Atomic and Nuclear Physics
Section 1: The Atom
(A historical overview…)
The Atom
The earliest references to the concept of atoms date back to ancient India in the 6th century BCE.
In approximately 450 BCE, Democritus coined the term átomos (Greek: ἄτομος), which means "uncuttable"
The ancient Greeks first developed the idea of atomic theory and thought of atoms as being the smallest building blocks of matter. They considered the idea of taking an amount of a substance, such as water, and dividing it into smaller portions. They knew that when a cup of water was poured into two smaller cups, the two smaller portions of water would have the same properties as the initial cup – it would still be the same substance. However, they thought that there would be a limit to how many times you could go on dividing the water. Eventually, they concluded, you would have the smallest amount of water possible that could not be divided any further while still having the properties of water
The Atom
In 1661, natural philosopher Robert Boyle suggested that matter was composed of various combinations of different "corpuscles" or atoms, rather than the classical elements of air, earth, fire and water.
In 1789 the term element was defined by the French nobleman and scientific researcher Antoine Lavoisier to mean basic substances that could not be further broken down by the methods of chemistry.
Robert Boyle (1627-1691)
Antoine Lavoisier
(1743-1794)
A replica of Lavoisier's laboratory at the Deutsches Museum in Munich, Germany. The large lens in the center of the picture was used to focus sunlight in order to ignite samples during combustion studies.
The Atom
Dalton used the concept of atoms to explain why elements always react in a ratio of small whole numbers - the law of multiple proportions - and why certain gases dissolve better in water than others.
John Dalton (1627-1691)
In 1803, English instructor and natural philosopher John Dalton proposed that each element consists of atoms of a single, unique type, and that these atoms can join together to form chemical compounds.
In 1897, he physicist J. J. Thomson, through his work on cathode rays, discovered the electron and its subatomic nature, which destroyed the concept of atoms as being indivisible units.
Thomson believed that the electrons were distributed throughout the atom, with their charge balanced by the presence of a uniform sea of positive charge named the the plum pudding model.
J. J. Thomson
(1856-1940)
The idea of a model being perfectly correct or not does not really matter, since the purpose of a model is to simplify a concept to make it easier to understand.
The Gold-Leaf Experiment
1909
However, in 1909, Geiger and Marsden, two researchers under the direction of physicist Ernest Rutherford, bombarded a sheet of gold foil with helium ions and discovered that a small percentage were deflected through much larger angles than was predicted using Thomson's proposal.
Ernest Rutherford
(1871-1937)
Hans Geiger
(1882-1945)
Ernest Marsden
(1889-1970)
The model of the day
How can we see ‘inside’ something when it is too small to see?
How do charged particles pass through a solid?
Sitting and waiting, looking for any green flashing light…
“It is almost as incredible as if you had fired a fifteen-inch shell at a sheet of tissue paper, and it came back and hit you.”
Rutherford
CERN: still using the same ideas
Consequences of the Rutherford’s experiment
All of an atom's positive charge and most of its mass is concentrated in a tiny core. Rutherford called this the nucleus.
The electrons surround the nucleus, but they are at relatively large distances from it.
The atom is mainly empty space!
What next?
Relative size of the nucleus and electron cloud
Rutherford’s model of the atom
Using this model Rutherford calculated that the diameter of the gold nucleus could not be larger than 10-15 m.
Other experiments confirmed the existence of a nucleus inside the atom – a small, massive object carrying the positive charge of the atom.
The force that would keep the electrons in orbit was the electrical force between electrons and the positive nuclear charge – Coulomb’s force.
Can you see any problems here?
Rutherford’s model of the atom
According to the electromagnetic theory an accelerated charge would radiate electromagnetic waves and thus lose energy.
The electrons move in circular paths around the nucleus. But if they radiate and lose energy, then they would fall towards the nucleus. Because of this, Rutherford’s model cannot explain way matter is stable, i.e., why atoms exist.
Bohr’s model of the atom
The first attempt to solve the problem with Rutherford’s model came from Niels Bohr, a Danish physicist, in 1911.
Bohr revised Rutherford's model by suggesting that the electrons were confined into clearly defined orbits, and could jump between these, but could not freely spiral inward or outward in intermediate states.
An electron must absorb or emit specific amounts of energy to transition between these fixed orbits.
