the
following equation:
k 2f  f  k ic k f  f  k f
which is rearranged to give:
k f k f  f  k ic f  1  0
For this equation to be correct, one must require that
k f  f  k ic f  1  0
This is rearranged to the following:
k f  f  1  k ic f
But, by the starting equation this is just the quantum yield. Substituting into this
equation gives the result requested:
  1  k inc f  1  5  10 8 s 1 1  10 10 s   0.95
Problem 26.28
%O2
I0/I
0
12
20
47
100
1
3.6
4.8
7.8
12.2
The slope of the line is ratio of kQ to kf.
-1
kQ =
18779.7 s
The fact that the graph is not linear
led the authors to postulate a second
process.
14
12
y = 0.1061x + 2.0811
R2 = 0.968
10
I/I0
8
6
4
2
0
0
20
40
60
Percent O2
80
100
120
Problem 26.30
First, one must calibrate the fluorescence intensity. Since 7500 counts
corresponds to an efficiency of 0.5, one can calculate the efficiency
corresponding to the other two counts.
Eff 1

5000
0.5  0.333
7500
Eff 2

10000
0.5  0.667
7500
Knowing the distance in the uncharged complex is 5.0 nm, one can then obtain
the distance in the charged complex from the efficiency:
r06
r06  r 6
Eff
This can be rearranged to
r 
6
1  Eff
r0
Eff
Substituting from above gives the value for the two cases:
r1
 5.6 nm
r2
 4.5 nm
So the fluctuation in the distance is
r
 r1
 r2
 5.6 nm  4.5 nm  1.1 nm
Problem 26.33
k (s-1)
2.10E+08
2.01E+07
2.07E+05
2.04E+02
d (nm)
1.4
1.7
2.3
3.2
ln(k/s-1)
19.16262
16.81623
12.24047
5.31812
By plotting the logarithm, one may determine

from the negative of the slope.
 =
7.6821
25
nm-1
y = -7.6821x + 29.901
R2 = 1
ln(k/s -1)
20
15
10
5
0
1
1.5
2
2.5
3
3.5
d (nm )
Using the rate constant for 1.7 nm, one may substitute to find the value of  that makes
the rate constant at 2.3 nm lower by a factor of 10. Using the formula gives
k (1.7)
k (2.3)

exp  1.7 nm 
 exp  1.7  2.3nm   10
exp  2.3 nm 
Taking the logarithm of both sides gives
ln(10)  0.6nm
Finally, solving this equation gives the result

 3.84 nm 1
Problem 12.2
The average speed is vrms

3kT
. The magnitude of the average
m
momentum is then
3kT

3kTm
m
According to de Broglie’s relation, the wavelength and the momentum are related
and the ratio is Planck’s constant. So,
h
h
 
.

p
3kTm
An example calculation for He at 100 K:
| pave |  mvave
 
h
3kTm
 m
6.6260693  10 34 J s

3(1.3806505  10  23 JK 1 )(100 K )(
0.00402602kg mole 1
)
6.022114115  10 23 mole 1
 1.259  10 10 m  0.1259 nm
Similar calculations give the results in the table:
He
Ar
At 100 K
0.126 nm
0.040 nm
At 500 K
0.056 nm
0.018 nm
Problem 12.6
a. The average power is the energy delivered divided by the time over which the
energy is delivered. For this laser, it is
E
0.10 J
J
P 

 1  10 7
 1  10 7 W  10 MW
8
t
s
1  10 s
b. First, calculate the energy of a 1000-nm photon.
6.6260693  10 34 J s 2.99792458  10 8 m s 1
hc
E photon 


(1  10 6 m)
19
 1.986  10 J
Now, this energy is divided into the total energy to give the number of photons:
0 .1 J
E

