# Chemistry exam

Instructions: You may use a pencil or pen. You will not need your calculator. No other materials are allowed. Write all answers on the exam in the spaces provided, or clearly continued on the backs of the pages and labeled unambiguously. When you are asked to explain something, do so in complete English sentences, with equations where appropriate.

There are three pages of equations distributed with the exam. Be sure to look at them before you begin the exam, since you may be able to make use of them to shorten the work of some of the problems.

 Problem Score 1 / 10 2. / 15 3. / 20 4. / 14 5. / 06 6. / 15 7. / 20 8. / 08 Total /108

## 1.    (10 points) Commutators of quantum mechanical operators

The x component of the angular momentum operator can be written in terms of the

position and momentum operators as

Lˆx  = y pˆz  z pˆy .

## Evaluate the commutator

éëLˆx , Lˆz ùû

You can use the basic commutators:

[x, pˆx ] = i, (same for y and z) and [y, pˆx ] = 0 (and

the obvious similar commutators that equal zero).

1. (15 points, first order perturbation theory) The motion of a particle of mass    is described by the Hamiltonian

 +
 2
 2

Hˆ    =-    d      k x2

H (1)

(x)

 2

2m dx     2

where

H (x) = ì cx    x ³ 0

 î

1                í-cx   x < 0          x

(dashed line in figure)

## Treating H (1)(x)as a perturbation, calculate the ground state energy of this system.

æ km ö1/2

 2

w =                a = ç     ÷

# è  ø

and the coefficient c.

Hinst: (1) Formulas for the harmonic oscillator and some useful integrals are given on the equation sheets distributed with this exam. The integral you need is among the “Gaussian integrals.” (2) Remember to add the unperturbed energy to the perturbation correction to get the complete approximation to the ground state energy.

workspace for problem 2

1. (20 points, Linear variational method, Stark effect) The Hamiltonian for the rigid rotor molecule with a dipole moment in an electric field of strength                                                   is

The angle q is the angle the rotor makes with the field, which is pointing along the z axis.

In presence of an electric field, two spherical harmonics, Y1,1 ( J = 1, m = 1 ) and Y2,1

( J = 2, m = 1) will be mixed to produce two new energy eigenstates.

Use the variational trial functionf c1Y11 (q ,j ) + c2Y21 (q ,j ) to estimate the energy of the two states.

The following integrals will be useful.

2p p

ò òY1,1 (q ,j ) cosq Y1,1 (q ,j )sinq dq dj = 0

0  0

2p p

ò òY2,1 (q ,j ) cosq Y2,1 (q ,j )sinq dq dj = 0

0  0

2p p

ò òY1,1 (q ,j ) cosq Y2,1 (q ,j )sinq dq dj =

0 0

workspace for problem 3

## 4.    (14 points: Atomic Orbitals of Hydrogenic Species)

Consider Li2+ ion, which is a hydrogenic (hydrogen-like) species.

• (10 pts) Calculate the average distance of the electron from the nucleus if the electron is in the 1s orbital. Express your answer in terms of the Bohr radius (a0).

• (4 pts) Express the distance where the radial node of the 2s orbital occurs, in terms of a0.

1. (6 points, Slater Determinant) Consider the excited state wave function for He atom given by the following Slater

F   (1, 2) = 1

y 2,1,1a (1)

y 3,2,-2a (1)

He                           2 y 2,1,1a (2) y 3,2,-2a (2)

Here y 2,1,1 and y 3,2,2 are hydrogenic wave functions (with 𝑍 = 2, see the equation sheet).

Show that FHe (1, 2) is an eigenfunction of

Lˆz  = Lˆz ,1  + Lˆz ,2 . What is the eigenvalue?

Lˆz ,1,  Lˆz ,2 ,  and Lˆz

are the z-components of the orbital angular momentum operators for

electrons 1 and 2, and the z-component of the total orbital angular momentum operator, respectively.

## 6.    (15 points) Atomic Term symbols

What atomic term symbols arise from the excited state configuration 1s13d1 for the lithium atom? Include the J quantum number subscripts in your list of all the term symbols.

Identify the lowest energy state and its degeneracy.

## Microstate Table

 M S 1 0 -1 M L 2 (2+ , 0+ ) (2+ , 0– ),(2– , 0+ ) (2– , 0– ) 1 (1+ , 0+ ) (1+ , 0– ), (1– , 0+ ) (1– , 0– ) 0 (0+ , 0+ ) (0+ , 0– ),(0– , 0+ ) (0– , 0– ) -1 (-1+ , 0+ ) (-1+ , 0– ),(-1– , 0+ ) (-1– , 0– ) -2 (-2+ , 0+ ) (-2+ , 0– ),(-2– , 0+ ) (-2–, 0– )

1. (20 points) Molecular Orbitals of H2+

Consider an H2+ ion. In the figure below, HA, HB, and e represent the two nuclei and the electron, respectively, and all the relevant distances are defined. You must use atomic units for this problem.

