University of Washington Department of Chemistry Chemistry ...

University of Washington

Department of Chemistry

Chemistry 453

Winter Quarter 2010

Homework Assignment 3; New Due Date 01/25/10

1) The exact ground state energy for a particle in a box is E1 =

h2

. And the exact

8mL2

2

¦Ðx

. Consider an approximate particle in the box wave

sin

L

L

function: ¦× 1 ( x ) = x ( L ? x ) .

wave function is ¦× 1 ( x ) =

a) Show that the approximate wave function satisfies the boundary conditions.

Solution: ¦× 1 ( 0 ) = 0 ( L ? 0 ) = 0 and ¦× 1 ( L ) = x ( L ? L ) = 0

b) There is a mathematical method called the Variational Method, that allows calculation

of the energy using an approximate wave function, assuming the wave function fits the

boundary conditions. The equation for the energy is:

L

E=

¡Ò¦× ( x ) H?¦× ( x )dx

0

L

¡Ò¦× ( x )dx

2

0

where H? is the Hamiltonian operator for the particle in the box. Calculate the energy

using the approximate wave function.

Solution:

L

E=

¡Ò¦× ( x ) H?¦× ( x )dx

0

L

¡Ò¦×

2

( x )dx

L

=

=2

d2

? xL ? x 2 ??dx

x

L

x

?

?

(

)

2 ?

¡Ò

dx

2m 0

L

¡Ò ( xL ? x ) dx

0

2 2

0

= 2 ? L3 ?

m?6?

= L 0

=L

= 5 ?5 ? 5 =

L L L

2 2

?

xL

x

(

) dx ¡Ò ( x 2 L2 ? 2 Lx 3 + x 4 )dx 3 ? 2 + 5

¡Ò0

0

2 L

=

m

¡Ò ( xL ? x )dx

2

= ? L L3 ?

?

m ?? 2 3 ??

2

3

= 2 ? L3 ?

m? 6?

=2 5

h2

h2

0.127

= 5 ?5 ? 5 =

=

¡Á

=

L L L

m L2

mL 7.89mL2

? +

3 2 5

c)According to the Variational Method, the more the approximate wave function

resembles the real wave function, the closer the energy matches the real energy. Graph

the approximate wave function and the real particle-in-the-box wave function. How

different are the energies? Calculate the percent difference.

For the purpose of doing the graph let L=1. Then the real wave function for n=1

is¦× 1,real ( x ) = 2 sin ¦Ð x and the trial wave function is ¦× 1,trial ( x ) = x (1 ? x )

The graph is:

1.6

1.4

Amplitude

1.2

1

0.8

x(1-x)

sinpi*x

0.6

0.4

0.2

0

-0.2 0

0.5

1

1.5

L

Here I have normalized the wave functions to be equal in amplitude at the maximum

x=L/2. This is a good idea because the trial wave function is not normalized. The

normalization is necessary because otherwise the trial wave function is not valid. When

this is done the plot above results and we find the wave functions have a relative

difference of 4.1%, approximately. Not a bad guess.

2) Generally the quantization of translational motion is not significant for atoms because

of their mass. However, this conclusion depends upon the size of the space to which the

atoms are confined. Zeolites are structures with small cubic pores with edge length 1 nm.

Calculate the energy of a H2 molecule confined to a zeolite pore, assuming nx=ny=nz=10.

Compare this energy to kBT assuming T=300K. Are quantization effects important in this

case?

Solution:

6.63 ¡Á 10?34 Js ) ( 6.02 ¡Á 1023 mol-1 ) 2

(

h2

2

2

2

E=

10 + 102 + 102 )

( nx + n y + nz ) =

2 (

?9

-1

8ML2

8 ( 0.002kg mol )(1.00 ¡Á 10 m )

2

( 44.0 ¡Á10 J s )( 6.02 ¡Á10

( 0.016kg mol )(1.00 ¡Á10

?68

=

2 2

-1

23

mol-1 )

?18

m

2

)

( 300 ) = 4.97 ¡Á106?68+ 23+18 J = 4.97 ¡Á10?21 J

Now k BT = (1.38 ¡Á 10?23 JK ?1 ) ( 300 K ) = 4.14 ¡Á10?21 J

3) Carbon monoxide absorbs radiation at a wave number ¦Í = 2144cm ?1 because of its

internal vibration.

a) Calculate the vibrational energy level spacing for CO. Assume the vibration of

CO can be modeled as a harmonic oscillation.

