What’s wrong with this picture?

The attractive Coulomb force between the positive nucleus and the orbiting electron could provide the attractive force which keeps the electron in it’s orbit, much as the planets orbit the sun with gravity providing the centripetal force. rm

r

vmF e

elcentripeta

20

20 ω==

Accelerating charges radiate. Could this electromagnetic radiation be the source of the spectral lines?

No. This radiation must come at the expense of the kinetic energy of the orbiting electron!

It will eventually spiral into the nucleus. The atom would be unstable!

But what does all of this mean??

...5,4,32

)cm(

:seriesBalmer

22

2

2 =⎟⎟⎠

⎞⎜⎜⎝

⎛

−= n

nn

Cλ

...6,5,43

)cm(

:seriesPaschen

22

2

4 =⎟⎟⎠

⎞⎜⎜⎝

⎛

−= n

nn

Cλ

...4,3,21

)cm(

:seriesLyman

22

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛

−= n

nn

Cλ

It also turns out that C1=C2=C3…

•Classical electromagnetism does not hold for atom sized systems.

•Used Planck’s energy quantization ideas to postulate that electrons orbit in fixed, stable, nonradiating states, given by

•Used Einstein’s concept of a photon to define the frequency of radiation emitted when an electron jumps from one state to another. The photon energy is just the energy difference between states, i.e.,

•Used classical mechanics to calculate the orbit of the electron.

hvEE if =−

€

mevr = nh n =1,2,3...

Coulomb force keeps electrons in orbit:

2

2

1vmKE e=

r

ekU

2

−=2

22

r

e

r

mvmaF ===

But we know that the allowed values of the angular momentum are:

€

mevr = nh n =1,2,3...

(Are you wondering why these are the allowed values? You should be.)

r

kevmKE e 22

1 22 ==

r

ke

r

ke

r

keUKEE

22

222

−=−=+=

€

mevr = nh ⇒ v =nh

mer

€

KE =1

2me

nh

mer

⎛

⎝ ⎜

⎞

⎠ ⎟

2

=ke2

2r⇒ rn =

n2h2

meke2

€

a0 =h2

meke2

= 0.0529 nm

For n=1, this gives the “Bohr radius”:

which was in good agreement with experimental values.

For other values of n: 02anrn =

Independent of the orbital angular momentum of the electron, the

frequency of a photon emitted is:

hvEE if =−

3,... 2, 1,1

2 20

2

=⎟⎠

⎞⎜⎝

⎛−= nna

keEn

energy quantization therefore follows from angular momentum conservation.

therefore the ionization energy is 13.6 eV. It takes 13.6 eV to liberate an electron in the ground state. At n=infinity, it is removed.

This also agreed with experiment.

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−=

220

2 11

2 if

fi

nnha

ke

h

EEν

1. The laser in its non-lasing state

2. The flash tube fires and injects light into the ruby rod. The light excites atoms in the ruby.

3. Some of these atoms emit photons.

4. Some of these photons run in a direction parallel to the ruby's axis, so they bounce back and forth off the mirrors. As they pass through the crystal, they stimulate emission in other atoms.

5. Monochromatic, single-phase, collimated light leaves the ruby through the half-silvered mirror -- laser light!

Note that the difference in energies of allowed orbits becomes smaller as n becomes larger.

In the limit of large quantum numbers, the frequencies and the intensities of radiation calculated from classical

theory must agree with quantum theory.

In the limit of large quantum numbers, the frequencies and the intensities of radiation calculated from classical

theory must agree with quantum theory.

At the beginning of class, I said that Bohr was mostly right…so

where did he go wrong?

• Failed to account for why some spectral lines are stronger than others. (To determine transition probabilities, you need QUANTUM MECHANICS!) Auugh!

• Treats an electron like a miniature planet…but is an electron a particle…or a wave?