# IB Physics/Quantum and Nuclear Physics
## 13.1 Atoms and their constituents
### 13.1.1

Millikan's oil drop experiment involved first producing small droplets of oil with an atomiser. Some of these then fell through a small hole, and into a region between two fields. Using a variable resistance, the strength of the field was adjusted until the upward force of the field (the positive plate was above, negative below, but could be reversed for drops of opposite charge) equals the downward force of gravity. After it was balanced, the potential difference was recorded, and then the drop was allowed to fall. After falling some distance, the drop falls at a constant speed (where the force of gravity is equaled by the air resistance). This terminal speed is measured, and allows the mass to be found using Stoke's law.
Then, we can equate F$_g$ and F$_f$, as follows : qE = mg therefore q = $^{mg}$/$_E$.
Since the mass of the drop can be found, and both g and E are known, we can find the charge of the drop. By graphing these charge values, we can find that the smallest difference between them is e, 1.6 x 10$^{-19}$C.

### 13.1.2

Thompson's experiment is based around the counteracting effects of electric and magnetic fields.
First, the two fields are adjusted so an electron beam passes undeflected between them, meaning the deflection effects of each cancel out. Therefore, F$_e$ = F$_b$, and so Eq = Bqv. From this, v = $^E$/$_B$.
The electric field is then removed, meaning the beam is deflected by the magnetic field. Since the force from a magnetic field = evB, from F$_c$ = $^{mv2}$ / $_r$ we get evB = $^{mv2}$ / $_r$. This simplifies down to $^e$/$_m$= $^v$/$_{Br}$. We can measure the radius of curvature, we know the velocity from above, and we know the strength of the magnetic field (B), and so the value of $^e$/$_m$can be found.

### 13.1.3

The distance of closest approach of a particle to the nucleus can be found by the conservation of energy. The initial energy of the particle is defined by E$_k$ = $^1$/$_2$mv$^2$. This energy is converted into potential energy as the charge approaches the nucleus. The potential energy at a given point is equal to the work done to move the charge to that point, so W = q$_α$V. V for a radial field is $^1$ / $_{4xπxEo}$ x $^{qnucleus}$ / $_d$, so the potential energy is $^{qαqnucleus}$ / $_{4xπxEoxd}$. When this is equated with the kinetic energy, all the terms are known except the distance, which can therefore be found.

## 13.2 Nuclei and their constituents
### 13.2.1

A mass spectrometer works as follows : First we start with a source of ions, all with a +1 charge. These are accelerated through an electric field. These ions then enter a velocity selector, an area with both magnetic and electric fields applying a force in opposite directions, so as to cancel each other out for a particular mass. As above the two forces equal out, and we get F$_e$ = F$_b$, and so Eq = Bqv, therefore v = $^E$/$_B$. Thus, only ions with a particular speed are allowed through to the next stage.
The electric field ends, and the ions are deflected through a magnetic field, resulting in a circular path. This means there is a centripetal force supplied by the field, and so $^{mv2}$ / $_r$ = Bqv, which rearranges to m = $^{Bqr}$/$_v$. Assuming the B field is varied to keep the radius constant, mass is proportional to magnetic field strength, as everything else is constant.

### 13.2.2

Chadwick discovered the neutron by bombardment of Beryllium with alpha particles. A wax block (with its chain of hydrocarbons it serves as a proton source) is placed after a short distance, and protons are detected after the wax block. Thus there must be some way to 'knock out' the proton. From the energy the protons are in, the Compton effect due to possible gamma ray emission is too low to compensate. If the particle that alpha knocked out is a neutron though, it fits perfectly both by the conservation of energy and the conservation of momentum (since the proton and neutron have approximately same mass, we would expect the 'pool players' result') yet these particles are indeed uncharged. Further more, when these are to collide with protons in cloud chambers, a right angle track would result.

### 13.2.3

The mechanisms for beta decay :
* p -> n + e$^+$ + a neutrino (Greek letter nu, lowercase) and
* n -> p + e$^-$ + an antineutrino (Greek letter nu, lowercase, with a bar on top)

### 13.2.4

For decay chains, everything can be written as a series of nuclear equations as in the following example.
$^A$$_Z$X + $^0$$_{-1}$e -> $^A$$_{Z-1}$Y. Note that the totals must be kept constant on both sides, but that should be fairly easy.

### 13.2.5

The equation N = N$_o$ x e$^{-λxt}$ is in the data book, while $^{ΔN}$/$_{Δt}$ = -λ x N is not.
Both those equations can be used to find decay information. N is the number of radioactive nuclei, N$_o$ is the starting number and λ is defined as $^{ln2}$/$_{halflife}$ (that's in the data book). $^{ΔN}$/$_{Δt}$ is the rate of decay.

