Particles, Antiparticles, and Photons

Key Words

Wave-Particle Duality, Photoelectric Effect, Photon, Quanta, Threshold Frequency, eV, Particles, antiparticles

In the exam you are expected to know :

  • That for every type of particle, there is a corresponding antiparticle. You should know that the positron, the antiproton, the antineutron and the antineutrino are the antiparticles of the electron, the proton, the neutron and the neutrino respectively.

  • Comparison of particle and antiparticle masses, charge and rest energy in MeV.

  • Photon model of electromagnetic radiation, the Planck constant, E = hf = λ/hc

  • Knowledge of annihilation and pair production processes and the respective energies involved. The use of E = mc2 is not required in calculations.

 

Particles and Antiparticles

Particle physics is concerned with fundamental particles, which means that the particles can't be broken down any further..  It used to be thought that protons, neutrons and electrons were the fundamental particles of matter.  However it has been found that nucleons (proton and neutron) are made up of smaller particles, so nucleons are now not fundamental. 

 

 

Each particle has an antiparticle.  However, antiparticles are not found in normal matter, but arise in:

We should note the following:

Antiparticles can be made in large quantities in accelerators, resulting from high-energy collisions.  They have short lifetimes, about 10-10 s because when they meet their equivalent particle, they annihilate each other in a burst of energy.   It is even possible to make simple antiatoms. 

It is thought that there is more matter than antimatter in the Universe.   It is not impossible that antimatter exists in large quantities somewhere, and that there are antimatter stars and planets.  None have yet been detected. 

Question 1

 Complete this table.  Some have been done as an example:                         

Particle

Mass compared with proton

Charge

Electron

1/1800

-1e

Positron

   

Proton

   

Antiproton

   

Neutron

   

Antineutron

   

ANSWER

 

Observing Smaller and Smaller Objects

Whenever we observe something, we need three different pieces of apparatus:

If you read a book, your eyes detect the changes caused by the effect of ink on paper (the object) in response to light (radiation). 

However the light is limited by its wavelength to resolving objects about 1 mm across.  Much less than that, then diffraction becomes important.  Waves will not travel through a gap less than a wavelength.  Light wavelengths cannot resolve atoms.  Shorter wavelengths can be used but the eye cannot detect these.

X-rays can be used to resolve individual atoms by X-ray crystallography. 

The wave properties of electrons can be harnessed in electron microscopy.  We can resolve the structure of individual molecules, but not the structure of the atoms themselves.

  The key point to bear in mind at the microscopic level is that:

E = hf

E - energy (J), f - frequency (Hz), and h - Planck's constant (6.63 × 10-34 Js)

To resolve the structure of atoms we need a very powerful microscope, several metres long.  This gives us very high-energy particles with a very short de Broglie wavelength.  Louis de Broglie argued that if light and other radiations could be considered as waves with particle properties, it was entirely reasonable to state that particles could have wave properties.  His experimental work led to the combination of the wave equation and Planck's equation:

We will discuss this more below.

 

The two pictures show a light microscope and an electron microscope.

Energies of particles are expressed in electron-volts (eV) where:

1 eV = 1.6 x 10-19 J

To see various levels of detail requires the following kinds of particle energies

Therefore to resolve parts of the nucleus needs very high particle energies to gain short de Broglie wavelengths.  However, there's a problem.  At these energies the particles have an unfortunate habit in smashing up the nuclei; a bit like asking a bull for a commentary on fine antique porcelain.

Question 2

What is the energy in joules of the following electron energies? 

ANSWER

From these energies in joules, we can work out the speeds at which the electrons travel using v2 = 2 Ek/m.  Mass of an electron = 9.11 × 10-31 kg.

Question 3 

What is the speed of an electron at an energy of 100 eV and at 10 GeV? 

ANSWER

The last answer gives us a speed of an electron that is faster than the speed of light.  In fact we cannot go faster than the speed of light.  A different (and more complex) equation is needed as the speed of the electron gets towards the speed of light.

We can also get short de Broglie wavelengths using heavier particles like alpha particles.

 

 

 

Light as a Particle?

Historically there had been a lot of controversy about the wave nature of light, as proposed by the Dutch physicist Hans Huygens, against the corpuscular model as proposed by the headstrong Isaac Newton.   The concept of wave-particle duality was the start of modern physics in the middle to late Nineteenth Century.

  We know that light shows wave properties such as:

·        Reflection

·        Refraction

·        Diffraction

·        Polarisation

 

However it can also be shown to have particulate properties as well.  Consider this model:

If we spray just a short burst, we get just a few spots on the screen:

The longer we spray, the more spots appear until the whole area is covered in paint:

When using a spray can, we don’t notice any diffraction effects as the particles pass through the stencil.  Hardly surprising as the paint droplets are particles, not waves.

  Now, if we expose a piece of photographic paper to a short burst of light we will see:

The intensity of the image on a photographic plate increases the longer the paper is exposed for.  That intensity is determined by the number of silver grains deposited.  We see that the pattern of silver grains deposited is random.  It seems that the light that deposited the grains was actually made of particles.

The debate raged on until the discovery in the late nineteenth century with the discovery of the photoelectric effect.

 

The Photoelectric Effect.

The concept of wave-particle duality was the start of modern physics in the middle to late Nineteenth Century.   We can show the photoelectric effect with apparatus like this:

1.      We charge the electroscope with a negative charge.

2.      We expose the reactive metal to light of a long wavelength, e.g. red.

3.      We observe that there is no effect, however bright the light.

4.      We then expose the metal to short wavelength light, e.g. UV.

5.      This time we see that the gold leaf drops down, showing that the electroscope is losing charge.

6.      It does not matter how bright or dim the UV light is.

7.      No effect was observed when the electroscope was positively charged.

The results were:

Metal

X-rays

Ultra-Violet

Blue Light

Red Light

Magnesium

P

O

O

O

Zinc

P

P

O

O

Sodium

P

P

P

O

Caesium

P

P

P

P

This led to the conclusion that:

Question 4

Why do these results suggest that light is not a wave?   

ANSWER

These findings led to the notion of light being tiny little packets of wave energy called photons.

Further work by Max Planck in 1900 produced the Photon Model of Electromagnetic Radiation.  We can sum this up in the following points:

  E = hf

[E – energy in J; h – Planck’s Constant; f – frequency of the radiation in Hz]

The constant h is Planck’s Constant with the value 6.6 ´ 10–34 Js (joule seconds, NOT joules per second).

We can combine the equation above with the wave equation:

                                    E = hf              and             c = fl

 

Þ            E = hc

                                                     l

 

Question 5

What is the photon energy of light wavelength 350 nm?

ANSWER

The joule is the SI unit for energy.  However atomic physicists find the joule far too big and clumsy.  (You would not measure the width of your desk in kilometres.)  So they use a unit called the electron volt (eV). The electron volt is the amount of energy used when a charge of electronic charge passes through a potential difference of 1 volt.

The charge on an electron is 1.6 × 10-19 C, so 1 eV = 1.6 × 10-19 J.

Question 6

Convert your answer to Question 2 to electron volts.   

ANSWER

  Electron volts are almost always used in atomic and nuclear physics, but before using equations like E = hf, the energies MUST be converted to joules.  This is a very common bear trap.

Question 7

A photon has an energy of 10.3 eV.  What is its wavelength?  Where on the electromagnetic spectrum would this be? 

ANSWER

     

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