CHM320 FINAL EXAM USEFUL INFORMATION

Constants

mass of electron: me = 9.11× 10−31 kg.

Rydberg constant: RH = 109737.35 cm−1 = 2.1798× 10−18 J.

speed of light: c = 3.00× 108 m/s

Planck constant: 6.626× 10−34 Js

κ = 1

4πǫ0= 8.99× 109 kgm3/c2s2

charge of electron: e = 1.6× 10−19 C

Bohr radius: a0 = 52.9 pm

Avogadro’s number: NA = 6.02× 1023 mol−1

Unit Conversion

1 J = 1 kgm2/s2

1 eV = 1.6× 10−19 J.

1 J = 5.03× 1022 cm−1

1 amu = 1.67× 10−27 kg.

1 A= 1.0× 10−10 m.

1 N (newton) = 1 kg·m/s2

Formula

(1) Energy of a particle in classical mechanics: E = kinetic (T) + potential (V)

E = T + V =p2

2m+ V

where p is the linear momentum (p = mv). Force is related to the potential energy as F =−∇V . For one-dimensional system, F = −dV

dx.

1

(2) Electromagnetic wave (classical view):

E(x, t) = E0sin(kx − ωt)

where k = 2πλ

(wavenumber) and ω = 2πν (angular frequency). E0 is called the amplitude

of wave. The period of oscillation is Γ = 1

ν. λ is the wavelength and ν is the frequency of

oscillation. The frequency and wavelength of light are related through the speed of light (c) as

ν =c

λ

(3) Photoelectric Effect:

Te = Ephoton − Φ

where Te is the kinetic energy of ejected electron and Φ is the work function. The photonenergy is given by

Ephoton = hν = hc

λ= hcν

where c is the speed of light and ν = 1

λ.

(4) Rydberg formula for hydrogen atom emission spectra:

ν(cm−1) =1

λ= RH

(

1

n21

− 1

n22

)

where n2 > n1 and RH is the Rydberg constant in cm−1. Lymann, Balmer and Paschen seriesof emission lines involve n1 = 1, 2, and 3, respectively.

(5) de Broglie hypothesis:

λ =h

p

(6) Classical physics of angular motion:

1. angular momentum: l = pr.

2. kinetic energy: T = l2

2mr2

3. centripetal force: Fcent =l2

mr3

2

(7) Bohr model of hydrogen atom:

Postulate: Angular momentum is quantized, l = nh

En = −(

κmee4

2h2

)

1

n2= −RH

n2, n = 1, 2, 3, ..

where κ = 1

4πǫ0and RH is the Rydberg constant.

(8) Uncertainty Principle:

∆x∆p = h

(9) Schrodinger Equation (1-dimensional):

− h2

2m

d2ψ(x)

dx2+ V (x)ψ(x) = Eψ(x)

(10) Probability:

Probability to find the particle with wavefunction ψ(x) between x = a and x = b is given by

P =∫ b

a|ψ(x)|2dx

Normalization condition: Since the particle should be found somewhere, sum of all proba-bility must be equal to 1.

∫ ∞

−∞|ψ(x)|2dx = 1

(11) Particle in a box (1-dimensional, 0 < x < a)

ψ(x) =

√

2

asin

(

nπx

a

)

En =

(

h2

8ma2

)

n2

where a is the box length.

3

(12) Harmonic Oscillator (1-dimensional, classical)

Harmonic oscillator experiences Hooke’s law force:

F = −kx

where x is the displacement from the rest position and k is force constant. The potentialenergy corresponding to Hooke’s law force is

V (x) =1

2kx2

Newton’s equation for harmonic oscillator:

d2x(t)

dt2+k

mx(t) = 0

General solution:

x(t) = Acos(ωt + φ)

where A and φ are constants that depend on the initial condition. ω =√

km

is (angular)frequency of oscillation. Note that ω = 2πν.