• (3 points) Write down the Schrödinger equation for the electrons and nuclei in the H2+ ion, specifying all the terms in the

Hamiltonian.

• (3 points) Write down the Born-Oppenheimer Hamiltonian for the electrons only, with the internuclear distance fixed, for the H2+ Explain the physical basis ofthe Born-Oppenheimer approximation.

The nuclei are much heavier than an electron, therefore the motion of nuclei is

much slower.

• (4 points) The lowest energy molecular orbital is expressed in terms of the linear combination of two 1s atomic orbitals centered on the two nuclear positions as:

y +  = N+ (y1sA +y1sB )

Determine the normalization constant N+ . Show the work and express the answer in terms of S .

S y1sA y1sB

= òall space dry1sA (r)y1sB (r)

(Problem continues in the next page)

• (10 points) Find an expression for the total energy of the ground state H2+ in termsof S and the following integrals. Show the work. Explain what each term in your result

1sA

1s

1sA

1s

= òall space

=

dry

dry

1sA

(r) 1 y

rB

 r
• 1 y

1sA

(r)

(r)

A                   B                      òall space

1sA

1sB

A

## 8 (8 points) Molecular Orbitals

Consider the ground state electron configuration of the N2- diatomic ion.

(1s

)2 (1s

)2 (2s

)2 (2s

)2 (3s

)2 (1p

)4 (1p  )1

g                 u                   g                   u                  g                 u                   g

• (3 points) List all the occupied bonding

1s g , 2s g , 3s g ,1pu

• (2 points) Calculate the bond order of N2-.

• (3 points) Sketch the shape of 1p g

only one.

orbital. If multiple orbitals are degenerate, draw

## Some Possibly Useful Equations

 ¶   +
 2
 2
 2

Operators

1D:

pˆx  = –i

KE =- 

Hˆ =- 

2

2       V (x)

x

3D:

2m x

2m x

3D volume element:

Uncertainty products: Da Db ³

éAˆ, Bˆù

where both operators are Hermitian.

ë       û

=     Y(t) éHˆ, Aˆù Y(t)   when operator Aˆ

itself does not depend on time

dt                            ë       û

## Constants and Units

 h

E = 27.211 eV = 4.3597 ´1018 J                                                    mass of electron = 9.109 ´1031kg

 0

h = 6.626 ´ 1034 J sec                                                                a = 5.29177 ´1011m = 0.529177 Å

charge of electron e = -1.602 ´1019 C, or in Gaussian units e = -4.803´1010 esu

atomic units:  = me

= a0

= e /

= 1, atomic unit of time = 24.189 ´1018 s

fine structure constant: a =     

me c a0

=      1

# 137.036

4pe2                         2

Bohr radius = a0  = 0     or in Gaussian units a0 =

 e
 e

m e2                           m e2

y (               æ npx ö

n2p 22

n2h2

## Particle in a 1D box:

n x) =

2

sinè  L

d2     k

ø                        En =

2

2mL2 = 8mL2

m m

Harmonic Oscillator:  H =-         +                                                             x

m =  1  2

 ( x) = p
 y

æ a ö1/ 4

0                  ç    ÷

è    ø

ea x2 /2

2m dx 2     2

m1

æ      1 ö

• m2

 3

y ( x) = æ 4a

ö1/ 4

xea x2 /2

E = çn +

 n

è

÷w

2 ø

1                  ç p ÷

è        ø

1/ 4

with                w =

y ( x) = æ a ö

(2a x2  -1)ea x2 / 2

2                  ç 4p ÷

æ km ö1/2

è      ø        a = ç        ÷

3 1/ 4

è 2 ø

 ç 9p ÷
 3

y ( x) = æ a  ö

è      ø

(2a x3  – 3x)ea x2 / 2

## Hydrogen and other one-electron atoms

Z 2 æ    e2   ö        Z 2

En =-   2 ç

# ÷ =-

2 Eh

2è a0 4pe0 ø      2n

## Angular momentum operators, Spherical Harmonics and Spin angular momentum

 è        ø

Y    = æ 5

ö1/ 2

(3cos2 J -1)