### 13.2.6

Deduce λ = $^{ln2}$/$_{halflife}$ as follows.
We start by taking a time after one half life, therefore N = N$_o$/2. N = N$_o$ x e$^{-λxt}$ becomes 2 = e$^{λxt}$. We then take the natural logarithm of both sides to get ln2 = λ x t$_{0.5}$, and this rearranges to the required expression.

### 13.2.7

By Coulomb's law, the protons in the nucleus should all repel each other and break it to bits. Clearly there must be another force holding it all together, which we call the strong nuclear force. This is a force which greatly outweighs the electromagnetic repulsion, but only acts over a very small distance (within the nucleus).

## 13.3 Energy changes within atoms
### 13.3.1

Atomic emission and absorption spectra result from the fact that electrons can move between energy levels when they have sufficient energy put in. They will then fall back, emitting a defined amount of energy as light. Emission spectra result from electrons being excited by electricity (or something), then emitting light as the electrons fall back. Absorption spectra result from electrons absorbing energy from electromagnetic radiation, and so effectively blocking it. In the emission case, there will be a series of thin bands representing the wavelengths of light produced, and for absorption, there will be a full spectrum with some lines cut out (the wavelengths that were absorbed).

### 13.3.2

Bohrs postulates :
* Some stable orbits exists (assumed circular)
* Electrons absorb/emit energy in changing orbits
* Quantum condition (rules for changing orbit) : $mvr=n{\frac {h}{2\pi }},\ where\ n=1,2,3...$
The last formula can be derived by equating Angular momentum L with $n\hbar$ where $\hbar ={\frac {h}{2\pi }}$ (pronounced h-bar). Historically it is based on experimental discovery. We'll call this equation *1 in subsequent sections.

### 13.3.3

Describe the spectrum of atomic hydrogen and account for it, using Bohrs' model. (This account should be slightly more detailed than that required by the IB.)
The charge in the nucleus = Ze Charge in electron = e
Assuming a circular path, centripetal force = $^{mv2}$ / $_{radius}$. Since this force is supported by the electronic attraction, $^{mv2}$ / $_r$ = k $^{(Ze)(e)}$ / $_{r2}$ Simplifying we obtain r = k $^{(Z)(e2)}$ / $_{m(v2)}$. We'll call this equation *2
From above *1, mvr = n (hbar), thus rearranging we obtain v = n $^{(hbar)}$ / $_{mr}$. Plugging into *2, simplify and you'll obtain a equation, which relates the orbital radius with (constant)(n$^2$). Energy of the electron = KE + PE = $^1$ / $_2$ m(v$^2$)+ k $^{Z(e2)}$ / $_r$. Substitute v by expression *1, we'll arrive at E at nth shell = $^{constant}$ / $_r$ at nth shell. Since r is proportional to (n$^2$), E = $^{constant}$ / $_{n2}$. (required by IB)
Knowing this and E = $^{hc}$ / $_λ$ (Planck's equation), you'll also need to know the Rydberg equation. This is shown in the syllabus, and can be easily derived from the above.

### 13.3.4

Evaluate the success and limitations of Bohr's model
Success :
* Explains why atoms emit and successfully predict emission for Hydrogen.
* Explains why atoms absorb
* Ensures the stability of atoms
* Predicts accurately the ionisation energy for Hydrogen.
Limitations :
* Not successful for multi-electron atoms.
* Can't explain fine structures (emission lines existing as 2 or more close lines)
* Can't explain bonding of atoms in molecules or solids & liquids
* Can't explain different intensity of spectral lines.

## 13.4 Energy changes within nuclei
### 13.4.1

Einstein's mass-energy equivalence : $\Delta E=\Delta mc^{2}$, E is in joules, m in kilograms, and c is the speed of light in a vacuum.

### 13.4.2

Unified mass unit : One twelfth the mass of a carbon 12 atom.
Mass defect : The amount of mass which is converted into energy in a nuclear reaction.
Binding energy : The energy equivalent of the difference in mass between the nucleus of an atom, and the masses of the individual protons and neutrons which make it up.

### 13.4.3

The binding energy can be calculated as described above, by finding the total mass defect between the individual nucleons, and the whole nucleus. The binding energy per nucleon is therefore, this divided by the number of nucleons.

### 13.4.4

The graph of atomic number vs binding energy per nucleon runs from Z=2 increasing rapidly (with 3 peaks I don't think we need to worry about) to about Z=20 where it becomes relatively flat at around 8 Mev per nucleon then begins to drop off after Z = 60. The higher an element is (i.e. the more binding energy it has) the more stable it will be, and so the most stable elements are those around Z = 20.

