2. •Limitations of classical mechanics in treating atomic
and molecular phenomena
• Schrodinger equation
• Particle in a box solution and their applications for
conjugated molecules and nanoparticles
•Molecular orbital treatment for homo-nuclear diatomic
molecules
• Bonding in Coordination Compounds: Crystal field
theory
3. Schrödinger Wave Equation
• Schrodinger wave equation is given by Erwin Schrödinger in 1926 and based on
dual nature of electron (Wave and Particle nature {Louis de Broglie 1924).
• According to him electron behaves like a three dimensional wave
• One dimensional wave propagating equation can be expressed
∂2f(x)/∂x2 = -4π2f(x)/λ2
• This is second order differential equation
• Extend this equation for 3 dimension replacing by Ψ (x,y,z), thus a 3
dimensional wave eq. will be
∂2Ψ/∂x2 + ∂2Ψ/∂y2 + ∂2Ψ/∂Z2= -4π2Ψ/λ2
• By putting the value of λ = h/mv (de Broglie eq.)
∂2Ψ/∂x2 + ∂2Ψ/∂y2 + ∂2Ψ/∂Z2= -4π2Ψm2v2/h2
• Total Energy = Kinetic Energy+ Potential Energy
4. Schrödinger Wave Equation
• Total Energy (E) = Kinetic Energy+ Potential Energy
E= ½ mv2+ V
2E = mv2+ 2V
mv2 = 2E - 2V = 2(E-V)
v2 = 2(E-V)/m
Put the value of v in below eq.
∂2Ψ/∂x2 + ∂2Ψ/∂y2 + ∂2Ψ/∂Z2= -4π2Ψm2v2/h2
∂2Ψ/∂x2 + ∂2Ψ/∂y2 + ∂2Ψ/∂Z2= -4π2Ψm2 2(E-V)/mh2
∂2Ψ/∂x2 + ∂2Ψ/∂y2 + ∂2Ψ/∂Z2 = -8π2m(E-V) Ψ/h2
∂2Ψ/∂x2 + ∂2Ψ/∂y2 + ∂2Ψ/∂Z2+8π2m(E-V) Ψ/h2 = 0 (1)
The probability of finding an electron at any point around the nucleus can be
determined by this Schrödinger Wave Equation (1)
Where x,y, and z are the 3 space co-ordinates, m = mass of electron, h = Planck’s
constant, E = Total energy, V = potential energy of electron,Ψ = amplitude of
wave also called as wave function, ∂ = for an infinitesimal change.
5. Schrödinger Wave Equation
• The Schrodinger wave equation can also be
written as,
∇2Ψ + (8π2m/h2) (E-V) Ψ = 0
∂2/∂x2 + ∂2/∂y2 + ∂2/∂Z2 = ∇2
Where ∇2 = Laplacian operator.
6. Physical significance of Ψ and Ψ2
(i) The wave function Ψ represents the amplitude of the
electron wave.
(ii) For a single particle, the square of the wave function (Ψ2) at
any point is proportional to the probability of finding the
particle at that point.
(iii) If Ψ2 is maximum than probability of finding e- is maximum
around nucleus and the place where probability of finding e-
is maximum is called electron density, electron cloud or an
atomic orbital. It is different from the Bohr’s orbit.
(iv)The solution of this equation provides a set of number
called quantum numbers which describe specific or definite
energy state of the electron in atom and information about
the shapes and orientations of the most probable
distribution of electrons around the nucleus.
7. Solution of Schrödinger Equation
Particle in one dimensional box
• A particle in a 1-dimensional box is a fundamental
quantum mechanical approximation describing the
translational motion of a single particle confined inside
an infinitely deep well from which it cannot escape.
• The solutions to the problem give possible values of
Energy that the particle can possess. E represents
allowed energy values and wave function, which when
squared gives us the probability of locating the particle
at a certain position within the box at a given energy
level.
8. Particle in one dimensional box
• To solve the problem for a particle in a 1-
dimensional box, following steps are followed:
• Define the Potential Energy, V
• Solve the Schrödinger Equation
• Define the wave function
• Define the allowed energies
9. Step 1: Define the Potential Energy V
A particle in a 1D infinite potential well of
dimension L or a.