Solution: Calculate the reduced mass for 12C16O in kilograms.

.012 ¡Á .016

1

= 1.14 ¡Á 10?26 kg

.012 + .016 6.04 ¡Á 1023

Calculate the oscillator frequency for 12C16O.

?=

¦Í=

1

2¦Ð

1.902 ¡Á 103 Nm ?1

= 6.5 ¡Á 1013 s ?1

?26

1.14 ¡Á 10 kg

?E = h¦Í = 6.63 ¡Á 10?34 Js ¡Á 6.5 ¡Á 1013 s ?1 = 4.31 ¡Á 10?20 J

b) Calculate kBT for T=300K. Based on this value and your answer in part a, will

CO vibrations be thermally excited at T=300K? Explain.

k BT = 1.38 ¡Á 10?23 JK ?1 ¡Á 300 K = 4.14 ¡Á 10?21 J . k BT  ?E so no.

c) At what temperature will thermal excitation of CO vibrations become

important?

?E

4.31 ¡Á 10?20 J

Solution: k BT  ?E ? T 

=

= 3.12 ¡Á 103 K

?23

?1

k B 1.38 ¡Á 10 JK

d) Calculate the vibrational heat capacity of 1 mole of CO at the temperature

calculated in part c. Assume first the expression for the vibrational heat

capacity based on the equipartition theorem. Compare this result to the

2

h¦Í

N AkB

e ? h¦Í / k B T

kBT

quantum expression for the heat capacity CV =

2

1 ? e ? h¦Í / k B T

c

FG IJ

H K

h

Solution: Based on the equipartition theorem, for one mole the vibrational heat

capacity is R=8.31JK-1mol-1. Using the quantum mechanical equation above:

h¦Í

4.31 ¡Á 10?20 J

=

= 1.00

k BT 1.38 ¡Á 10?23 JK ?1 ¡Á 3120 K

2

? h¦Í ? ? h¦Í / kBT

N Ak B ?

e

k BT ??

R ¡Á e ?1

8.31JK ?1 ¡Á 0.368 3.06 JK ?1

?

CV =

=

=

=

= 7.65 JK ?1

2

?1 2

? h¦Í / k BT 2

0.399

(1 ? 0.368)

(1 ? e )

(1 ? e

)

The two values are comparable.

4) The fundamental vibration frequency of gaseous 14N16O is 1904 cm-1.

a) Calculate the force constant using formula for a simple harmonic oscillator.

Solution: ¦Í =

?=

=

¦Í

c

? ¦Í = c¦Í = ( 3 ¡Á 1010 cm s ?1 )(1904cm ?1 ) = 5.71¡Á 1013 s ?1

mN mO

1 M N MO

=

mN + mO N A M N + M O

( 0.014kg )( 0.016kg ) = 2.24 ¡Á10?27 kg = 1.24 ¡Á10?26 kg

1

6.02 ¡Á1023 mol ?1 0.014kg + 0.016kg

0.181

¦Ê

¦Ê

. Therefore ¦Ø 2 = 4¦Ð 2¦Í 2 = and so 4¦Ð 2¦Í 2 ? = ¦Ê

?

?