 5.03  1017
N photons 
19
E photon
1.986  10 J


Problem 12.11
-19
Wavelength (nm) Kinetic Energy (10 J) Frequency (Hz)
250
4.49
1.19917E+15
300
3.09
9.99308E+14
350
1.89
8.5655E+14
400
1.34
7.49481E+14
450
0.7
6.66205E+14
500
0.311
5.99585E+14
slope =
intercept =
7.01554E-34 J s
-3.95831E-19 J
According to the equation, the work function is the negative of the intercept
and the slope is Planck's constant.
h=
b=
7.01554E-34 J s
3.95831E-19 J
5
Kinetic Energy (10-19 J)
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
1.40E+15
1.20E+15
1.00E+15
8.00E+14
6.00E+14
4.00E+14
2.00E+14
0.00E+00
Frequency (Hz)
Problem 12.14
To determine this emission, one must determine the surface area of the Sun.
Treating the Sun as a sphere, the surface area is
A  4r 2
where r is the radius of the sphere.
Substitution gives the surface area of the Sun:

A  4 7.00  10 8 m

2
 6.158  1018 m 2
The emissivity per unit area of a blackbody at 6000 K is given by:
P  T 4
 5.67  10 8 W m 2 K 4 6000 K 
4
 7.35  10 7 W m 2
Multiplying this by the surface area of the Sun gives its power:
Ptotal

 P  A  7.35  10 7 W m 2 6.158  1018 m 2

 4.53  10 26 W
Problem 12.15
The equation in the problem shows that when n =  (1/n2 = 0), the energy of the
transition is the largest of that series. Thus, the maximum energy of each series
is given by the equation:
~ 
RH
n12
Substitution in the equation gives the results in the table, and the energy per
mole is found by multiplying this by N0hc.
Series
n1
~ (in cm 1 )
E (kJ/mole)
Lyman
Balmer
Paschen
1
2
3
109,677.
27,419.3
12,186.3
13.120
3.280
1.458
Problem 13.7
In each case, one needs to convert to the polar representation. This requires
finding r and the angle , which can be found by the following formulas:
y
when z  x  iy
z  re i if r 
x 2  y 2 and   arctan
x
a.
x  2
y  4  r 
4
  arctan    0.352
 2 
z 
20 e i 0.352
2 2  (4) 2

20
b.
x  6
y  0  r 
0
  arctan   0
6
z  6 e i 0  6(1)  6
62

36
 6
c. In this case, one first as to get the function into the proper form by dividing.
3i
3 1
1
3i




z 
4i
4i 4
4
4
2
2
1
3
10
1 5
1  3
 r 
x 
y  

     
4
4
16
2 2
4  4
3


 
  arctan 4   arctan 3   0.398
 1 


 4 
1 5 i 0.398
z 
e
2 2
d. In this case, one has to reduce the problem again to the proper form.
8  i 2  4i   16  4  i32  2  12  34i  0.6  1.7i
8i

z 
2  4i 2  4i 
2  4i
4  16
20
2
2
x  0.6
y  1.7  r 
0.6  (1.7) 
3.25  1.803
1
.
7


  arctan
  0.392
 0.6 
z  1.803 e i 0.392
Problem 13.8
This problem is the reverse of problem 13.7. One uses the following formulas to
change the forms, something called Euler’s reduction.
z  x  iy if
x  r cos 
and
y  r sin 
a.
z  2e i / 2
z  2i
 x  2 cos  / 2  0
y  2 sin  / 2  2 
b.
z  e i  x  (1) cos    1
z  1
y  sin   0 
c.
 x  2 5 cos  / 2  0
z  2 5e i / 2
z   2 5i
y  2 5 sin   / 2   2 5 
d.
z 
z 
3 2
5 3
3
5 3
e i / 4
1
 x
 i
3 2
5 3
cos  / 4 
3
5 3
y
3 2
5 3
sin  / 4 
3
5 3

Problem 13.9
a.
cos 2   sin 2 

1 i
( e  e  i ) 2
4

1 i 2
e  2e i (  )  e i 2
4
1 i
( e  e  i ) 2 \
4




1 i 2
e  2e i (  )  e i 2
4


1 i 2
e  2  e i 2  e i 2  2  e i 2
4



4
4
 1
b.
d
cos 
d

d  1 i

e  e  i  

d  2


i i
e
2

i 4 i
e  e  i
3
2i
 

1  de i de i 



2  d
d 


i  i
e
2

i i
e  e i
2






  sin 
1 i
e  e  i
2i

1 i
e  e  i
3
2i


c.


