Lecture V
Download
1 / 28

Lecture V Hydrogen Atom dr hab. Ewa Popko - PowerPoint PPT Presentation


  • 130 Views
  • Uploaded on

Lecture V Hydrogen Atom dr hab. Ewa Popko. Niels Bohr 1885 - 1962. Bohr Model of the Atom. Bohr made three assumptions (postulates) 1. The electrons move only in certain circular orbits, called STATIONARY STATES. This motion can be described classically

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' Lecture V Hydrogen Atom dr hab. Ewa Popko' - carys


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

Lecture V

Hydrogen Atom

dr hab. Ewa Popko


Niels Bohr1885 - 1962


Bohr Model of the Atom

  • Bohr made three assumptions (postulates)

  • 1. The electrons move only in certain circular orbits, called STATIONARY STATES. This motion can be described classically

  • 2. Radiation only occurs when an electron goes from one allowed state to another of lower energy.

  • The radiated frequency is given by

  • hf = Em - En

  • where Em and En are the energies of the two states

  • 3. The angular momentum of the electron is restricted to integer multiples of h/ (2p) =

  • mevr =n(1)


The hydrogen atom

The Schrödinger equation

Partial differential equation with three independent variables

The potential energy in

spherical coordinates

(The potential energy function is spherically symmetric.)


The spherical coordinates

(alternative to rectangular coordinates)


The hydrogen atom

For all spherically symmetric potential-energy functions:

( the solutions are obtained by a method called separation of variables)

Radial function

Angular function of q and f

Thus the partial differential equation with three independent variables

three separate ordinary differential equations

The functions q and f are the same for every spherically symmetric potential-energy function.


The solution
The solution

The solution is determined by boundary conditions:

- R(r) must approach zero at large r (bound state - electron localized near the nucleus);

- Q(q) and F(f) must be periodic:

(r,q,f) and(r,q,f+2p) describe the same point, so

F(f)=F(f+2p);

- Q(q) and F(f) must be finite.

Quantum numbers:

n - principal l – orbital ml - magnetic


Principal quantum number: n

Ionized atom

n = 2

- 3.4 eV

n = 1

E = - 13.6 eV

The energy En is determined by n = 1,2,3,4,5,…;

n = 3

m - reduced mass


Quantization of the orbital angular momentum.

The possible values of the magnitude L of the orbital angular momentum L are determined by the requirement, that the Q(q) function must be finite at q=0 and q=p.

Orbital quantum number

There are n different possible values of L for the n th energy level!



Quantum numbers momentum: n, l, m

n – principal quantum number

n – determines permitted values of the energy

n = 1,2,3,4...

l– orbital quantum number

l -determines permitted values of the orbital

angular momentum

l = 0,1,2,…n-1;

ml - magnetic quantum number

ml – determines permitted values of the z-component of the orbital angular momentum


Wave functions

l momentum= 0

n= 1

n= 2

l = 0,1

l = 1

m = ±1

n= 3

l = 0,1,2

yn,l,m

Wave functions

Q(q) - polynomial

F(f) ~


Quantum number notation
Quantum number notation momentum

Degeneracy : one energy level En has different quantum numbers l and ml

l = 0 : s states n=1 K shell

l = 1: p states n=2 L shell

l = 2 : d states n=3 M shell

l = 3 : f states n=4 N shell

l = 4: g states n=5 O shell

. .

. .


Electron states momentum

3s

K

1s

3p

2s

M

L

3d

2p




Spin angular momentum and magnetic moment
Spin angular momentum and magnetic moment momentum

Electron posseses spin angular momentum Ls. With this momentum magnetic momentum is connected:

where ge is the gyromagnetic ratio

For free electron ge=2


Własny moment pędu - spin momentum

Spin angular momentum and magnetic moment

spin quantum number s = ½

Allowed values of the spin angular momentum are quantized :

The z – component of the spin angular momentum:



Many – electron atoms and the exclusion principle atom, 4 quantum numbers are need:

  • Central field approximation:

    - Electron is moving in the total electric field due to the nucleus and averaged – out cloud of all the other electrons.

  • - There is a corresponding spherically symmetric potential – energy function U( r).

  • Solving the Schrodinger equation the same 4 quantum numbers are obtained. However wave functions are different. Energy levels depend on both n and l.

  • In the ground state of a complex atom the electrons cannot all be in the lowest energy state.

  • Pauli’s exclusion principle states that no two electrons can occupy the same quantum – mechanical state. That is, no two electrons in an atom can have the same values of all four quantum numbers (n, l, ml and ms)


Shells and orbitals atom, 4 quantum numbers are need:

l

shell

n

Nmax

orbital

2

1

K

0

s

2

2

L

0

s

6

L

1

p

2

3

M

0

s

6

M

1

p

10

M

2

d

s

2

0

N

4

p

1

6

N

10

2

d

N

N

f

14

3

Nmax - maximum number of electrons occupying given orbital


Shells K, L, M atom, 4 quantum numbers are need:

­¯

­¯

­­

­¯

­¯

­¯­­

state with ms = +1/2

1s22s22p2

1s22s22p4

carbon

oxygen

N : number of allowed states

state with ms = -1/2

Hund’s rule - electrons occupying given shell initially set up their spins paralelly


The periodic table of elements atom, 4 quantum numbers are need:


Atoms of helium lithium and sodium
Atoms of helium, lithium and sodium atom, 4 quantum numbers are need:

3s

n =3,  = 0

2p

n =2,  = 1

n =2,  = 1

2s

n =2,  = 0

n =2,  = 0

n =2,  = 0

1s

n =1,  = 0

n =1,  = 0

n =1,  = 0

Sodium (Z= 11)

Helium (Z = 2)

Lithium(Z = 3)


Electron configuration – the occupying of orbitals atom, 4 quantum numbers are need:

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d105p6 6s2 4f14 5d10 6p6 7s2 6d10 5f14


Total angular momentum j
Total angular momentum - J atom, 4 quantum numbers are need:

Possible two magnitudes ofj :

Example:l = 1, s = ½

j = 3/2

j = 1/2


Nmr nuclear magnetic resonance
NMR ( nuclear magnetic resonance) atom, 4 quantum numbers are need:

Like electrons, protons also posses magnetic moment due to orbital angular momentum and spin ( they are also spin-1/2 particles) angular momentum.

Spin flip experiment:

Protons, the nuclei of hydrogen atoms in the tissue under study, normally have random spin orientations. In the presence of a strong magnetic field, they become aligned with a component paralell to the field. A brief radio signal flips the spins; as their components reorient paralell to the field, they emit signals that are picked up by sensitive detectors. The differing magnetic environment in various regions permits reconstruction of an image showing the types of tissue present.


ad