Niels Bohr (1885-1962)
Bohr’s postulates
Modelling the electrons as standing waves shows us why there are distinct shells. Standing waves must start and end with zero amplitude (assuming fixed ends). Our model is of a string in which the two fixed ends are joined up.
Bohr’s postulates
Light is Quantized
Photons of light can only have certain ______________ values of energy
discrete
Energy of a Photon
Planck’s Constant | h | 6.63 × 10-34 J s |
Energy
[J]
Frequency
[Hz]
Energy of a Photon
Bohr’s postulates
The energy level diagram of the hydrogen atom according to the Bohr model
Energy eV
-13.6
0
Electron can’t have less energy than this
n = 1 (the ground state)
n = 2
n = 3
n = 4
n = 5
High energy n levels are very close to each other
Bohr’s postulates
An electron in a higher state than the ground state is called an excited electron.
High energy n levels are very close to each other
n = 1 (the ground state)
n = 2
n = 3
n = 4
n = 5
-13.6
Energy eV
0
electron
Bohr’s postulates
An electron in an excited state can make a transition to a lower state. Thus an atom in state n = 2 can go to n = 1 (an electron jumps from orbit n = 2 to n = 1)
n = 1 (the ground state)
n = 2
n = 3
n = 4
n = 5
-13.6
Energy eV
0
electron
Wheeee!
Bohr’s postulates
When an electron drops to a lower level, a single photon of light is emitted.
n = 1 (the ground state)
n = 2
n = 3
n = 4
n = 5
-13.6
Energy eV
0
electron
Bohr’s postulates
The energy of the photon is equal to the difference in energy (ΔE) between the two states. It is equal to hf. E = hf
n = 1 (the ground state)
n = 2
n = 3
n = 4
n = 5
-13.6
Energy eV
0
electron
E = hf
Bohr’s postulates
The OPPOSITE happens when a photon of light is absorbed.
n = 1 (the ground state)
n = 2
n = 3
n = 4
n = 5
-13.6
Energy eV
0
electron
E = hf
Where is the evidence for Bohr’s ideas?
This is an example of white light being split into all wavelengths it contains, but some light is made of only specific wavelengths.
Where is the evidence for Bohr’s ideas?
If you split light from glowing hydrogen gas you see the following wavelengths - why?
Where is the evidence for Bohr’s ideas?
Kirchoff spotted a clue in that the emission and absorption lines were clearly related - why?
Practice question
Quick Recap of eV
Unit of Energy
Try This…
Calculate the energy carried by one photon of microwaves of wavelength 9 cm (as might be used in wifi signals) in J and eV
0.09 m
Energy Levels
1
2
3
4
5
6
7
Electrons in an atom exist at discrete energy levels
Absorbed
Emitted
Energy Levels
E4
E3
E2
E1
A photon is emitted whenever an electron transitions from one energy level down to a lower energy level
How many different transitions are possible between these four energy levels?
6
Energy Levels
n = 1
-13.6 eV
n = 2
-3.40 eV
n = 3
-1.51 eV
n = 4
-0.85 eV
n = 5
-0.54 eV
n = ∞
0.00 eV
Excited States
Ground State
Energy Transitions
n = 1
-13.6 eV
n = 2
-3.40 eV
n = 3
-1.51 eV
n = 4
-0.85 eV
n = 5
-0.54 eV
n = ∞
0.00 eV
Lyman Series (UV)
Balmer Series (Visible)
Paschen Series (IR)
Different Energy transitions result in different energies (wavelengths) of light that are absorbed or emitted
Absorption Spectrum
The emission and absorption spectra are negative images of each other
Calculating Wavelength Emitted
n = 1
-13.6 eV
n = 2
-3.40 eV
n = 3
-1.51 eV
n = 4
-0.85 eV
n = 5
-0.54 eV
n = ∞
0.00 eV
What is the wavelength emitted?
Try This…
n = 1
-13.6 eV
n = 2
-3.40 eV
n = 3
-1.51 eV
n = 4
-0.85 eV
n = 5
-0.54 eV
n = ∞
0.00 eV
What is the wavelength emitted?
Working Backwards…
What is the energy in eV for a 434 nm blue emission line?
434 × 10-9 m
Working Backwards…
n = 1
-13.6 eV
n = 2
-3.40 eV
n = 3
-1.51 eV
n = 4
-0.85 eV
n = 5
-0.54 eV
n = ∞
0.00 eV
Draw in the Energy Transition for a 434 nm blue emission line?
What transition has an energy difference of 2.86 eV?