sin    
2


1 i (  / 2 )
e
 e i (  / 2 )
2i


1 i i / 2
e e
 e i e i / 2
2i





1 i
e cos  / 2  i sin  / 2   e i cos  / 2  i sin  / 2
2i

1 i
e 0  i   e i 0  i 
2i

1 i
e  e  i
2




 cos 


i i
e  e  i
2i


Problem 13.10
This problem involves an understanding of the concept of a complex conjugate
and the multiple definitions of various numbers.
2
*
 mn d
2
 c

0

im
cn e in d
me
*
0
2
 cm* cn  e i ( m  n ) d
0

cm* cn
e i ( n  m )
i ( n  m)

2
 cm* cn  e im e in d

2
0
0

cm* cn
e i ( n  m ) 2  e i 0
i ( n  m)




For orthogonality, this must be zero. For that condition, the term in parentheses
must go to zero. That gives the requirement for orthogonality:
e i ( n  m ) 2
 1
Since n and m are integers other than 0, n-m is always an integer, k
e ik 2
 cos 2k
 i sin 2k
For any integer k, the sine term must be zero. For any integer, the cosine term is
1, so it is proved.
Although the authors do not request this, consider the situation that n = m. In
that case,
2
   d
*
n n
0
2

 c e  c e
in *
n
n
in
d
2
 cc
*
n n
0
2
 cc
*
n n
 (1)d
e
 in
in
e d
2
 cc
0
 2 cn* cn
0
In this case, the integral is NOT zero unless cn is zero.
*
n n
e
0

cm* cn
e i ( n  m ) 2  1
i ( n  m)
i ( n  n )
d
Problem 14.2
Remember that one can show that the free-particle wave function is an
eigenfunction of the momentum operator:
pˆ x e ikx
  i
de ikx
dx
  i (ik )e ikx
 k e ikx
with an eigenvalue k . Therefore, the eigenvalue of the energy may be rewritten
as
E 
k 2 2
2m

p x2
2m
where, in this case, the px stands for the eigenvalue of the momentum when in
this state.
But, the momentum and the speed are related by the equation
px
 mv x .
Substitution of this relationship into the equation, gives
E 
p x2
2m

m 2 v x2
2m

1 2
mv x
2
Thus, one sees that the free particle expression is equivalent to the classical
expression.
Problem 14.4
Let’s evaluate this integral over the box. The integral gives this equation, from
which one can determine the value of N.
a
N
0
2
 nx 
sin 2 
dx  1
 a 
Substitution, as suggested, gives
1
 2nx  
2
2  nx 
2 1
0 N sin  a dx  N 0  2  2 cos a  da
a
a
N
2

2 a
 dx 
0

N2
a  0 
2

N 2a
2

N2
 2nx 
cos
dx

2 0
 a 
a
N2 a
2 2 n
  2na 

sin  a   sin 0

 

N 2a
sin 2n   0
4 n
Because n is restricted to being an integer, the argument of the sine function is
always some integral multiple of 2. But the value of the sine at this value of the
argument is always zero. So, the second term does not contribute to the righthand side.
Now, the integral must be normalized, according to the first equation, so setting it
equal to 1 gives this equation
N2

2
a
Assuming N is a real, positive number gives the value for the normalization
constant of this function
N

2
a
Note that this exercise was for any value of n. This exercise has shown that the
normalization constant for this wave function does not depend on the quantum
number of the state.
Problem 14.12
a.
One may solve this by substitution.
n y 
n y 

n x
n x
 2 
  2
 2  2  N sin x sin y   E  N sin x sin y 

a
b 
a
b 
2m  x
y 

2
Distributing the derivatives gives
n y y   2  2 
n y y 
n y y 

n x x
n x x
n x x
2 2 

N
N
E
N


sin
sin
sin
sin
sin
sin






a
b  2m y 2 
a
b 
a
b 
2m x 2 

The derivative operators act only on functions of the variable on which they are
based. Functions of other variables are treated as constants. This gives
n y y  2  n x x 
n y y 
n y y 

n x x  2 
n x x
2
2
N
2m
sin
b
sin
sin

  E  N sin
 N sin
 N
sin
2m
a 
a y 2 
b 
a
b 
x 2 

Taking the second derivatives gives
n y y   2 n y2  2 
n x x
 2 n x2  2 
2m a
n y y 
n y y 

