Light and Atoms

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# Light and Atoms - PowerPoint PPT Presentation

Light and Atoms. When an atom gains a photon, it enters an excited state. This state has too much energy - the atom must lose it and return back down to its ground state, the most stable state for the atom. An energy level diagram is used to represent these changes.

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Presentation Transcript
Light and Atoms
• When an atom gains a photon, it enters an excited state.
• This state has too much energy - the atom must lose it and return back down to its ground state, the most stable state for the atom.
• An energy level diagram is used to represent these changes.
Energy Level Diagram
• Energy

Excited States

photon’s path

Ground State

Light

Emission

Light

Emission

Light

Emission

Emission Energetics - I

Problem: A sodium vapor light street light emits bright yellow light of

wavelength = 589 nm. What is the energy change for a sodium atom

involved in this emission? How much energy is emitted per mole of

sodium atoms?

Plan: Calculate the energy of the photon from the wavelength, then

calculate the energy per mole of photons.

Solution:

( 6.626 x 10 -34J s)( 3.00 x 10 8m/s)

h x c

wavelength

Ephoton = hv = =

589 x 10 -9m

Ephoton = 3.37 x 10 -19J

Energy per mole requires that we multiply by Avogadro’s number.

Emole = 3.37 x 10 -19J/atom x 6.022 x 1023 atoms/mole = 2.03 x 105 J/mol

Emole = 203 kJ / mol

Emission Energetics - II

Problem: A compact disc player uses light with a frequency of

3.85 x 1014 per second. What is this light’s wavelength? What portion of

the electromagnetic spectrum does this wavelength fall? What is the

energy of one mole of photons of this frequency?

Plan: Calculate the energy of a photon of the light using E=hv, and

wavelength x C = v . Then compare the frequency with the

electromagnetic spectrum to see what kind of light we have. To get the

energy per mole, multiply by Avogadro’s number.

Solution:

3.00 x 108m/s

3.85 x 1014/s

wavelength = c / v = = 7.78 x 10 -7 m = 778 nm

778 nm is in the Infrared region of the electromagnetic spectrum

Ephoton = hv = (6.626 x 10 -34Js) x ( 3.85 x 1014 /s) = 2.55 x 10 -19 J

Emole = (2.55 x 10 -19J) x (6.022 x 1023 / mole) = 1.54 x 105 J/mole

1

n12

1

n22

1

2 2

1

4 2

Using the Rydberg Equation

Problem: Find the energy change when an electron changes from the

n=4 level to the n=2 level in the hydrogen atom? What is the wavelength

of this photon?

Plan: Use the Rydberg equation to calculate the energy change, then

calculate the wavelength using the relationship of the speed of light.

Solution:

Ephoton = -2.18 x10 -18J - =

Ephoton = -2.18 x 10 -18J - = - 4.09 x 10 -19J

h x c

E

(6.626 x 10 -34Js)( 3.00 x 108 m/s)

wavelength = = =

4.09 x 10 -19J

wavelength = 4.87 x 10 -7 m = 487 nm

The de Broglie Wavelengths of Several Objects

Substance Mass (g) Speed (m/s)  (m)

Slow electron 9 x 10 - 28 1.0 7 x 10 - 4

Fast electron 9 x 10 - 28 5.9 x 106 1 x 10 -10

Alpha particle 6.6 x 10 - 24 1.5 x 107 7 x 10 -15

One-gram mass 1.0 0.01 7 x 10 - 29

Baseball 142 25.0 2 x 10 - 34

Earth 6.0 x 1027 3.0 x 104 4 x 10 - 63

Table 7.1 (p. 274)

Figure 7.19: Probability of finding an electron in a spherical shell about the nucleus.
Figure 7.20: Scanning tunneling microscope of benzene molecules on a metal surface.Photo courtesy of IBM Almaden Research Center.
Figure 7.22: Quantum corral.Photo courtesy of IBM Almaden Research Center; research done by Dr. Don Eigler and co-workers.
Figure 7.24: Cross-sectional representations of the probability distributions of S orbitals.
Light Has Momentum
• momentum = p = mu = mass x velocity
• p = Plank’s constant / wavelength
• or: p = mu = h/wavelength
• wavelength = h / mu de Broglie’s equation
• de Broglie’s expression gives the wavelength relationship of a particle traveling a velocity = u !!

de Broglie Wavelength Calc. - I

Problem: Calculate the wavelength of an electron traveling 1% of the

speed of light ( 3.00 x 108m/s).

Plan: Use the de Broglie relationship with the mass of the electron, and

its speed. Express the wavelength in meters and nanometers.