2,0

Y2,±1

ç 16p ÷

æ 15 ö1/ 2

 8p

=  ç     ÷

è      ø

sinJcosJe±ij

Lˆx

Lˆy

Lˆ

= ypˆz

= zpˆ x

= xpˆ

• zpˆy
• xpˆz
• ypˆ

 8p

Y     =  æ

3 ö1/ 2

sin2 J e±2ij

z               y               x

2,±2             ç      ÷

è      ø

 1

Hˆ rigid rotor  =

2I

Lˆ2

Sˆ 2a = (3/ 4) 2a

Sˆ 2b = (3/ 4) 2b

Sˆza = (1/2) a

Sˆzb = -(1/2) b

## Hydrogenic Radial Functions, Rn,l (r)                                                              (Z = atomic number)

æ Z ö 3/ 2

• Zr

1 æ Z

ö 3/ 2 Zr

• Z r

(r) = 2 ç    ÷

e a0

R  (r) =       ç       ÷

e 2 a0

1,0

è a0 ø

2,1

3 è 2 a0 ø      a0

æ Z ö 3/ 2 æ

Zr ö – Zr

4 2 æ Z ö 3/ 2 Zr æ

Zr ö – Z r

R   (r) = 2ç      ÷     ç1-

÷ e 2a0

R  (r) =         ç       ÷

ç1-

÷ e 3 a0

2,0

è 2 a0 ø    è

2 a0 ø

3,1

9   è 3a0 ø

a0 è

6 a0 ø

æ  ö 3/ 2 æ      2Zr

2Z 2 r2 ö – Zr

2 2 æ Z ö 3/ 2 æ Z rö 2

• Z r

R   (r) = 2ç      ÷     ç1-        +

÷ e 3a0

R (r) =

ç       ÷     ç

÷ e 3 a0

 0

3,0

è 3 a0 ø    è

3a0

27 a2 ø

3,2

27 5 è 3a0 ø

è a0 ø

sin2æ a ö = 1 cosa

## Trigonometric identities

cos2æa ö = 1+ cosa

# ç   ÷                                      ç   ÷

è 2 ø         2                           è 2 ø               2

sin x sin y 1 cos(x – y) – 1 cos(x y)

2 2

cos x cos y = 1 cos(x y) + 1 cos(x + y)

2 2

sin x cos y 1 sin(x – y) + 1 sin(x y)

e± ix

= cos x ± i sin x                sin x =

2

eix – eix

2i

2

cos x =

eix eix

2

## Integrals

ò sin2(ax)dx = x  1 sin(2ax)

## Indefinite Integrals

ò sin(ax) cos(ax)dx = 1 sin2(ax)

2    4a

 ò

x     1

2a

 ò              å

m               r        mxm – r

cos2(ax)dx =    +

# 2

sin(2ax)          xmeaxdx eax    (-1)

4 a                                           r = 0

(m r)!ar+1

ò sin2

a

(Cxdx =

## Definite integrals

2(b a)C + sin(2aC) – sin(2bC) 4C

L      æ np x ö     æ mp x ö

L       æ np x ö      æ mp x ö        L

ò sinç        ÷sinç        ÷dx = ò cosç                 ÷cosç

÷dx =

dn,m

0      è  L  ø     è L ø

0       è L ø

è L ø 2

L      æ np x ö        æ mp x ö          L2        2((-1)m+n -1) mnL2

ò sin ç

L ÷ x sin ç

L ÷ dx =

if m = n, or

4

(m2n2 )2p 2

if m ¹ n

0      è        ø        è         ø

L      æ np x ö          æ mp x ö

L3 æ         3    ö

4(-1)m+n mnL3

ò sin ç        ÷ x2 sin ç         ÷ dx =

ç 2 – 2 2 ÷ if m = n, or

2           2 2 2

if m ¹ n

0      è   L  ø          è   L ø

12 è       n p ø

(m n ) p

L      æ np x ö d       æ mp x ö

2mn

ò sinç        ÷     sinç         ÷dx = 2

if n m is odd, or 0 if n m is even

0      è   L  ø dx      è L ø

n m

simple exponential integrals

¥

ò xneaxdx =

0

n! an+1

Some Gaussian integrals

¥

ò xeaxdx = 0

 ¥

xeaxdx =  1

¥

ò eaxdx =

 ¥
 2 a

x 2ne– ax 2 dx 1· 3· 5…(2n 1)

 0

ò            2a

¥

ò                      n n

 ò                     n +1 n

¥

 ò                  n +1

x 2n +1eax 2 dx = n!

0                                            2a

x 2ne– ax 2 dx 1· 3· 5…(2n 1)

0                                                      2    a

mistry exam