n x x
n x x
sin
sin
sin

 N sin
  E  N sin
 N sin

2
a
b  2m b 
a
b 
a
b 


2
Factoring this expression gives
2
n y y 
n x x
 2 2  n x2 n y 
n y y 

n x

sin
 2  N sin
 E  N sin x sin


2
2m  a
a
b 
a
b 
b 

Finally, multiplying by 2 gives this result
2
n y y 

n x x
h 2  n x2 n y 
8m  a
2

n y y 
n x
sin
N sin
 E  N sin x sin



a
b 
a
b 
b 

2
By equating coefficients, one obtains the form
E
2
h 2  n x2 n y 


8m  a 2 b 2 
b. Counting nodes in each direction allows one to the determine the quantum
numbers
(a) nx = 1; ny = 1
(b) nx = 2; ny = 3
(c) nx = 3; ny = 1
(d) nx = 2; ny = 2
(e) nx = 1; ny = 5
(f) nx = 2; ny = 1
Problem 14.22
The first thing one needs to do is consider the “length” of the box. The drawing
shows the “box”.
The box should be considered to be (135 + 154 + 135) pm = 424 pm long.
The energy difference between the state with n =1 and n = 2 is

h2

2 2  12
2
8ma
E

3(6.6260693  10 34 J s ) 2

8(9.1093826  10 31 kg )(424  10 12 m) 2
 1.00536  10 18 J
This wavelength corresponding to a photon with energy equivalent to this energy
difference is
 
hc
E

(6.6260693  10 34 J s )(2.99792458  10 8 m s 1 )
1.00536  10 18 J
 1.9759  10 7 m  197.59 nm
The calculated wavelength is shorter than the observed wavelength of 290 nm.
Thus, the calculated energy difference is larger than the experimentally
determined energy separation. An examination of the equation for the energy
shows that, to make the calculated energy smaller (i.e. closer to agreeing with
the observed value) one should make the apparent box length larger than that
determined by adding the lengths together.
Problem 14.27
The ground state wave function of the harmonic oscillator has the form
2
0 ( x)  A0 e x / 2
where A0  ( /  )1 / 4 is the normalization coefficient. Schroedinger’s equation
for the harmonic oscillator in this lowest excited state is
Hˆ 0  E 0 0
 0
0
2
 2 d 2
 
2
 2m dx

k 2
x 0
2 
 2 d 2
 
2
 2m dx

2
k 2
x  A0 e x / 2
2 
 2 d 2 e x

2m dx 2

2
/2
 2 d  xe x

2m
dx


2
/2


k 2 x 2 / 2
x e
2


2
 0
A0 e x / 2
2
 0 x 2 / 2
e
2
k 2 x 2 / 2
x e
2

2
2
2
 e x / 2   2 x 2 e x / 2
2m
 2 x 2 / 2
e
2m

 k  2 2
  
2m
2



 0 x 2 / 2
e
2
k 2 x 2 / 2
x e
2
 2 x 2 / 2
 x e



 0 x 2 / 2
e
2
 0 x 2 / 2
e
2
Now, assume that the coefficient of the second term on the left side is zero:
 2 2
km
 
k 
or
m
2
The first equation reduces to:
 0
 0
 2
 2 km
or


2
2m
2
2m 
2
Reducing the left-hand side, one has
 0
 k
k

or
 0
m
2 m
2
But this last equation is just the definition of the fundamental frequency in terms
of the force constant and mass. So, this has shown that substitution of the wave
function into Schroedinger’s equation gives something that we know is true.
Thus, assumption that this wave function satisfies the equation is true.
Problem 14.32
Only do the n = 0 and n = 1 states.
For n = 0, the wave function is 0
 
  
 
1/ 4
e x
2
/2
. The expectation value of x2
is found by the integral:

 
 x    0 x 0 dx   
 

2
1 / 2 
2
 
 2 
 
1 / 2 
e
x 2
0
e
x 2 / 2
2
x e
x 2 / 2

 
x dx  2 
 
1/ 2
2
 
dx   
 


1
4
1 / 2 
e
x 2
x 2 dx

1
2

1/ 4
2
 4 3 
 xe x / 2 . Substitution into the
For n = 1, the wave function is: `  
  
formula gives an equation for the expectation value

 4 3 

 x    1 x 1 dx  
  

2
2
 4 3 

 2
  
1 / 2 
e
0
x 2
1 / 2 
 xe
x 2 / 2
x xe
x 2 / 2

 4 3 

x dx  2
  
4
2
1/ 2
3
8
2


 4 3 

dx  
  

3
2
1 / 2 
e

x 2
x 4 dx
Problem 14.36
In order to calculate the vibrational frequency, one has to extract the reduced
mass of the HCl molecule. In problem 14.35, the bond length of the HCl
molecule is given as 127 pm. By the definition of the moment of inertia, the
reduced mass is:
 
