Section 12 - Nuclear Physics Flashcards
(229 cards)
Describe how ideas 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
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 is fired at thin gold foil
- Circular 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°)
If the plum pudding model of atomic structure were true, what would you expect to see in the Rutherford scattering experiment?
The alpha particles 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
• Most of the mass must be in the nucleus, since positively-charged alpha particles were repelled and deflected by a large angle
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
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 fire with has been converted into potential energy)
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
- 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 (m)
(NOTE: Not given in exam)
What is the charge of a nucleus?
+Ze
Where:
• Z = Proton number
• 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.
Why can electron beams be diffracted?
They show wave-particle duality and have a de Broglie wavelength.
What is the equation for the de Broglie wavelength of electrons AT HIGH SPEEDS?
λ = hc / E
Where: • λ = de Broglie wavelength (m) • h = Planck constant = 6.63 x 10^-34 • c = Speed of light in a vacuum (m/s) • E = Electron energy (J)
(Note: Not given in exam, but can be derived!)
Derive the equation for the de Broglie wavelength of electrons at high speeds.
- The speed of high-energy electrons is almost the speed of light, c.
- So λ = h / mv = h / mc
- Since E = mc²:
- λ = hc / E
In order to use electron diffraction to determine nuclear radius, what must the electrons’ energy be and why?
High, because the wavelength must be very small in order for diffraction to be observed due to the tiny nucleus.