Our spectrum tubes
Is the new model correct?
Atomic and Nuclear Physics
Probing the nucleus
Nuclear notation
To their great surprise, they found that some alpha particles (1 in 20 000) had very large scattering angles
Geiger-Marsden Alpha Scattering
Alpha particles
Thin gold foil
Small-angle scattering
Large-angle scattering
The results suggested that the positive (repulsive) charge must be concentrated at the centre. Most alpha particles pass undisturbed, only alpha particles passing very close to this small nucleus get repelled (the nucleus must also be very massive for this to happen).
Explaining the results...
Remember on this scale, if the nucleus is 2 cm wide, the atom would be 200 m wide!
The results suggested that the positive (repulsive) charge must be concentrated at the centre. Most alpha particles pass undisturbed, only alpha particles passing very close to this small nucleus get repelled (the nucleus must also be very massive for this to happen).
Explaining the results...
Remember on this scale, if the nucleus is 2 cm wide, the atom would be 200 m wide!
what is the percentage that are deflected?
Rutherfords calculations...
Rutherford calculated theoretically the number of alpha particles that should be scattered at different angles (using Coulomb’s law). He found agreement with the experimental results if he assumed the atomic nucleus was confined to a diameter of about 10-15 metres.
Using the idea of energy conservation, it is possible to calculate the closest an alpha particle could get to the nucleus during a head-on collision.
Closest Approach method and what this tells us
Alpha particle
nucleus
Initially, the alpha particle has kinetic energy = ½mu2
Closest Approach 2
K.E. = ½mu2
At the point of closest approach, the particle reaches a distance b from the nucleus and comes momentarily to rest.
Closest Approach 3
b
K.E. = 0
All the initial kinetic energy has been transformed to electrical potential energy.
Closest Approach 4
K.E. = 0
b
Using the formula for electrical potential energy which is derived from Coulomb’s law
Kinetic energy = Electrical potential energy (VQ)
Closest Approach 5
K.E. = 0
b
For an alpha particle, m = 6.7 x 10-27 kg, q1 = 2 x (1.6 x 10-19 C) and u is around 2 x 107 m.s-1. If the foil is made of gold, q2 is 79 x (1.6 x 10-19 C).
What is r?
Closest Approach 6
For an alpha particle, m = 6.7 x 10-27 kg, q1 = 2 x (1.6 x 10-19 C) and u is around 2 x 107 m.s-1. If the foil is made of gold, q2 is 79 x (1.6 x 10-19 C).
Therefore r (or b) = 2.7 x 10-14 m
Closest Approach 6
Determine the closest distance that an alpha particle of energy 1.37 MeV could approach to:
Why were alpha particles were used in the Geiger and Marsden experiment? What would have been observed if electrons were used? or neutrons?
Questions
R = R0A⅓
R = nucleus radius
R0 = Fermi radius = 1.2 x 10-15 m
A = nucleon number
using this find the density of a helium atom
Radius of Atomic Nucleus
Estimate the radius of:
Questions
Questions
(c) Nuclear density is significantly greater than atomic density. What three things does this suggest about the structure of an atom?
At high energies the scattered intensity departs from predictions using the Coulomb force as the principal scattering force. Why?
High Energy Rutherford Scattering - corrections to the closest approach method
At high energies the alpha particles are getting close enough to the nucleus for the strong nuclear force to begin to have an effect.
High Energy Rutherford Scattering - corrections to the closest approach method
In order to investigate the size of the nucleus in more detail, high energy electrons are used instead.
High Energy Rutherford Scattering - corrections to the closest approach method
High Energy Rutherford Scattering - corrections to the closest approach method
sinθ ≈ λ/d
High Energy Rutherford Scattering - corrections to the closest approach method
sinθ ≈ λ/d
High Energy Rutherford Scattering - corrections to the closest approach method
We will return to this after quantum mechanics
If the electrons have an energy of 400 MeV, λ = hc/E (from de Broglie) = 3.1 x 10-15m. If first minimum is at 35°, what is the radius of the nucleus?
High Energy Rutherford Scattering - corrections to the closest approach method
If the electrons have an energy of 400 MeV, λ = hc/E (from de Broglie) = 3.1 x 10-15m. If first minimum is at 35°,
d ≈ λ/sinθ ≈ 5.4 x 10-15. So radius ≈ 2.7 x 10-15 m.
High Energy Electron Scattering