I
re2

2.644  10 47 kg m 2
1.27  10
10
m

2
 1.64  10 27 kg
The transition of lowest energy goes from the state with n = 0 to n = 1. This
transition has an energy h   .
The frequency can be calculated from the force constant and the reduced mass
by the formula
 
k


516 N m 1
1.64  10 27 kg

3.146  10 29 s  2
 5.61  1014 rad s 1
The equivalent circular frequency is



2

5.61  1014 rad s 1
2 rad s 1 Hz 1
 8.93  1013 Hz
This number is expressed in another way as 89.3 THz (terahertz). The infrared
region of the spectrum runs from about 1 THz to 430 THz. So, this transition
would be observed in the INFRARED region of the spectrum.
Energies are expressed in different units in various areas of the spectrum. In
infrared spectra, the energy is often expressed in cm-1. One can easily convert
the frequency to this unit. The result is 2979 cm-1.
Theoretically one could express the energy of transition in joules by converting.
Doing that gives the energy of transition as 5.9210-19 joule – a very small energy
indeed. Of course, this is the energy per molecule. Considering a mole of
molecules, the energy would be 35.6 kJ mole-1.
Problem 15.1
(a) Assuming the hydrogen atom to be a sphere and its nucleus to be a sphere,
one needs the formula for the volume of a sphere:

Vsphere
4 3
r
3
The volume of the nucleus is

Vnucleus
4
 (1.2  10 15 m) 3
3
 7.238  10  45 m 3
The volume of the atom is

Vatom
4
 (25  10 12 m) 3
3
 6.545  10 32 m 3
Taking the ratio of these numbers gives the fraction of the volume occupied by
the nucleus.
f

Vnucleus
Vatom

7.238  10 45 m 3
6.545  10 32 m 3
 1.106  10 13
This is a tiny fraction of the atomic volume that is occupied by the nucleus.
(b) To find the fraction of the mass that is in the nucleus, one needs the masses
of the proton and the electron. These are found in your Handbook:
Particle
Proton
Electron
Mass ( in kilograms)
1.6726217110-27
9.109382610-31
The fractional mass in the nucleus is
fm

mp
m p  me

1.6762171  10 27 kg
1.6762171  10  27 kg  9.1093826  10 31 kg
Virtually all of the mass resides in the nucleus.
 0.9995
Problem 15.2
The equation is

2
2me
 1   2  
 
 
e2
1
r


  E
sin





 r 2 r
2
  4 0 r
 r  r sin   

Substituting the function into this equation gives the following equation

2
2me
 1   2 e  r / a0
 r
 2

r
r
r



 
e  r / a0
1
  2
 sin 

 r sin   

e2
 
e  r / a0
 4 0 r
 Ee  r / a0
where the normalization constant has been factored out and cancelled out in each term.
The next thing is to take the derivatives. Let’s do this is succession:

2
2me
2
 1   r 2  r / a0 


1
  2
  e
sin  0  e e r / a0
 2
 r sin  
 r r  a 0
 4 0 r
 Ee  r / a0
Now the next derivative

2
2me
1
 2
 r
 2 r  r / a0 r 2  r / a0  
e2
 e


e

e  r / a0

2
 a

r
4

a
0
0
 0
 
 Ee  r / a0
Next, factor out the exponentials and divide them out to give an equation of the form