(13) Vibration of diatomic molecule:

Small amplitude vibration can be modeled by harmonic oscillator. The mass of the oscillatorshould be the reduced mass of diatomic molecule.

µ =mAmB

mA +mB

where mA and mB are the masses of two atoms of diatomic molecule.

(i) Schrodinger equation of molecular vibration:

− h2

2µ

d2ψ(x)

dx2+

1

2µω2x2ψ(x) = Eψ(x)

where ω =√

kµis the frequency of vibration and k is the force constant.

(ii) Wavefunctions:

ψ0(x) =

(

β

π

)1/4

e−βx2/2, ψ1(x) =

(

4β3

π

)1/4

xe−βx2/2, ψ2(x) =

(

β

4π

)1/4

(2βx2 − 1)e−βx2/2

where β = µωh.

(iii) Energy levels:

En = hω(

n+1

2

)

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(14) Infra-red spectrum of diatomic molecule

Under the harmonic oscillator model, IR absorption frequency should be

νobs =1

2π

√

k

µ=

ω

2π

Note that

Ephoton = hνobs = hc

λobs= hcνobs

(15) Quantum mechanical operators:

(i) kinetic (T) and potential (V) energies

T = − h2

2m

d2

dx2, V = V (x)

(ii) position (x) and momentum (p)

x = x, p = −ih ddx

(16) Free particle:

For a free particle (V=0) in space with the mass m and the energy E, the unnormalized

wavefunctions are

ψ(x) = eikx, and ψ(x) = e−ikx

where k =√2mEh

, or E = h2k2

2m.

(17) Quantum mechanical average (Expectation value):

If a system is described by a normalized wavefunction, Ψ, then the average values of anobservable corresponding to the operator A is given by

〈a〉 =∫

allspaceΨ∗AΨdτ

Note that dτ is the volume element (dx for 1D, dxdydz for 3D, for example)

(18) Orthogonality of wavefunctions:

Eigenfunctions of an operator are orthogonal to each other.∫ ∞

−∞φ∗n(x)φm(x)dx = 0, (n 6= m)

where φn(x) and φm(x) are eigenfunctions of given operator.

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(19) Commutator of two operators, A and B:[

A, B]

= AB − BA

(20) Single particle in 3-dimension ((Cartesian coordinates):

(i) Hamiltonian and Schrodinger Equation

H = − h2

2m∇2 + V = − h2

2m

(

d2

dx2+

d2

dy2+

d2

dz2

)

+ V (x, y, z)

Hψ(x, y, z) = Eψ(x, y, z)

(ii) Separable Hamiltonian

H = − h2

2m∇2 + V = − h2

2m

(

d2

dx2+

d2

dy2+

d2

dz2

)

+ Vx(x) + Vy(y) + Vz(z) = hx + hy + hz

where

hx = − h2

2m

d2

dx2+ Vx(x), hy = − h2

2m

d2

dy2+ Vy(y), hz = − h2

2m

d2

dz2+ Vz(z),

(iii) Wavefunction and energy for a separable Hamiltonian

ψ(x, y, z) = φx(x)φy(y)φz(z)

where φx(x), φy(y) and φz(z) are the wavefunctions for the 1-dimensional Hamiltonian definedin (ii) with energy value Ex, Ey, and Ez, respectively. Then, the total energy is given by

E = Ex + Ey + Ez

(21) 2-dimensional rotation (classical):

(i) Angular velocity: ω = 2πν, where ν is the frequency of roation.

(ii) Linear velocity: v = rω where r is the radius of rotation.

(iii) Moment of inertia: I = mr2 where m is the mass of rotating body.

(iv) Angular momentum: L = Iω = mrv.

