THE STRUCTURE OF ATOMS
(Focus 8)
(Focus 8)
THE STRUCTURE OF ATOMS
• Description of the arrangement of electrons in
atoms.
• Basis for understanding the structures and properties
of elements and their compounds.
For the Hydrogenic atoms (H, He+, Li2+, C5+, U91+ etc), the
Schrödinger equation can be solved exactly.
These concepts are used to describe structures of many-
electron atoms and molecules.
Solution of Hydrogenic atom problem
Two-particles, 3D system
Potential energy (V) =
Solution of Hydrogenic atom problem
Schrodinger eq. for this problem is better solved in
spherical polar coordinate system with proper
boundary conditions.
r ≥ 0,
0° ≤ θ ≤ 180° (π rad),
0° ≤ φ < 360° (2π rad)
Eigenvalues of Hydrogenic atom
Allowed energy levels:
h cRZ 2
En
n2
n=1,2,…principal quantum no
e4 mem N
h cR
3 2 2 0 2
2 me mN
µ is the reduced mass
En values are all negative, i.e.
electron in an atom has lower
energy than when it is free.
For a transition:
hcR hcR
E E 2 E1 2 2 Ionization energy:
n1 n2 I=hcR
This energy is carried away by a photon.
Eigenfunctions of Hydrogenic atom
The wavefunctions (called atomic orbitals) can be written as
the product of radial and angular functions:
n ,l ,m l
( r , , ) Y l ,m l ( , ) R n ,l (r )
n, l and ml are quantum numbers arises due to restrictions
(boundary conditions)
•Principal Quantum number (n), n = 1, 2, …
Quantization of energy
•Orbital angular momentum Quantum number (l), l = 0, 1,…,n-1
Quantization of angular momentum magnitude (L)
L l(l 1) l = 0,1, 2,…, n-1
•Magnetic Quantum number (ml), ml = l, l-1, … , -l
Quantization of angular momentum direction
L z m l m l l , l 1 , l 2 ,............, l (2l 1)
values
Shell Structure of Hydrogenic atom
S-Orbitals
The solution of Schrödinger equation gives ( n 1, l 0 , m l 0 )
1/ 2
1 4 r / a0 4 0 2
1s 3 e Bohr radius a 0
4
1/ 2
a0 mee 2
r
1
1
e a0
a
0
3 2
Ψ1s decays exponentially towards
zero from its maximum at the
nucleus
s-orbitals are
spherically symmetrical
Probability of finding electron
Hydrogenic atom n ,l ,m l
( r , , ) Y l ,m l ( , ) R n ,l (r)
What is the probability of finding the electron in
a region of space with its coordinates lie in the
ranges r to r + dr, θ to θ + dθ, and φ to φ + dφ?
Probability =
Now, What is the probability of finding the electron in
a thin spherical shell centered at the origin, of inner
radius r and outer radius r + dr ?
The probability =
Surface area= 4πr2,
thickness= dr
So Prob =
Radial distribution function P(r)
Probability of finding the electron between two spherical
shells of thickness r and r+dr
P ( r ) dr Where, p ( r ) r 2 R ( r ) 2
For spherically symmetric orbitals, p ( r ) 4 r
2 2
The RDF gives the
probability that an
electron will be
found anywhere in a
shell of radius r and
thickness Δr
regardless of angle.
Other Hydrogenic Wavefunctions
n ,l ,m ( r , , ) Y l ,m l ( , ) R n ,l (r )
The p-orbitals (l = 1): l
1/ 2
1 1
1/ 2
3 r r /2a 0
2pz cos 3 e
4 2 6a0 a0
1/ 2
1
5
r co s e r / 2 a 0
3 2 a 0
2px → sinθ cosφ 2py → sinθ sinφ
The d-orbitals (l = 2):
See
Table
8A.1 for
more
orbitals
Other Hydrogenic Radial Wavefunctions
Orbitals have (n - l -1) radial nodes.
Electron spin
Stern & Gerlach (1921) shot a
beam of silver atoms through an
inhomogeneous magnetic field.
The observation seems to conflict
with one of the predictions of QM.
The observed angular momentum was due to the motion of the
electron about its own axis (and not due to orbital motion).
This intrinsic angular momentum of the electron is called its spin.
It is a quantum mechanical phenomenon -no classical
counterpart.
Electron spin is described by spin quantum number (s = 1/2).
With magnitude =
With components:
Electron spin
•Two allowed spin states of electron
ms (+1/2 or -1/2)
•The magnitude of the spin angular
momentum is (31/2/2)ħ
• Directions of spin are opposite
ms=+1/2 α electron ms=-1/2 β electron
Spin can be clockwise or
counterclockwise
Spectrum of Atomic Hydrogen
Discrete lines
electron energy in H-atom
is quantized
Bohr frequency
1 1 n1= 1, 2, condition
R H 2 2
Rydberg :
n2= n1+1, n1+2,….
n1 n 2 RH=Rydberg constant ΔE hν hcv
n1 =1 Lyman, n1 =2 Balmer, n1=3 Paschen,
n1=4 Brackett, n1=5 Pfund
During a transition, a photon (with its one unit of angular
momentum) is generated.
Selection rules for transitions
To compensate for the angular momentum carried away by
photon, the angular momentum of electron must change by
one unit.
l 1 m l 0,1 d(l=2) s(l=0) Not allowed
s(l=0) s(l=1) Allowed
Transition dipole moment
s f id
μs ≠ 0, transition is allowed
The Grotrian Diagram μs = 0, transition is not allowed