2
2me
1
 2
 r
 2r r 2
 
 2
 a0 a0

e2


 4 0 r



 E
This can be reduced by multiplication to
e2
2
2


me a0 r 2me a 02 4 0 r
 2
e2


 me a 0 4 0
1

r

 E
2
2me a 02
 E
This equation depends on the distance, so the left side is not a constant unless the
coefficient of the first tem is zero. Let’s substitute in this coefficient the definition of the
Bohr radius.
  2 me e 2
e2

 m  h 2  4
0
 e 0
1

r

 e2
e2


 4 0 4 0

1

r

2
2me a 02
2
2me a 02
or
E
 
2
2me a 02
 
2
2me a 0
 E
 E
 me e 2

2
  0h

1
  
4 0

 e2

 2a 0



An examination of this equation shows that it is consistent with the definition of the
energy if one chooses the quantum number, n, to be equal to 1. So, this is the groundstate hydrogenic wave function.
Problem 15.9
(a) Start with the definition of the integral, followed by integration by parts. To do this,
one needs the differential of a function shown below.
du  e  r /  dr

u   e  r / 
v  r2

dv  2rdr
Using these, one may show that
2 r / 
2 r / 
   e  r /  2r dr   r 2 e  r /   2  re  r /  dr
 r e dr   r e


Once again, one applies the integration by parts to the second term, this time with the
following definitions
du  e  r /  dr

u   e  r / 
v  r

dv  dr
Substitution into the second term gives the following equation
2 r / 
2 r / 
 2  re  r /  dr
 r e dr   r e
  r 2 e  r / 

 2   re  r /     e  r /  dr


  r 2 e  r / 
 2 2 re  r / 
 2 2  e r /  dr
  r 2 e  r / 
 2 2 re  r / 
 2 3 e  r / 






  r 2  2 2 r  2 3 e  r / 
This is the result that the question asks to be demonstrated.
(b) First form the probability of having the particle in this region:
P (0, r ) 
r 
2
0 0
0
2
*
2
  sin d  d100 100 r dr
2
 1  r
r / a0 2 2 2

 
sin
d
d
e
r dr






3/ 2 
a

0
0

 00
2
r
 1 
2
  3   sin d  d  e  2 r / a0 r 2 dr
 a 0  0
0
0
 4  r 2 r / a0 22
 4 r
  3   e
r dr   3   e  2 r / a0 r 2 dr
 a 0  0
 a0  0
The integral is the one solved in part (a), provided one identifies 
this substitution gives the result


2
2
 4  a 0 r 2
 a0 
 a 0    2 r / a0
P (0, r )    3 
 2  r  2  e
 2
 2  
 a 0  2
 2 r 2 2 r   2 r / a0
 1   2 
 1e
a
a
0
 0

r
0
 a 0 / 2 . Making
(c) To evaluate the probability at these various radii, one must plug into the equation.
 20.1a 0 2 2 * 0.1a 0

P (0,0.1a 0 )  1  

 1e  2*0.1*a0 / a0
2
a0
a0


 1  0.02  0.2  1e 0.2
 0.001148

 2a 0 2 2 * a0
P (0, a 0 )  1  

 1e  2 a0 / a0
2
a0

 a0
 1  2  2  1e  2
 0.323
  2*4 a0 / a0
 24a 0 2 2 * 4a 0
e
P (0, a 0 )  1  
1


2

a
a
0
0


8
 1  8  8  1e
 0.986
The answers in the back of your book seem to be incorrect. The above answers are more
logical because as the portion of space one considers becomes larger, the probability
should also become larger (reaching 1 when the radius tends to infinity).
Problem 15.23
The average value on any quality, V, is given by the following integral.
  2
V  
    Vˆd sin d r
*
2
dr
0 0 0
Substituting the ground-state wave function gives the following integral to
evaluate:
  2
1
e2
1
 r / a0
V    
e
e  r / a0 d sin d r 2 dr
3
3
4 0 r a 0
a 0
0 0 0

2


1 e2
e2
 2 r / a0



sin
d
d
e
rdr

e  2 r / a0 rdr
3
3




a 0 4 0 0
a 0  0 0
0
0
Let x = 2r/a0. The integral can then be converted to an integral over x.
2
e 2  a0 
V  
   e  x xdx
3
a 0  0  2  0
This integral is given in the integral table in your Handbook. Substitution gives
2
e 2  a0 
1 e2
V  
  (1) 
4 0 a 0
a 03 0  2 

Problem 15.27
The shortest wave length occurs when the function on the right is its largest
value. That largest value happens when 1/n2 is the smallest. That will be when
n = , in which case 1/n2 = 0 and
RH
1
 1 
 RH  2  