10. • The potential energy is 0 inside the box (V=0
for 0<x<L) and goes to infinity at the walls of
the box (V=∞ for x<0 or x>L).
• We assume the walls have infinite potential
energy to ensure that the particle has zero
probability of being at the walls or outside the
box. These boundary conditions will be
used when solving the Schrödinger Equation.
Particle in one dimensional box
11. Particle in one dimensional box
• Step 2: Solve the Schrödinger Equation
∇2Ψ + (8π2m/h2) (E-V) Ψ = 0
∇2Ψ + (8π2mEΨ/h2) - (8π2mVΨ/h2)= 0
(8π2mVΨ/h2)- ∇2Ψ = 8π2mEΨ/h2
(8π2mV/h2- ∇2)Ψ = 8π2mEΨ/h2
Put the value of h = ℏ2π
ℏ is the reduced Planck Constant
(8π2mV/ ℏ24π2- ∇2)Ψ = 8π2mEΨ/ℏ24π2
(2mV/ ℏ2- ∇2)Ψ = 2mEΨ/ℏ2
2m/ℏ2 (V- ∇2/2m/ℏ2)Ψ = 2mEΨ/ℏ2
(V- ∇2ℏ2/2m)Ψ = EΨ
(- ℏ2∇2/2m+V)Ψ = EΨ
12. Particle in one dimensional box
• Step 2: Solve the Schrödinger Equation
(- ℏ2∇2/2m+V)Ψ = EΨ
• The time-independent Schrödinger equation for a particle of
mass m moving in one direction (x) with energy E is
(- ℏ2(d2/dx2)/2m+Vx)Ψx = EΨx
• Ist case
• Vx=0 (particle moving inside the box)
(- ℏ2(d2/dx2)/2m+0)Ψ = Eψ
(- ℏ2(d2Ψ/dx2)/2m) = EΨ
d2Ψ/dx2 = - 2mEΨ/ℏ2
d2Ψ/dx2 + 2mEΨ/ℏ2 = 0
K2 = 2mE/ℏ2
d2Ψ/dx2 + K2 Ψ = 0
This is second order differential equation which has the solution of the
form
Ψ = A cos Kx + B sin Kx
Where A and B are the constants
14. Step 3: Define the wave function
• From the second order differential equation
Ψ = A cos Kx + B sin Kx
• (1) when x = 0 i.e. Ψx = 0
• Since Ψ is the function of x, Ψ is zero at wall thus
Ψ = A cos Kx + B sin Kx
0 = A cos 0 + B sin 0
0 = A X1 + B X 0
A = 0
(2) when x = a, i.e. it is also at the wall Ψa = 0
Ψ = A cos Kx + B sin Kx
0 = A cos Ka + B sin Ka
A=0 thus
B sin Ka = 0
(B is a constant) thus
sin Ka = 0
(when sin x = o it is x=nπ) thus
Ka=nπ
15. Solution for K
where n is integer and π is radial angle
K = nπ/a
We know that K2 = 2mE/ℏ2 where ℏ = h/2 π
n2π2/a2 = 2mE/ h2 /4 π2
n2π2/a2 = 8 π2 mE/ h2
n2/a2 = 8mE/ h2
E = n2 h2 /8ma2
Where n is integer no or quantum no.
This eq. represent energy of particle in 1 dimensional box
16. Step 4: Determine the Allowed
Energies
• Solving for E results in the allowed energies for a
particle in a box:
• or
• This is an important result that tells us:
• The energy of a particle is quantized
• The lowest possible energy of a particle is NOT zero.
This is called the zero-point energy and means the
particle can never be at rest because it always has
some kinetic energy.
17. What does all this mean?
• The wave function for a particle in a box at
the n=1 and n=2 energy levels look like this:
18. • The probability of finding a particle a certain
spot in the box is determined by squaring ψ.
The probability distribution for a particle in a
box at the n=1 and n=2 energy levels looks like:
19. Application of Particle in box
• The particle in box model can also be applied to the electronic
spectrum of linear polyenes.
• The simplest case is the 1,3-butadiene molecule. The four π-
electrons are assumed to move freely over the four carbon
framework of single bonds. Now, in the lowest energy state of
butadiene, the 4 delocalized electrons will fill the two lowest
molecular orbital's.