(v) Kinetic energy: T = 1

2mv2 = 1

2Iω2

(22) Spherical polar coordinates:

(i) From spherical polar coordinates to Cartesian coordinates

x = rsinθcosφ, y = rsinθsingφ, z = rcosθ

6

(ii) From Cartesian coordinates to spherical polar coordinates

r =√

x2 + y2 + z2, θ = cos−1

[

z√x2 + y2 + z2

]

, φ = tan−1

(

y

x

)

(iii) volume element

dxdydz = r2sinθdrdθdφ

(23) Rigid Rotor

(i) Hamiltonian in spherical polar coordinate:

H = T = − h2

2I

[

1

sinθ

∂

∂θ

(

sinθ∂

∂θ

)

+1

sin2θ

(

∂2

∂φ2

)]

where I = µr2eq is the moment of inertia of the rigid rotor with a fixed bond length req. µ isthe reduced mass of the rigid rotor.

(ii) Rotational energies:

El =h2

2Il(l + 1)

where l = 0, 1, 2, .... Note that each energy level has 2l + 1 degeneracy.

(iii) Rotational wavefunctions:

The wavefunctions of rigid rotor are spherical harmonics, which depend on two quantum num-bers, l and m.

HYl,m(θ, φ) =h2

2Il(l + 1)Yl,m(θ, φ)

(iv) Spherical harmonics:

Yl,m(θ, φ) = P|m|l (θ)eimφ

where P|m|l (θ) is called associated Legendre function.

Y0,0 =1√4π, Y1,0 =

(

3

4π

)1/2

cosθ, Y1,±1 =(

3

8π

)1/2

sinθe±iφ

Y2,0 =(

5

16π

)1/2

(3cos2θ − 1), Y2,±1 =(

15

8π

)1/2

sinθcosθe±iφ, Y2,±2 =(

15

32π

)1/2

sin2θe±2iφ

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(24) Angular momentum operator (L)

(i) L2 operator in spherical polar coordinates

L2 = −h2[

1

sinθ

∂

∂θ

(

sinθ∂

∂θ

)

+1

sin2θ

(

∂2

∂φ2

)]

(ii) Eigenfunctions of L2 are spherical harmonics (see (23)-(iv)).

L2Yl,m = h2l(l + 1)Yl,m

(iii) Quantum operator for the z-component of L

Lz = −ih ddφ

Eigenfunctions and Eigenvalues:

Lz

(

eimφ)

= mh(

eimφ)

where m = −1,−l + 1,−l + 2, ...,+l.

(iv) Angular momentum operator in Cartesian coordinates:

L = Lxi + Ly j + Lz j

Lx = yPz − zPy = −ih[

y∂

∂z− z

∂

∂y

]

Ly = zPx − xPz = −ih[

z∂

∂x− x

∂

∂z

]

Lz = xPy − yPx = −ih[

x∂

∂y− y

∂

∂x

]

(v) Commutator relations:[

L2, Lx

]

=[

L2, Ly

]

=[

L2, Lz

]

= 0

[

Lx, Ly

]

= ihLz,[

Ly, Lz

]

= ihLx,[

Lz, Lx

]

= ihLy

8

(25) Microwave spectroscopy

(i) Rotational constant:

B =h2

2I

where I = µr2eq is the moment of inertia of diatomic molecule.

(ii) Rotatonal constant in cm−1:

B =B

hc=

h

8π2cI

where c = 3.0× 1010cm/s. Other quantities should be in SI unit.

(iii) Rotational energy levels in terms of rotational constant:

EJ = BJ(J + 1)

where J = 0, 1, 2, ... is the rotational quantum number (i.e l in (23)-(ii)).

(iv) Rotatonal transition (J → J + 1):

∆Erot = 2B(J + 1)

(26) Hydrogen atom

(i) Hamiltonian operator (in spherical polar coordinates):

H = T + V = Tr +L2

2mer2− κ

e2

r

Tr = − h2

2me

[

1

r2∂

∂r

(

r2∂

∂r

)]

where r is the distance of electron from the center (nucleus) and κ = 1

4πǫ0.

