The Bohr model: success

and failure

Applying the photon model to

electronic structure

The emergence of the

quantum world

Learning objectives

Describe the basic principles of the Bohr

model

Distinguish between the “classical” view

and the “quantum” view of matter

Describe Heisenberg Uncertainty principle

and deBroglie wave-particle duality

Calculate wavelengths of particles

Bohr’s theory of the atom: applying

photons to electronic structureElectrons occupy specific levels (orbits) and no others

Orbits have energy and size

Electron excited to higher level by absorbing photon

Electron relaxes to lower level by emitting photon

Photon energy (hν) exactly equals gap between levels

– Gap ↑, ν ↑

Larger orbits are at higher energy – larger radius

Size of energy gap determines

photon energy

Small energy gap, low

frequency, long

wavelength (red shift)

High energy gap, high

frequency, short

wavelength (blue shift)

Each set of lines in the H spectrum comes from transitions from all the higher levels to a particular level.

The lines in the visible are transitions to the second level

2 2

1 2

1 1 1HR

n n

The full spectrum of lines for H

Successes and shortcomings of Bohr

Could not explain why these levels were allowed

Only successful agreement with experiment was with the H atom

Introduced connection between spectra and electron structure

Concept of allowed orbits is developed further with new knowledge

Nonetheless, an important contribution, worthy of the Nobel prize

Waves, particles and diffraction

Electrons are waves too!

Life at the electron level is very different

Key to unlocking the low door to the secret garden of the atom lay in accepting the wave properties of electrons

De Broglie wave-particle duality

All particles have a wavelength –wavelike nature.

– Significant only for very small particles – like electrons

– As mass increases, wavelength decreases

Electrons have wavelengths about the size of an atom

– Electrons are used for studying matter – electron microscopy

De Broglie relation

E = mc2 m = E/c2

But... E = hc/λ, so m = h/cλ– h = Planck’s constant = 6.626 x 10-34 m2kg/s (Js)

– For electron: m = 9 x 10-31 kg, v = 2 x 106 ms-1

– λ = 3 x 10-10 m (0.3 nm)

The electron’s wavelength is of the order

of the atomic diameter (0.1 – 0.5 nm)

mv

h

Electron microscopes can peer

within and provide resolution where

visible light fails

Wavelengths of large objects

Should we be concerned about the wave-

particle nature of large objects?

Consider a baseball pitched at 100 mph.

What is the wavelength of the ball?

– Use m = 100 g, v = 50 m/s, h = 7 x 10-34 Js

– λ = 10-34 m

For normal size objects, the wavelength

will be immeasurably and irrelevantly small

mv

h

Quantum effects: when should we

care?

The Correspondence Principle states that

quantum effects disappear when Planck’s

constant is small compared to other

physical quantities

Relating this to Bohr:

Standing waves and strings

Strings of fixed length

can only support

certain wavelengths.

These are standing

waves.

The Bohr orbits revisited

The allowed orbits have a circumference equal

to a fixed number of wavelengths

All others disappear via destructive interference

Orbit has

exact

number of

wavelengths

OK

Orbit has

inexact

number of

wavelengths

BAD

Heisenberg Uncertainty Principle:

the illusive electronWe can exactly predict the motion of a ball– Newton’s laws are deterministic

But not an electron

Heisenberg Uncertainty Principle

The position and momentum of a particle

cannot be measured simultaneously to

unlimited accuracy

Δx Δp > 0

Locating the electron: catching a

goldfish in a bowl

The act of “seeing” an electron using

photons changes electron’s energy,

thereby changing its position

As the object increases in size, the impact

of the photon decreases

Limits precision of determining position

and momentum

Heisenberg Uncertainty Relation

In mathematical terms,

If the position is known precisely, Δx is small and

the uncertainty in momentum is large

If the velocity is known precisely, there is a high

uncertainty in the position

The electron will appear as a blur rather than a

sharp point

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hx m v

Implications for the electron

m = 9 x 10-31 kg, v = 2 x 106 m/s

If uncertainty in v is 10%,

– mΔv = (9 x 10-31) x (2 x 105) kgm/s

– = 18 x 10-26 kgm/s

– Δx ≥ h/(4π•18 x 10-26) m

– ≥ 6 x 10-34/(4π•18 x 10-26) m

– ≥ 3 x 10-10 m or 300 pm

Diameter of the H atom is about 100 pm

vm

hx

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