6. Atomic and Nuclear Physics Chapter 6.4 Interactions of matter with energy.

6. Atomic and Nuclear Physics

Chapter 6.4 Interactions of matter with energy

Heinrich Hertz first observed this photoelectric effect in 1887. This, too, was one of those handful of phenomena that

Classical Physics could not explain. Hertz had observed that, under the right conditions, when light

is shined on a metal, electrons are released.

The Photoelectric effect

Photosurface

Photoelectrons

The photoelectric effect consists on the emission of electrons from a metallic surface by absorption of light (electromagnetic radiation).

An apparatus to investigate the photoelectric effect was set by Millikan and it allowed him to determine the charge of the electron.


When light falls on the surface, the electrons are removed from the metal’s atoms and move towards the positive cathode completing the circuit and thus creating a current.

http://phet.colorado.edu/simulations/sims.php?sim=Photoelectric_Effect

Whether the photoelectric effect occurs or not depends only on: The nature of the photosurface


The frequency of the radiation

When the intensity of the light source increases so does the current.

Current and intensity are directly proportional

A high current can be due to: Electrons with high speed Large number of electrons being emitted

To determine what exactly happens we need to be able to determine the energy of the emitted electrons.

This is done by connecting a battery between the photosurface and the collecting plate.


When the battery supplies a p.d. the charge of the collecting plate will be negative.

This means that the negatively charged electrons can be stopped if a sufficiently negative p.d. is applied to the electrodes.


G

collecting plate

evacuated tube

light

photosurface

electron


The electrons leave the photosurface with a certain amount of kinetic energy – EK.

To stop the electrons we must supply a potential difference (called stopping voltage) so that:

eVs = EK


The stopping voltage stays the same no matter what the intensity of the light source is.

This means that: The intensity of light affects the number of electrons emitted but not their

energy The energy of the electrons depends on the nature of light: the larger the

frequency, the larger the energy of the emitted electrons and thus the larger the stopping voltage

Critical of threshold frequency

The two graphs represent the EK of

the electrons versus frequency

These graphs tell us that: there is a minimum frequency fc, called

critical or threshold frequency, such

that no electrons are emitted.

if the frequency of the light source is less

than fc then the photoelectric effect does

not occur

the threshold frequency only depends on

the nature of the photosurface

The Photoelectric effect - Observations

1. The intensity of the incident light does not affect the

energy of the emitted electrons (only their number)

2. The electron energy depends on the frequency of the

incident light, and there is a certain minimum frequency

below which no electrons are emitted.

3. Electrons are emitted with no time delay – instantaneous

effect.


Problem:

According to Classical Physics, the electron should be

able to absorb the energy from light waves and

accumulate it until it is enough to be emitted.

Solution:

Einstein suggested that light could be considered particles

of light, photons, packets of energy and momentum or

quanta. The energy of such quantum is give by:

E = h f

where: f is the frequency of the e-m radiation

h =6.63x10-34J (constant known as Planck constant)


When a photon hits a photosurface, an electron will absorb

that energy.

However, part of that energy will be used to pull the

electron from the nucleus.

That energy is called the work function and represented

by Ф.

The remaining energy will be the kinetic energy of the free

electron.

So,

Ek = hf - Ф


Recalling that

EK = eVs

So,

eVs = hf – Ф

that is:

ef

e

hVs


When a photon hits a photosurface, 3 things can happen: The energy of the photon is not enough to remove the

electron nothing happens

The energy of the photon is just enough to remove the

electron: the photon’s energy equals the ionization energy

the electron leaves the atom without any Ek

The energy of the photon is larger than the ionization

energy the electron leaves the atom with Ek

Exercise:

1. What is the work function for the photosurface (in joules)?

2. What is the energy of the “green” photoelectron?

3. What is the speed of the “green” photoelectron?

4. What is the energy of the “blue” photoelectron?

5. What is the speed of the “blue” photoelectron?

Light: wave or particle

The photon has an energy given by E = hf

But if it is considered a particle it also carries momentum

p = m v

According to Einstein

E = m c2 ↔ m = E /c2

So,

p = (E /c2) c ↔ p = E / c

p = hf /c

p = h /λ

Light: wave or particle

Light can behave as a particle and the photoelectric effect is

evidence for that fact.

But if we do Young’s double slit experiment so that make

photons of light go through the slits one at each time, the

photon will produce an interference pattern.

Somehow, even when light behaves like a particle it

conserves its wave properties.

So, we talk about wave-particle duality.

De Bröglie’s wavelength

In 1923, Louis de Bröglie suggested that if

light can behave as a particle then particles

could have a wave associated to them.Louis de Broglie

The wave-particle duality or duality of matter can be applied

to matter and energy.

All particles have a wave associated to them so that

λ = h / p

Big particles have a wavelength so small that it can’t be

measured.

But small particles, like electrons, would have a wavelength

that is possible to be measured.

The electron as a wave

To have an obstacle with this size we must look at the structure of

crystals. The typical distance between atoms in a crystal is of the order

of 10-8m. When electrons are made to pass through crystals, they do

diffract thus proving its wave nature.

To prove that the electron behaves like a wave it must have wave properties,

such diffraction. To make an electron diffract around an obstacle of size d, its

wavelength λ must be comparable to or bigger that d.

An electron of mass 9.1x10-31kg and speed of 105 m/s will have a wavelength λ

= 7.2x10-9m.

Davisson and Germer experiment

In this experiment, electrons of

kinetic energy 54eV were

directed at a surface ok nickel

where a single crystal had been

grown and were scattered by it.

Using the Bragg formula and

the known separation of the

crystal atoms allowed the

determination of the

wavelength which has then

seen to agree with the De

Broglie formula.

Structural analysis by electron diffraction

6. Atomic and Nuclear Physics Chapter 6.4 Interactions of matter with energy.

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