• The lowest unoccupied molecular orbital (LUMO) in butadiene
corresponds to the n = 3 particle-in-a box state. Neglecting
electron-electron interaction, the longest-wavelength (lowest-
energy) electronic transition should occur from n = 2, the highest
occupied molecular orbital (HOMO).
20. • Application of Particle in box
The energy difference is given by ∆E = E3 − E2 = (32 − 2 2 ) h 2 /8mL2
Here m represents the mass of an electron 9.1×10−31 Kg, and L is the effective
length of the box, 5.6 × 10−10 m.
By using ∆E = hν = hc/λ
The wavelength is predicted to be 210nm. This compares well with the
experimental maximum of the first electronic absorption band, λmax ≈ 210 nm, in
the ultraviolet region.
21. Application of Particle in box
Hexa-triene
• In this case there are 6 electrons
The Energy difference is given by ∆E = E4 − E3 = (42 − 32) h2/8ma2
a= 8.37 × 10−10 m
Wavelength = 303 nm
22. Nano-Particles
• Quantum Dots Quantum dots are semiconductor
nanocrystals having dimensions typically
between ~1-10 nm. Thus a quantum dot can be
seen in analogy to the “particle in a box” model.
26. Numerical Problems
Q1. Calculate the lowest three energy levels of a particle of mass
10-26 Kg in a box of length L = 10-9 m. Plank constant (6.625 × 10–34 m2kg/sec)
Q2 Calculate the magnitude of the energy of the photon (or quantum) associated with
light of wavelength 6057.8 Å. (Å = 10–8 cm)
Q3. Calculate the wavelength for excitation of the electron from LUMO to HOMO in
hexatriene the mass of an electron 9.1×10−31 Kg, and L is the effective length of the
box, 1× 10−10 m.
Q4 Calculate zero point energy of electron having the mass of an electron 9.1×10−31
Kg, and L is the effective length of the box, 1× 10−10 m.
Q5
• Hint (a) Calculation of Frequency : ν =c/ λ
• C= 3x1010 cm/sec
• ν = 4.952 × 1014 sec–1
• (b) Calculation of Energy :
• E = hν = (6.625 × 10–27 erg sec) (4.952 × 1014 sec–1 )
• = 3.281 × 10–12 erg
27. Q1. Calculate the lowest three energy levels of a
particle of mass 10-26 Kg in a box of length L =
10-9 m. Plank constant (6.625 × 10–34
m2kg/sec)
Ans Hint. E = n2 h2 /8ma2
n=1,2,3
E = 12 (6.625 × 10–34 )2 /8 (1x 10-26)(10-9)2
28. Molecular Orbital diagrams of
diatomic molecules
• The smallest molecule, hydrogen gas exists as dihydrogen (H-H)
with a single covalent bond between two hydrogen atoms.
• As each hydrogen atom has a single 1s atomic orbital for
its electron, the bond forms by overlap of these two atomic
orbitals.
• In the figure the two atomic orbitals are depicted on the left and on
the right.
• The vertical axis always represents the orbital energies. Each atomic
orbital is singly occupied with an up or down arrow representing an
electron.
• For a stable bond, the bond order must be positive, defined as
• Bond order = (No. of electron in bonding MOs - No. of electron in anti bonding MOs)/2
29.
30. • Dihelium (He-He) is a hypothetical molecule
and MO theory helps to explain why dihelium
does not exist in nature.
32. • Up to nitrogen the energy gap between the
sigma 2S orbital and sigma 2Px orbital is small,
thus the mixing effect is high which push the
out- phase mixing high as they have same
symmetry also. Molecular orbitals do not
retain the pure s and p character. Due to s-p
mixing the energy of all these orbital
redistributed.
36. PROBLEMS
Q1. Arrange following in order to increase
Bond order, bond length, bond energy
• O2, O2
-
, O2
2-
, O2
+
• N2, N2
-
, N2
2-
, N2
+
Q2. Calculate magnetic moment of following using
spin only formula (μ= n+2)
• O2, O2
-
, O2
2-
, O2
+
• N2, N2
-
, N2
2-
, N2
+
