AtomicStructure

Model A: The plum pudding model

J.J. Thompson

Negative chargeslike raisins in plumb pudding

Positive charge is spreadout like the pudding

detector

Fire a beam of alpha particles (He nucleii with charge +2) at a thingold foil. Watch what happens.

detector

What would Thompson’smodel predict?

Collisions? Sticking?Bouncing?

What was observed:

1) most alpha particles passed throughundeflected by the gold foil.

2) some were deflected at large angles

3) some were deflected backwards!

Rutherford’s Solution:The atom is composed of a concentratedregion of positive charge at its centerknown as the nucleus.Negative charges (electrons) followedcircular orbits around the nucleus.

New problem:According to Maxwell, an acceleratingelectron will radiate energy. Since theelectrons have centripetal acceleration, theyshould radiate.

In addition, the frequency of the emittedlight will match the frequency of the orbitof the electron around the nucleus.

What did you find in the lab when youlooked at the Hydrogen spectrum?

This implies a continuous spectrum ofemitted radiation should be observed.

Only a few colors (wavelengths).

The emission spectrum of your second source(Helium or Mercury) looked very differentthan the Hydrogen source!

Each element has its own unique emissionspectrum -- a spectral FINGERPRINT!

Experimentally, the observed lines in thevisible part of the Hydrogen spectrum arefound at wavelengths given by:

1 12

12 2

FHG IKJRnH n = 3, 4, 5,...

The constant RH is found to be:

R mH 10973732 107 1.

So we only observe emitted radiation ata discrete set of wavelengths.

Furthermore, atoms absorb radiation at onlya discrete set of wavelengths as well.

Balmer Series

1 12

12 2

FHG IKJRnH n = 3, 4, 5,...

These two sets (the emission spectrum andthe absorption spectrum) are identical.

The latter is known as the absorption spectrum.

Which is to say, if we shined light with uniformintensity across the entire spectrum through agas cloud of a single type of atom, radiationwould be missing from the resulting spectrumas this discrete set of wavelengths.

410.2

nm43

4.1nm

486.1

nm

656.3

nm

Emission Spectrum

Absorption Spectrum

So how do we explain the fact that onlydiscrete wavelengths are absorbed/emittedfrom the atoms?

Bohr’s Solution:1) Only certain electron orbits are stable2) Stable orbits have quantized angular momentum.3) Radiation is emitted when electrons “jump” from one orbital to another.4) The energy of the emitted radiation is proportional to the energy difference between the orbitals.

Recall from Physics 111 that the angularmomentum of a point mass in a circularorbit of radius r is given by

L = m v rBohr’s suggested that only certain valuesof L resulted in stable orbits. The orbitalswhich had

L nh

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

were stable.

To figure out the total energy of the electronsin these orbitals, recall

Total Energy = KE + PE

KE mv 12

2 PE ker

2

E mv ker

12

22

We also know from Newton’s Second Lawthat the electrical force of attraction betweenthe proton and the electron must balancethe centripital force the electron feels in orbit.

ker

mvr

2

2

2

Which means that the kinetic energy isgiven by:

KE mv ker

12

22

2

So the energy of the electron orbitals isgiven by:

E ker

ker

ker

2 2 2

2

Recall, however, that not all orbitalsare possible. Only the ones with quantizedangular momentum are allowed:

L mvr nh

2Solving for v andsquaring...

v n hm r

22 2

2 2 24

ker

mvr

2

2

2

Compare with theforce balance...

Solving for v2...

v kemr

22

Setting these twoequal to one another...

Solving for r tellsus the allowed orbitals.

n hm r

kemr

2 2

2 2 2

2

4

r n hmke

n a 2 2

2 22

04Where a0 is calledthe “Bohr Radius”

The smallestallowed orbital

a0 = 0.0529 nm

Plugging our quantized values of r into ourequation for the energy of the electronorbitals gives us:

E ken nm

eVnn

2

2 22 0 0529136

( . ).

An atom is said to be in the “ground state” when n = 1.

Now we can predict the energy of the emittedradiation when an electron “jumps” fromone orbital to another...

E eVn nf i

FHG

IKJ13 6 1 1

2 2.

And we know theenergy of a photon is

E hc

hc eVn nf i

FHG

IKJ136 1 1

2 2.

1 136 1 12 2

FHG

IKJ. eV

hc n nf i

Look familiar?

1 12

12 2

FHG IKJRnH

Empirical resultsfor the Balmer series...

136 10946775 107 1. .eV

hcm

R mH 10973732 107 1.

derivedresults

1 13 6 1 12 2

FHG

IKJ. eV

hc n nf i

A photon of the correct frequency interactswith the electron. The photons energy isabsorbed by the atom. The electron jumpsto a higher orbital.

The electron jumps to a lower orbital (lowerenergy state). A photon is emitted withexactly the amount of energy lost as theelectron falls to the lower orbital.