(ii) Schrodinger Equation:[

Tr +L2

2mer2− κ

e2

r

]

ψ(r, θ, φ) = Eψ(r, θ, φ)

Since the Hamiltonian is separable between r and (θ, φ), the exact wavefunction will be of thefollowing form.

ψ(r, θ, φ) = R(r)Y (θ, φ)

(iii) Angular solution:

L2Y (θ, φ) = βY (θ, φ)

The solution of this equation is spherical harmonics where β = h2l(l + 1) (see (23)-(iv) and(24)-(ii))

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(iv) Radial solution:[

Tr +h2l(l + 1)

2mer2− κ

e2

r

]

R(r) = ER(r)

The solutions of this equation are associated Laguerre function, Rnl(r) where l = 0, 1, 2, ..n−1.The corresponding energies are

En = −κ2mee

4

2h2n2= −RH

n2, n = 1, 2, 3, ...

where RH is Rydberg constant. First three Rnl(r) functions:

R1,0(r) =2

a3/20

e−r/a0 , R2,0(r) =1√8a

3/20

(

2− r

a0

)

e−r/2a0 R2,1(r) =1√

24a3/20

(

r

a0

)

e−r/2a0

where a0 = h2/meκe2 = 52.9 pm is the Bohr radius.

(27) Radial probability distribution (P (r)):

P (r) = r2|Rn,l(r)|2

where P (r)dr represents the probability to find the electron between r and r + dr.

(28) Wavefunctions of hydrogen-like atom (ψn,l,m):

Note that Z is the atomic number of atom (Z = 1 for hydrogen) and a0 is Bohr radius in thefollowing expressions. Subscripts refer to the three quantum numbers, n, l and m.

ψ1,0,0 =1√π

(

Z

a0

)3/2

e−Zr/a0 , ψ2,0,0 =1√32π

(

Z

a0

)3/2 (

2− Zr

a0

)

e−Zr/2a0

ψ2,1,0 =1√32π

(

Z

a0

)3/2 (Zr

a0

)

e−Zr/2a0cosθ, ψ2,1,±1 =1√64π

(

Z

a0

)3/2 (Zr

a0

)

e−Zr/2a0sinθe±iφ

ψ3,0,0 =1

81√3π

(

Z

a0

)3/2(

27− 18Zr

a0+ 2

(

Zr

a0

)2)

e−Zr/3a0

(29) Electron spin

(i) Magnetic moment due to orbiting electron:

−→M = − e

2me

−→L

10

where−→L is the angular moment of orbiting charge.

(ii) Magnetic moment due to electron spin:

−→M = γ

−→S

where γ = − ge2m

and g = 2 for electron (spin gyromagnetic ratio).

(iii) Interaction energy between electron spin and magnetic field:

E = −−→M · −→B = −γ−→S · −→B

For the magnetic field aligned along z-axis (i.e−→B = (0, 0, B)),

E = −γSzB,

where Sz is the z-component of the spin, which is either Sz =h2(up-spin) or Sz = − h

2(down-

spin).

(iv) Spin wavefunctions (ψα, ψβ) and eigenvalues

S2ψα/β = h2s(s + 1)ψα/β

Szψα/β = hmsψα/β,

where s = 1

2and ms = −1

2(down-spin, β) or 1

2(up-spin, α).

(30) Hamiltonian operator for multi-electron systems

(i) Atom with N electrons:

H =N∑

i=1

(

− h2

2me

∇2

i − κZe2

ri

)

+∑

i<j

κe2

rij

where κ = 1

4πǫ0and Z is the charge of nucleus. ri is the distance of electron i from the nucleus

and rij is the distance between electron i and j.

(ii) Molecule with N electrons and M nuclei:

H = −N∑

i=1

h2

2me

∇2

i −M∑

p=1

h2

2Mp

∇2

p −N∑

i=1

M∑

p=1

κZpe

2

|ri −Rp|+∑

i<j

κe2

|ri − rj|+∑

p<q

κZpZqe

2

|Rp −Rq|= Te + TN + VeN + Vee + VNN

where ri and Rp are the coordinates of electron i and nucleus p, respectively.