4
2 
That wavelength will be
4
 
RH
The Rydberg constant is given in your Hanbook as 10973731.568525 m-1.
Hence,
 
4
10973731.56525 m 1
 3.6450682  10 7 m  364.50682 nm
Problem 16.4
Consider the molecular-orbital occupancy of these two molecules in their ground
states. For He2, there are 4 electrons. These are put into molecular orbitals to
create the configuration:
(1 g ) 2 (1 u* ) 2
The net bond order of this molecule is:
BO 
1
2  2  0
2
There is no net bonding in this molecule.
Consider He2+. It has 3 electrons. These are put into molecular orbitals by the
Aufbau principle to create the configuration:
(1 g ) 2 (1 u* )1
The net bond order of this molecule is:
BO 
1
2  1 
2
1
2
So, the ion has a slight bonding structure. Hence, it would be considered to be
slightly stable, whereas the uncharged diatomic would be unstable.
Problem 16.6
The three diatomic molecules have different numbers of electrons that must be
put into molecular orbitals to form a configuration:
Molecule
Number of
electrons
O2
16
O2
17
O 2+
15
One can create a configuration for each of these:
O2 :
(1 g ) 2 (1 u* ) 2 (2 g ) 2 (2 u* ) 2 (1 u ) 4 (3 g ) 2 (1 g* ) 2
BO 
1
(10  6)  2
2
O2 :
(1 g ) 2 (1 u* ) 2 (2 g ) 2 (2 u* ) 2 (1 u ) 4 (3 g ) 2 (1 g* ) 3
BO 
1
(10  7)  1.5
2
O2 :
(1 g ) 2 (1 u* ) 2 (2 g ) 2 (2 u* ) 2 (1 u ) 4 (3 g ) 2 (1 g* )1
BO 
1
(10  5)  2.5
2
All of these have spin, which means that they all have unpaired spins in the
ground state. The O2 molecule is the classic example of where quantum
mechanics predicts this behavior that shows up in measurements.
Problem 16.8
One creates the configuration by filling molecular orbitals in order of energy to
the number of electrons:
N 2 : 13 e 
(1 g ) 2 (1 u* ) 2 (2 g ) 2 (2 u* ) 2 (1 u ) 4 (3 g )1
N 2 :15e 
(1 g ) 2 (1 u* ) 2 (2 g ) 2 (2 u* ) 2 (1 u ) 4 (3 g ) 2 (1 g* )1
O2 :15e 
(1 g ) 2 (1 u* ) 2 (2 g ) 2 (2 u* ) 2 (1 u ) 4 (3 g ) 2 (1 g* )1
O2 :17e 
(1 g ) 2 (1 u* ) 2 (2 g ) 2 (2 u* ) 2 (1 u ) 4 (3 g ) 2 (1 g* ) 3
Problem 16.11
In each case consider the configuration of the two states. The bond order will
help one decide on the bond length.
(a) Li2
Li 2 : 6 e 
(1 g ) 2 (1 u* ) 2 (2 g ) 2
BO 
1
4  2  1
2
1
3  3  0
2
The bond length will increase when the molecule is excited. In fact, it is
predicted to dissociate.
Excited : 6e 
(1 g ) 2 (1 u* ) 2 (2 g )1 (2 u* )1
BO 
(b) N2
N 2 : 14 e 
Excited :14e 
(1 g ) 2 (1 u* ) 2 (2 g ) 2 (2 u* ) 2 (1 u ) 4 (3 g ) 2
BO 
(1 g ) 2 (1 u* ) 2 (2 g ) 2 (2 u* ) 2 (1 u ) 4 (3 g )1 (1 g* )1 BO 
1
10  4  3
2
1
9  5  2
2
One would expect the bond length to increase in the excited state.
(c) Be2
Be2 : 8 e 
(1 g ) 2 (1 u* ) 2 (2 g ) 2 (2 u* ) 2
BO 
1
4  4  0
2
1
5  3  1
2
In this situation, one expects the low-energy state to be unstable, whereas the
excited configuration seems to be stable.
Excited :8e 
(1 g ) 2 (1 u* ) 2 (2 g ) 2 (2 u* )1 (1 u )1
BO 