Section 12 - Nuclear Physics Flashcards
(313 cards)
Describe how ideas about atoms have changed over time.
- The idea of atoms has been around since the time of Ancient Greeks -> Proposed by Democritus
- In 1804, John Dalton suggested that atoms couldn’t be broken up and each element was made of a different type of atom
- Nearly 100 years later, JJ Thomson showed that electrons could be removed from atoms
- Thomson suggested that that atoms were spheres of positive charge with negative electrons in them like a plum pudding
- Rutherford suggested the idea of a nucleus - that atoms did not have uniformly distributed charge and density
What was the original model for atom structure?
Plum pudding model
Describe the plum pudding model.
Atoms are made of positive charge with electrons stuck in them like plum pudding.
Who suggested an alternative to the plum pudding model?
Rutherford (and Marsden)
Which experiment showed the existence of a nucleus in atoms?
Rutherford scattering
Describe the Rutherford scattering experiment.
- Beam of alpha particles from radioactive source is fired at thin gold foil.
- Circular, fluorescent defector screen surrounding gold foil (and the alpha source) was used to detect alpha particles deflected at any angle.
- Most of the alpha particles went straight through the foil, but a small proportion were deflected by a large angle (up to 90°).
Why is the foil used very thin?
Ideally 1 atom thin.
Since the gold foil was very thin, it was thought that the alpha particles could pass straight through it, or possibly puncture the foil.
(So it doesn’t have many interactions.)
To prevent the alpha particles been absorbed by the gold and so that they are only scattered once.
If the plum pudding model of atomic structure were true, what would you expect to see in the Rutherford scattering experiment?
The flashes on the screen/detector should have been seen within a small angle of the beam.
This is because the alpha particles (positively charged) would be deflected by a small amount by the electrons.
Describe the main conclusions of the Rutherford scattering experiment.
Atoms must have a small, positively-charged nucleus at the centre:
• Most of the atoms must be empty space, since most of the alpha particles passed straight through the foil
• Nucleus must have a large positive charge, since positively-charged alpha particles were repelled and deflected by a large angle
• Nucleus must be small, since most of the alpha particles passed straight through the foil (very few deflected by > 90°)
• Most of the mass must be in the nucleus, since positively-charged alpha particles were repelled and deflected by a large angle by the nucleus.
What does the Rutherford scattering experiment tell us about the empty space in the atom?
Most of the atom must be empty space, since most of the alpha particles passed straight through the foil
What does the Rutherford scattering experiment tell us about the charge of the nucleus?
Nucleus must have a large positive charge, since positively-charged alpha particles were repelled and deflected by a large angle
What does the Rutherford scattering experiment tell us about the size of the nucleus?
The nucleus is small, since most of the alpha particles passed straight through the foil
What does the Rutherford scattering experiment tell us about the distribution of mass in the atom?
Most of the mass must be in the nucleus, since positively-charged alpha particles were repelled and deflected by a large angle
How did Rutherford and Kay discover EVIDENCE for the existence of a neutron?
Fired high energy alpha particles at different gases.
Thought there was only protons in the nucleus.
If there was only protons, you’d expect high mass (massive) nuclei to have very high charges (compared to lower mass nuclei).
But the charges observed were lower than expected.
Must be another part in the nucleus: he called in “proton-electron doublet” - it was actually the neutron.
When an alpha particle is fired at a nucleus, what can be assumed at the point at which it’s direction of travel is reversed?
Initial kinetic energy = Electric potential energy
(This is because all of the initial kinetic energy that the alpha particle was fired with has been converted into potential energy)
What does an alpha particle reaching it’s closest approach to the nucleus look like?
What is r?
R = shortest distance between nucleus and alpha particle
Describe how you can estimate the closest approach of a scattered particle to a nucleus, given the initial kinetic energy.
- Equate the initial kinetic energy that the particle was fired with with the potential energy of the particle at the turning point. This is from Coulombs law.
- Initial kinetic energy = Electric potential energy
- Ek = Qgold x Qalpha / 4πε₀r
- Calculate r
Give the equation used to find the closest approach of an alpha particle to the a gold nucleus.
Ek = Qgold x Qalpha / 4πε₀r
Where:
• Ek = Kinetic energy (J)
• Qgold = Charge of the gold nucleus (C)
• Qalpha = Charge of the alpha particle (C)
• ε₀ = 8.85 x 10^-12 F/m
• r = Distance from centre of nucleus or Distance of closest approach (m)
(NOTE: Not given in exam)
What is the charge of a nucleus?
+Ze
Where:
• Z = Proton number (Number of Protons)
• e = Size of charge of an electron
How can the radius of a nucleus be estimated using scattered particles?
- Calculate an estimate for the closest approach of an alpha particle to the nucleus
- This is the maximum possible radius
An alpha particle with initial kinetic energy of 6.0MeV is fired at a gold nucleus. Estimate the radius of the nucleus by finding the closest approach of the alpha particle to the nucleus.
- Initial kinetic energy = 6.0 x 10^6 MeV = 9.6 x 10^-13 J
- This equals electric potential energy, so:
- 9.6 x 10^-13 = Qgold x Qalpha / 4πε₀r
- 9.6 x 10^-13 = (79 x 1.60 x 10^-19) x (2 x 1.60 x 10^-19) / 4π x 8.85 x 10^-12 x r
- r = 3.8 x 10^-14 m
- This is a maximum estimate for the radius.
What are the two methods of estimating nuclear radius and which is better?
- Closest approach of scattered particle
- Electron diffraction
Electron diffraction gives more accurate values.
Why are electrons used to estimate nuclear radius?
They are leptons, so they do not interact with the strong nuclear force.
We know very little about the strong nuclear force.
Why can electron beams be diffracted?
Like other particles, they show wave-particle duality and have a de Broglie wavelength.
They are also used because they are lighter = better to accelerate