Successes of the Bohr model

Explains the Balmer formula and predicts the

empirical constant R using fundamental constants:

21

2'11

nnR

Explains the spectrum for other single electron atoms like singly ionized helium or double ionized lithium.

Predicts the approximate size of the hydrogen atom (orbit radius)

Sort of explains why atoms emit discrete spectral lines

Sort of explains why electrons don’t spiral into the nucleus.

• Why is angular momentum quantized?

• Why don’t electrons radiate when they are in fixed orbitals?

• How does electron know which level to jump to? (i.e. how to predict intensities of spectral lines)

• Can’t be generalized to more complex (multi-electron) atoms

• Shapes of molecular orbits and how bonds work

• Can’t explain doublet spectral lines

Remaining issue with the Bohr model

A. Photons have momentum of h/.

B. The angular momentum of atomic electrons is quantized

C. Matter particles have a wavelength

D. The position and momentum of an object cannot both be measured to arbitrary accuracy at the same time.

E. None of the above

Reading quiz (no discussion)

De Broglie proposed which of the following?

Compton showed that photons have momentum (A)

Bohr postulated quantization of angular momentum (B)

Heisenberg came up with the uncertainty principle (D)

de Broglie Waves• In 1923, French grad student Louis de Broglie

suggested that maybe electrons are actually little waves going around the nucleus.

• This seems plausible because…

– Standing waves have quantized frequencies, might be related to quantized energies.

– Einstein had shown that light, typically thought of as waves, have particle properties. Might not electrons, typically thought of as particles, have wave properties?

So going around the circle must take an integral number of wavelengths.

A. r = B. r = nC. r = nD. 2r = nE. 2r = /n

n = 1, 2, 3, …

Standing waves in a ring

Just like a standing wave on a string but now the two ends of the string are joined together.

What are the restrictions on the wavelength?

Circumference is 2r so the condition is 2r = n.

If you start at a peak and go around the circle, you must end up at a peak. Otherwise you will not have a standing wave.

Think of this as constructive interference of the wave with itself, which is needed for it to not destructively interfere to zero amplitude.

1

23 4

5

6

7

8910

What is n in this picture? How many wavelengths does it take to complete a loop?

n=1 n=2 n=3

n=10

= node = fixed point that doesn’t move.

The de Broglie wavelength:

What is the wavelength of an electron?!

For photons we know how to relate momentum and wavelength

de Broglie proposed the same relationship for massive particles

(momentum)

p

(wavelength)

ph /

Given the deBroglie wavelength (=h/p) and the condition for standing waves on a ring (2r = n), what can you say about the angular momentum L of an electron if it is a de Broglie wave?

A. L = n/rB. L = nC. L = n/2D. L = 2n/rE. L = n/2r

Angular momentum: prL

momentum: vmp

Remember = h/2

Substitute de Broglie wavelength =h/p into the standing wave equation 2r=n to get

pnhr 2

Can rearrange this as 2nhrp which is nL

So the de Broglie wavelength for an electron going around a nucleus is the same as (and basically explains) the quantization of angular momentum proposed by Bohr and therefore essentially explains quantization of energy.

de Broglie waves

So the idea of particles also being waves explains one of Bohr’s postulates (quantization of angular momentum)

But is this idea correct? How to tell?

To tell if something is a wave we look for interference.

Davisson-Germer experiment.

Davisson – Germer experiment

Davisson and Germer.

Two slit interference with light

Huygen’s Principle: waves spread as spherical waves.

Double-slit experiment

0.5 mm =d

r1

r2

H

L

d

r = d sin~ d = m H = L sin~ L

H = mL d

bright

bright

bright

r = r2-r1

r = m (where m=1,2,3…)

5 x10-4m = d

r1

r2

r = r2-r1

r = m (where m=1,2,3…)

r = m = d sin= d = m

H

L

H = L m = 1, = 500 nm, so angle to first bright = λ/d = 500 x 10-9/(5 x 10-4) = 0.001 radif L = 3 m, then H = 3 m x 0.001 = 3 mm.

Calculating the pattern for light

So what will the pattern look like with electrons?

Double-slit experiment

Massless particles (photons):/hcpcE

// hcEp

Ehcph //

Visible light photons:

m/s 103 8cvnm 500

Massive particles (electrons):

2

222

21

22 mh

mp

mvK

/2 hmKp

Low energy electrons:

Energy and momentum relationships

The lowest energy (useful) electrons are around 25 eV. We just found these electrons have a wavelength of 0.25 nm. If we use the same two slits as for visible light (d = 0.5 mm), how far apart are the m=0 and m=1 maxima on a screen 3 m away?

r = m = d sin= d = m H = L sin= L A. 3 mmB. 1.5 mmC.3 mD.1.5 mE. 3 nm

The wavelength of these electrons (0.25 nm) is 2000 times smaller than visible light (500 nm) so the angle and interference spacing is 2000 times smaller for the same slit spacing.

m 5.1mm 5.0

)nm 25.0)(1)(m 3( dLmH

This is too small to see. Need slits that are much closer.

Clue comes from X-ray diffraction…

Brilliant idea: Two slits are just two sources.

Hard to get two sources the size of an atom.

Using atoms for slits

Easy to get two objects that scatter electrons that are the size of an atom!

It is difficult to get just two atoms next to each other.

Just like reflection diffraction grating discussed for X-ray diffraction.

But multiple equally separated atoms are easy (crystal lattice) and work even better.

Using atoms for slits

Interference from electron scattering off very clean nickel surface.

ee

ee

e

e

e ee

e e

electrons scatter off nickel atoms

e det.

move detector around and see what angle electrons come off

Ni

Davisson – Germer experiment

Plot the results for number of electrons versus scattering angle and find…

ee

ee ee

e

e

e det.

Ni

# e’s

scatt. angle 5000

A peak!

So the probability of finding an electron at a particular angle is determined by the interference of de Broglie waves!

Davisson – Germer results

# e’s

scatt. angle 5000

1/p

To further prove the de Broglie wave hypothesis, they increased the electron energy. If de Broglie’s theory is correct, what will happen?

A. The peak will get largerB. The peak will get smallerC.The peak will shift to smaller angleD.The peak will shift to larger angleE. Nothing will happen.

ph / md

pdmh

dm

Increasing energy increases momentum which decreases the angle

Davisson – Germer tried this as well and it worked.

More on matter wavesTwo slit interference has been seen with electrons, protons, neutrons, atoms, and in just the last decade with Buckyballs which have 60 carbon atoms.

Electron diffraction, like X-ray diffraction can be used to determine the crystal structure of solids

Points come from a regular crystal. Rings come from many crystals randomly arranged.