Quantum Mechanics

Through the Looking Glass

This is how the model of the atom has developed so far:

Rutherford

ThomsonDemocritus Dalton

c = where c =3.00 x 108 m/s

Sample Problem: The yellow light given off by a sodium lamp has a wavelength of 589 nm. What is the frequency of this radiation?

c = , where c =3.00 x 108 m/s

3.00 x 108 m/s = 589 nm 1 m

1x109nm

= 5.08 x 1014 s 1-

Planck’s Theory: Energy is released incrementally as

individual packets of energy called quanta, where the change in energy of a system is E = h,

2h,…n h and h(plank’s constant) =

h = 6.63 x 10-34 J-s

we know from the previous problem: c = , that = 5.08 x 1014 s 1-

since E = h and h (plank’s constant) = 6.63 x 10-34 J-s

E = (6.63 x 10-34 J-s )(5.08 x 1014 s 1- )

E = 3.37 x 10-19 J

Sample Problem: Calculate the smallest increment of energy that an object can absorb from yellow light

whose wavelength is 589 nm

A Continuous Spectrum

Light is a form of ...

Electromagnetic Radiation

An Emission Spectrum...…is produced when a gas is placed under reduced pressure...

...as a high voltage is applied

Balmer’s Description of the Emission Spectrum of Hydrogen

= C 1

22

- 1

n2

where n = 3, 4, 5, 6… and C = 3.29 x 1015 s-1

Bohr’s Model of the Atom (1914)

Limited the path of electrons to circular orbits with discrete energy (quantum energy levels)

Explained the emission spectrum of hydrogen

0 Ao

2.12 Ao

4.77 Ao

n = 1

n = 2

n = 3

-2.18 x 10 -18 J

0

-0.545 x 10 -18 J

-0.242 x 10 -18 J

Radii and Energies of the Three Lowest Energy orbits in the Bohr Model

radius = n2 (5.3 x 10-11m)

0.53 A

En = -RH1n2 where RH = 2.18 x 10 -18J

En = -RH

1

2=

Hydrogen’s Spectrum is Produced When

Electrons are excitedfrom their ground state

Electrons appear in excited state

electrons transfer from an excited state

photons produced

Electrons return totheir ground state

energy is absorbed

Lyman Series

Balmer Series

Paschen Series

•Ultraviolet

•Visible and Ultraviolet

•Infrared

Explaining the Emission Spectrum of Hydrogen

since E = E f - E i

then E = -RH

1nf

2-

1ni

2

Sample Problem: Calculate the wavelength of light that corresponds to thetransition of the electron from the n = 4 to the n=2 state of the hydrogen atom.

RH1

ni2

-1

nf2

E =

2.18 x 10-18J1

42-

1

22E =

-4.09 x 10-19 JE =

= E h

= -4.09 x 10-19 J

6.63 x 10-34 J-s

= 6.17 x 1014 s -1

= c

= 3.00 x 108 m/s

6.17 x 1014 s -1

= 4.86 x 10-7 m = 486 nm (green)