Y2: Nuclear Physics Flashcards
What is the Rutherford scattering experiment
- A stream of (+) α-particles from a radioactive source are fired at a thin gold foil
- Alpha particles passed through the foil and struck a curved fluorescent screen (small, visible flash of light)
- The number of α-particles scattered at different angles was recorded
What would have been seen in the α scattering experiment if J.J.Thompson’s plum pudding model was correct
αparticles would have been deflected by small angles, as they were repelled by the electrons.
However, the experiment instead showed that most α-particles passed straight through, but some occasionally deflected at angles greater than 90 degrees, which can’t be explained by the plum pudding model.
What were the conclusions from Rutherford’s α-scattering experiment
- Most of the α-particles passed straight through the foil
∴ Most of the atom is empty space - Some α-particles are deflected at large angles
∴ Nucleus must have a large positive charge (to repel +α) - Only a few α-particles were deflected by large angles
∴ Nucleus must be tiny compared to the size of the atom - Fast moving α-particles (∴ high mv) are deflected
∴ Most of the mass of the atom must be in the nucleus
How were protons and neutrons discovered
- In 1919, protons were discovered by firing high-energy α-particles at different gasses
- ∴ It was though that the nucleus was made up of only protons, however, this theory didn’t work as the mass didn’t have the relationship with the charge that this would suggest (1:1)
- In 1920, Rutherford proposed a ‘proton-electron’ couplet to correct this error, but it was James Chadwick in 1932 that provided experimental evidence for the neutron to solve the issue.
How would you calculate the distance of closest approach to the nucleus of a scattered α-particle
Initial Ek = Qq/(4πεor)
(r: distance of closest approach)
- If an α-particle is deflected through 180 degrees by a nucleus, it will have stopped a short distance from the nucleus before rebounding
- At this point, all of the initial Ek will have been converted to electric potential energy
∴ Initial Ek = Final Ep = Qq/(4πεor)
∴ Initial Ek = Qq/(4πεor)
How can the distance of closest approach to the nucleus of a scattered α-particle be used to estimate the nuclear radius
Nuclear radius ≤ Distance of closest approach
(∴ r ≈ closest approach)
How can electron diffraction be used to determine an accurate value for the nuclear radius (R)
- Electrons are leptons, so don’t interact with the strong nuclear force
∴ They will get closer to the nucleus before they are diffracted, giving a more accurate value for R - The (De Broglie) wavelength of these electrons can be calculated: λ = hc/E
- To investigate nuclear radius, λ must be tiny (10^-15m)
∴ electrons must have high energy - If these electrons are directed at a thin film of material, a circular diffraction pattern will occur, the size of which can be used to determine R
(Sinθ = 1.22λ/2R)
What is the equation to calculate the nuclear radius by electron diffraction
Sinθ = 1.22λ/2R
(θ: Angle of the first minimum)
What is the radius of an atom
~0.05nm
What is the radius of the (smallest) nucleus
~1fm
What is the relationship between Radius of a nucleus and the number of nucleons within it
R = (Ro)A^1/3
R: nuclear radius
A: Nucleon number
Ro: Constant ≈ 1.4fm
What is the equation for nuclear density
p = 3m(nucleons) / 4π(Ro^3) = 1.45 x 10^17
The mass of all nucleons is roughly the same, given as: m(nucleon)
∴ Mass of nucleus = (A)m(nucleon)
Assuming the nucleus is spherical, V=(4/3)πR^3
p = m/V
∴ p = (A)m(nucleon) / (4/3)πR^3
R = (Ro)A^1/3
∴ p = (A)m(nucleon) / (4/3)π((Ro)A^1/3)^3
∴ p = 3m(nucleons) / 4π(Ro^3)
Why is the density of all nuclear matter constant
R = (Ro)A^1/3
∴ R^3 ∝ A
V = (4/3)πR^3
∴ R^3 ∝ V
∴ A ∝ V
∴ The volume of each nucleon is roughly the same
The mass of each nucleon is roughly the same
p = m/V
m = constant
V = constant
∴ p = constant
What is the value for nuclear density
p = 1.45 x 10^17 (kgm^-3)
What can be concluded from the difference between nuclear density and atomic density
Nuclear density (1.45 x 10^17)»_space; Atomic density (10^3-10^5)
∴ Most of the mass of an atom is in it’s nucleus
∴ Nucleus is small compared to the size of the atom
∴ Atom is mostly empty space
What is radioactive decay
The release of energy and/or particle from an unstable atomic nucleus until it becomes stable
(Individual decay is random and can’t be predicted)
What is Alpha radiation (α)
- Constituents: Helium nucleus (2p &2n)
- Relative charge: +2
- Relative mass: 4u
- Penetrating power: Stopped by Paper, skin, few cm of air
What is Beta-minus radiation (ꞵ-/ꞵ)
- Constituents: 1 Electron
- Relative charge: -1
- Relative mass: Negligible
- Penetrating power: Stopped by ~3mm of alluminium
What is Beta-plus radiation (ꞵ+)
- Constituents: 1 Positron
- Relative charge: +1
- Relative mass: Negligible
- Penetrating power: Positron almost instantly annihilates with an electron, so range = 0
What is Gamma radiation (𝛾)
- Constituents: Short wavelength, high energy EM wave
- Relative charge: 0
- Relative mass: 0
- Penetrating power: Stopped by several cm of lead or several m of concrete
How can you investigate the types of radiation emitted by a source (based on penetrating power)
- Using a Geiger-Müller tube and a Geiger counter, record the background count with no source present (to be subtracted from later counts)
- Place an unknown source near the tube (record count)
- Place a sheet of paper between tube and source (record count)
- Place a sheet of aluminium between tube and source (record count)
Depending on when the count rate significantly decreases, the type of radiation emitted can be determined
How can you investigate the types of radiation emitted by a source (with a magnetic field)
- Charged particles moving through a uniform, perpendicular magnetic field are deflected in circular paths
- Direction of deflection depends on charge
- Radius of circular path can show magnitude of charge and mass.