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(31) Variational Principle

For a given trial function, φ, the energy value,

Eφ =

∫

φ∗Hφdτ∫

φ∗φdτ,

is always greater than the true ground state energy.

(32) Anti-symmetry principle

When the coordinates of two electrons are exchanged, the wavefunction must change sign. Forexample, in case of two-electron system:

ψ(1, 2) = −ψ(2, 1)

(33) Orbital Approximation for many-electron atoms

(i) Wavefunction of N -electron atom (Hartree product):

Under orbital approximation, the wavefunction is expressed as a product of one-electron wave-functions (orbitals).

Ψ(r1, r2, · · ·, rN) = φ1(r1)φ2(r2) · · · φN(rN)

(ii) Wavefunctions of two electron system (incorporating anti-symmetry principle):

A. When two electrons occupy the same spatial orbital, φ(r)

Ψ(1, 2) =1√2φ(1)φ(2) [α(1)β(2)− α(2)β(1)]

where α(2) means electon #2 has up spin etc.

B. When two electrons occupy two different spatial orbitals, φ1(r) and φ2(r).

Ψ(1, 2) =1√2[φ1(1)φ2(2)− φ1(2)φ2(1)]α(1)α(2)

Ψ(1, 2) =1√2[φ1(1)φ2(2)− φ1(2)φ2(1)] β(1)β(2)

Ψ(1, 2) =1

2[φ1(1)φ2(2)− φ1(2)φ2(1)] [α(1)β(2) + α(2)β(1)]

Ψ(1, 2) =1

2[φ1(1)φ2(2) + φ1(2)φ2(1)] [α(1)β(2)− α(2)β(1)]

where φ1(2) means electron #2 occupies φ1.

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(34) Molecular Schrodinger equation

Total wavefunction under Born-Oppenheimer approximation:

Ψ(r,R) = ψel(r;R)ψnucl(R)

where r and R represent the coordinates of all electrons and all nuclei, respectively.

(i) Schrodinger equation for electrons (with fixed nuclei coordinates):

Helψel(r;R) = ǫ(R)ψel(r;R)

where the electronic Hamiltonian for a given nuclei configuration R is given by

Hel = Te + VeN + Vee

(ii) Schrodinger equation for nuclei:

Hnuclψnucl(R) = Eψnucl(R)

where E is the total energy of the molecule and the nuclear Hamiltonian is give by

Hnucl = TN + VNN + ǫ(R)

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Math formula

(1) Integral formula:∫

sin2axdx =x

2− sin(2ax)

4a,∫

cos2axdx =x

2+

sin(2ax)

4a,

(2) Integral formula:∫

x2sin2axdx =x3

6− (2a2x2 − 1)sin(2ax)

8a3− xcos(2ax)

4a2

∫

xsin2axdx =x2

4− cos(2ax)

8a2− xsin(2ax)

4a

(3) Trigonemetric relation:

sinαsinβ =1

2[cos(α− β)− cos(α + β)]

sin(A ± B) = sinAcosB ± cosAsinB

(4) Gaussian integral:∫ ∞

−∞e−αx2

dx =

√

π

α

In general,∫ ∞

−∞x2ne−ax2

dx =1× 3× 5 · · · (2n− 1)

2nan

√

π

a

(5) Integral formula:∫ ∞

0

xne−axdx =n!

an+1

∫

x2e−axdx = −(a2x2 + 2ax+ 2)e−ax

a3

∫

x4e−axdx = −(a4x4 + 4a3x3 + 12a2x2 + 24ax+ 24)e−ax

a5

(6) Integral formula:∫

1

0

xm(1 − x)ndx =m!n!

(m+ n+ 1)!

(7) Standard deviation of distribution s.

σs =√

〈s2〉 − 〈s〉2

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