### 13.4.5

Fission is the process by which an atom breaks up into smaller fragments. This is often caused by the addition of neutrons to the atoms, causing it to become unstable and eventually break up. This breaking up may, in some cases, produce more neutrons, and so these can then go on to produce more fission reactions, creating a chain reaction which perpetuates itself.

### 13.4.6

Fission is good because it provides a lot of energy from a source that is more viable long term than fossil fuels, and because it is relatively clean in terms of air pollution compared to fossil fuels. The down side is that it produces radioactive material which must be stored somewhere, and also, that it can be dangerous if not controlled properly (i.e. meltdowns).

### 13.4.7

Nuclear fusion occurs when two smaller nuclei fuse together to form one bigger, and more stable nucleus, and produce lots of energy in the process. Initiation of fusion requires a great deal of heat, because the nuclei must be given enough initial energy to overcome the Coulomb repulsion between them as they approach. Energy calculations can be made using E = mc$^2$, when the masses of the different fragments are given.

## 13.5 Interaction of matter and energy
### 13.5.1

The explanation of the photoelectric effect is that the energy carried by light is broken into discrete units, the size of which depends on the frequency of the radiation. The energy carried in each 'photon' is defined as E = hf (plank's constant x frequency). The atoms require a certain amount of energy to release an electron, W$_o$ = hf$_o$, where W$_o$ is called the work function. If there is more energy than this, then that may be given to the electron in the form of kinetic energy, and so E = hf = W$_o$ + $^1$/$_2$x m x v$_{max}$$^2$.

### 13.5.2

The photoelectric effect can be measured by applying a stopping voltage in the opposite direction to the current induced by the photoelectron emission. As the frequency of the light is increased, more energy will be required to stop these electrons. If the frequency is decreased, however, there is eventually a point where no emissions occur, and so no voltage is required.
hf = hf$_o$ + eV$_s$, where V$_s$ is the stopping voltage.

### 13.5.3

X-rays are produced by first placing an anode and cathode in a vacuum tube. Behind the anode is some type of photo-sensitive material, and between the two is a potential difference of about 150,000v. The cathode is heated to produce thermo-electrons. These electrons the accelerate towards the anode. When the electrons are deflected, by coming close to the nuclei, their kinetic energy changes. This change results in the production of an X-ray.
Since the electrons may come close or stay far away from the nucleus, the x-ray spectrum is continuous, not discrete. There are, however, peaks caused by inner shell electrons being excited by small energy loss. These peaks occur on the left side of the curve, which is generally an inverted parabola type shape. There is a shortest wavelength possible for the x-rays due to the fact that when electrons lose all their kinetic energy, there is no way to further increase the wave frequency.

### 13.5.4

When electrons lose some of their energy, X-rays are produced. The short wavelength limit is when they lose all of it so eV = $^{hc}$/$_λ$ = $^1$/$_2$ mv$^2$. Rearranging, we get λ$_{min}$ = $^{hc}$/$_{eV}$
Apparently there was some other stuff I meant to come back to, but no longer remember.

### 13.5.5

DeBroglie's equation is λ = $^h$/$_p$. From this, we can see that all mass has a wave equivalent, and any wave has a mass equivalent. p refers to the momentum of the particle, f its frequency, and λ its wavelength.

### 13.5.6

The velocity of an electron can be found from $^1$/$_2$mv$^2$ = eV, and this can then be used in the equation mv=$^h$/$_λ$ to find the electron's wavelength. This can be seen or verified by the diffraction of electrons through thin crystals, showing that electrons have a wave nature.

## 13.6 Particle physics
### 13.6.1

Linear accelerators are designed based on a series of 'tubes' through which the particles are pulled, and then pushed by electric fields. The lengths of the tubes become longer and longer so the frequency with which the electric fields must oscillate is constant.

### 13.6.2

Circular particle accelerators work on the basis of magnetic fields making the particles rotate, and when they cross between the two Ds, they are accelerated between them by a electric field. The radius inside is defined by r = $^{mv}$ / $_{Bq}$, and so as the velocity increases, the magnetic field must be increased to keep the radius constant.

### 13.6.3

When a particle having been accelerated collides with a fixed target, both usually break up into smaller fragments. These can sometimes be identified in a cloud chamber.

### 13.6.4

Particle anti-particle pairs are only really produced from interactions involving great amounts of energy. When two such particles collide, they completely annihilate, producing only energy.

### 13.6.5

Particle / anti-particle pairs :
* electrons - positrons (positrons are uncommon).
* proton - antiproton (these are the same except for charge)
* photon - photon (same particle)