Now ¦Ø = 2¦Ð¦Í =

We obtain:

¦Ê = 4¦Ð 2¦Í 2 ? = ( 6.28 ) ( 5.71¡Á 1013 s ?1 ) (1.24 ¡Á10?26 kg )

2

2

= 1595kg s ?1 = 1595 Nm ?1

b) Using the force constant in part a, calculate the fundamental vibration frequency of

gaseous 15N16O.

m m

1 M N MO

?= N O =

mN + mO N A M N + M O

=

( 0.015kg )( 0.016kg ) = 2.40 ¡Á10?27 kg = 1.28 ¡Á10?26 kg

1

6.02 ¡Á1023 mol ?1 0.015kg + 0.016kg

0.187

(

¦Í(

¦Í

15

14

N16O )

N O)

16

=

? ( 14 N16O )

? ( N O)

15

16

?¦Í

(

15

N16O ) = ¦Í

(

14

N16O )

? ( 14 N16O )

? ( N O)

15

16

= (1904cm ?1 )

1.24

= 1874cm ?1

1.28

c) When 14N16O is bound to hemoglobin the oxygen is anchored to the protein so its

effective mass is much greater (i.e. essentially infinite) compared to the mass of the

nitrogen atom. Using the force constant from part a, calculate the fundamental vibration

frequency of NO bound to hemoglobin.

0.014kgmol ?1

? ¡Ö mN =

= 2.33 ¡Á 10?26 kg

23

?1

6.02 ¡Á10 mol

¦Í =

¦Ê

1

1

1595 Nm ?1

=

= 1389cm ?1

?1

?26

10

2¦Ð c ? ( 6.28 ) 3 ¡Á 10 cms

2.33 ¡Á10 kg

1

Or¡­¦Í

(

14

N¡Þ ) = ¦Í

(

14

N O)

16

? ( 14 N16O )

mN

= (1904cm ?1 )

1.24

= 1389cm ?1

2.33

5) The vibrational frequency of 1H35Cl is 8.963x1013 s-1.

a) Calculate the bond force constant ¦Ê and the reduced mass ?.

m m

1 M H M Cl

? = H Cl =

mH + mCl N A M H + M Cl

=

( 0.001kg )( 0.0355kg ) = 3.55 ¡Á10?28 kg = 1.61¡Á10?27 kg

1

6.02 ¡Á1023 mol ?1 0.001kg + 0.0355kg

.220

As in Problem 4a¡­

¦Ê = 4¦Ð 2¦Í 2 ? = ( 6.28 ) ( 8.963x1013 s ?1 ) (1.61¡Á 10?27 kg )

2

2

= 510kg s ?1 = 510 Nm ?1

b) Assume 1H35Cl is in the n=0 quantized vibrational state. Calculate the average of the

square of the vibrational amplitude about the equilibrium position x 2

x

¦Á=

q2

?¦Ø

¡à x2

¡Þ

¡Þ

¡Þ

2

2

1

? 1 ? ? q2

=

= ¡Ò ¦× 0 ( q ) q ¦× 0 ( q ) dq = ¡Ò q ?

e dq =

q2e? q =

?

¡Ò

2¦Á

¦Á

¦Á ?¡Þ

¦Á ?¡Þ ? ¦Ð ?

¦Á ¦Ð0

2

=

1

2

1

2

( 2¦Ð ) (8.963 ¡Á1013 s ?1 )(1.61¡Á10?27 kg )

= 8.55 ¡Á 1021 m ?2

1.06 ¡Á10?34 Js

1

1

=

=

= 5.85 ¡Á 10?23 m 2

2¦Á ( 2 ) ( 8.55 ¡Á1021 m ?2 )

=

c) Use you results from parts a and b to calculate the average potential energy V ( x ) .

¦Ê

510 Nm ?1

5.85 ¡Á 10?23 m 2 ) = 1.49 ¡Á 10?20 J

V =

x =

(

2

2

d) Starting with the definition for the momentum operator for a one dimensional

d

harmonic oscillator p? = ?i= , calculate the average squared momentum p 2 for

dx

1 35

H Cl in the n=0 quantized vibrational state. Use this result and the results of part a to

calculate the average kinetic energy K for 1H35Cl in the n=0 quantized vibrational

state.

2

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