Solution:

electron mass = 9.11 x 10 -31 kg

velocity = 0.01 x 3.00 x 108 m/s = 3.00 x 106 m/s

h

m x u

6.626 x 10 - 34Js

( 9.11 x 10 - 31kg )( 3.00 x 106 m/s )

wavelength = = =

kg m2

s2

J =

therefore :

wavelength = 0.24244420 x 10 - 9 m = 2.42 x 10 -10 m = 0.242 nm

Heisenberg Uncertainty Principle
• It is impossible to know simultaneously both the position and momentum (mass X velocity) and the position of a particle with certainty !

Quantum Numbers

Allowed Values

n

1 2 3 4

L

0 0 1 0 1 2 0 1 2 3

mL

0 0 -1 0 +1 0 -1 0 +1 0 -1 0 +1

-2 -1 0 +1 +2 -2 -1 0 +1 +2

-3 -2 -1 0 +1 +2 +3

Problem: What values of the azimuthal (L) and magnetic (m) quantum

numbers are allowed for a principal quantum number (n) of 4? How

many orbitals are allowed for n=4?

Plan: We determine the allowable quantum numbers by the rules given

in the text.

Solution: The L values go from 0 to (n-1), and for n=3 they are:

L = 0,1,2,3. The values for m go from -L to zero to +L

For L = 0, mL = 0

L = 1, mL = -1, 0, +1

L = 2, mL = -2, -1, 0, +1, +2

L = 3, mL = -3, -2, -1, 0, +1, +2, +3

There are 16 mL values, so there are 16 orbitals for n=4!

as a check, the total number of orbitals for a given value of n is n2, so for

n = 4 there are 42 or 16 orbitals!

Radial probability “Accurate” “Stylized” Combined area

distribution representation of the 2p of the three 2p

of the 2p distribution orbitals: 2px, 2py

distribution and 2pz orbitals

Fig. 7.17

Quantum Numbers - I
• 1) Principal Quantum Number = n
• Also called the “energy “ quantum number, indicates the approximate distance from the nucleus .
• Denotes the electron energy shells around the atom, and is derived directly from the Schrodinger equation.
• The higher the value of “n” , the greater the energy of the orbital, and hence the energy of electrons in that orbital.
• Positive integer values of n = 1 , 2 , 3 , etc.
Quantum Numbers - II
• 2) Azimuthal
• Denotes the different energy sublevels within the main level “n”
• Also indicates the shape of the orbitals around the nucleus.
• Positive interger values of L are : 0 ( n-1 )
• n = 1 , L = 0 n = 2 , L = 0 and 1

n = 3 , L = 0 , 1 , 2

Quantum Numbers - III
• 3) Magnetic Quantum Number - mL Also called the orbital orientation quantum #
• denotes the direction or orientation in a magnetic field - or it denotes the different magnetic geometriesound the nucleus - three dimensional space
• values can be positive and negative (-L 0 +L)
• L = 0 , mL = 0 L =1 , mL = -1,0,+1

L = 2 , mL = -2,-1,0,1,2

Quantum Numbers Noble Gases

Electron OrbitalsNumber of ElectronsElement

1s2 2 He

1s2 2s22p6 10 Ne

1s2 2s22p6 3s23p6 18 Ar

1s2 2s22p6 3s23p6 4s23d104p6 36 Kr

1s2 2s22p6 3s23p6 4s23d104p6 5s24d105p6 54 Xe

1s2 2s22p6 3s23p6 4s23d104p6 5s24d105p6 6s24f14 5d106p6 86 Rn

1s2 2s22p6 3s23p6 4s23d104p6 5s24d105p6 6s24f145d106p6 7s25f146d10?

The Periodic Table of the Elements

Electronic Structure

H

He

Li

Be

B

C

N

O

F

Ne

Ar

Na

Mg

Al

Si

P

S

Cl

Kr

K

Ca

Sc

Ti

V

Cr

Mn

Fe

Co

Ni

Cu

Zn

Ga

Ge

As

Se

Br

Xe

Rb

Sr

Y

Zr

Nb

Mo

Tc

Ru

Rh

Pd

Ag

Cd

In

Sn

Sb

Te

I

Rn

Cs

Ba

La

Hf

Ta

W

Re

Os

Ir

Pt

Au

Hg

Tl

Pb

Bi

Po

At

Fr

Ra

Ac

Rf

Ha

Sg

Ce

Pr

Nd

Pm

Sm

Eu

Gd

Tb

Dy

Ho

Er

Tm

Yb

Lu

Th

Pa

U

Np

Pu

Am

Cm

Bk

Cf

Es

Fm

Md

No

Lr

“ S” Orbitals

“ P” Orbitals

“ f ” Orbitals

“ d” Orbitals