∴ Type of radiation can be determined based on it’s path in a magnetic field
What is an example of an application for α radiation
Smoke alarms:
- α-particles are strongly positive so can easily ionise atoms (Knock off electrons)
- This ionisation transfers energy from the α to the atoms
- In air, α-particles quickly ionise (~10000mm^-1), so quickly loose all their energy
∴ α-particles allow a current to flow, but don’t travel very far (won’t release radiation)
∴ If smoke is present, α-particles can’t pass through the air, stopping the current, setting off the alarm
What is an example of an application for ꞵ radiation
Material thickness controller:
- (ꞵ-)-particles have a lower mass and charge than α-particles, but have a higher speed
∴ (ꞵ-)-particles will ionise atoms in air (~100mm^-1)
- When a material is fed through rollers to be flattened, a ꞵ- source is placed on one side, with a detector on the other
- Radiation is absorbed by the material, so a thicker material causes the count to decrease more
∴ If too much is absorbed, rollers move closer together to make the material thinner
∴ If too little is absorbed, rollers move further apart to make the material thicker
What are some examples of applications for 𝛾 radiation
Radioactive tracers:
- 𝛾 is less ionising than ꞵ-, so does less damage to body tissue
∴ Radioactive source can be ingested/injected into a patient and traced within the body
(eg. PET scanner: Positron emission tomograph - positrons emitted by source annihilate with e- to produce detectable 𝛾)
Cancer treatment:
- Radiation damages cells, so can destroy cancerous tissue
- May damage healthy cells, so risks of use must be evaluated
(Shielding/rotating beams may be used to reduce damage to healthy cells)
What is background radiation
Nuclear radiation from sources other than the one being measured (In the ‘background’)
This background count must be measured and subtracted from any recorded values
What are some potential sources of background radiation
- Geological (in the air)
- Radioactive radon gas is released by rocks (α)
- Conc. varies, usually the largest source of background
- The ground/buildings
- Nearly all rocks contain radioactive material
- Cosmic radiation
- High energy photons from space
- Collide with particle in atmosphere to release radiation
- Living things
- All plants/animals contain carbon (some C-14)
- Man-made radiation
- Medical or industrial sources
- In most places, this is a tiny proportion of background
What is the equation for the intensity of radiation, and what does it show about the relationship between intensity and distance
I = k/(x^2)
∴ I ∝ 1/x^2 (inverse square law)
- Intensity is the amount of radiation per unit area
- A 𝛾 source will emit radiation in all directions, spreading out further from the source
∴ Radiation will spread out as a sphere
∴ SA = 4πr^2, (r=x)
∴ SA ∝ x^2
Intensity will decrease as this area increases,
∴ I ∝ 1/SA
∴ I ∝ 1/x^2
∴ I = k/(x^2)
(k: constant)
How would you investigate the relationship between the intensity of radiation and the distance from the source
- Set up a Geiger-Müller tube at the start of a measuring scape
- Record background count
- Place a radioactive source a distance d from the tube
- Take at least 3 count rate readings and record an average (subtract background)
- Move the source to distances 2d, 3d, 4d, etc… and record the average count rate at each
- Plot a graph of count against distance
- This will show an inverse-square relationship between count and distance (count ∝ 1/x^2)
- The area of the Geiger-Müller tube is constant, so count ∝ Intensity
∴ I ∝ 1/x^2
∴ I = k/(x^2)
What are the main safety measures when working with radioactive sources
- Keep as far away from the source as possible
- I ∝ 1/x^2, so intensity and ∴ exposure decreases with distance
- Hold source away from body (+ long handle tongs), and stand back when possible
- Keep source in lead box when no in use
- Lead will absorb the radiation
- Only remove from the box for as short a time as possible