Atomic structure and particle accelerators flashcards

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Proton number Z
The number of protons in a nucleus. It fixes the element.
Nucleon number A
The total number of protons and neutrons in a nucleus.
Isotopes
Nuclei of the same element (same proton number) with different numbers of neutrons, so different nucleon numbers.
Thermionic emission
The release of electrons from the surface of a metal when it is heated.
Electronvolt (eV)
The energy gained by an electron accelerated through a p.d. of 1 V: 1 eV = 1.60 × 10⁻¹⁹ J.
Drift tube (linac)
A hollow metal tube with no electric field inside, in which a particle moves at constant speed between accelerating gaps.
Dee (cyclotron)
One of two hollow D-shaped electrodes; inside it there is no electric field and the magnetic field turns the particle through a semicircle.
Number of neutrons in a nucleus
N = A − Z (No — learn it)
Speed of an electron accelerated from rest through p.d. V
eV = ½mv², so v = √(2eV/m) (No — derive it from V = W/Q and Ek = ½mv²)
Radius of the path of a charged particle in a uniform magnetic field
r = p/BQ (from BQv = mv²/r) (Yes)
Cyclotron frequency
f = BQ/2πm — independent of the particle's speed (while v ≪ c). (No — derive it)
Length of a linac drift tube
L = v/2f: the particle must spend half a period of the supply in each tube. (No — derive it)
Head-on distance of closest approach of an alpha particle
Ek = (2e)(Ze)/4πε₀r — all kinetic energy becomes electric potential energy. (No — derive it)
Alpha scattering: how does the number detected change with angle?
It falls very steeply: most at small angles, very few beyond 90°, a tiny number near 180°.
Linac diagram: why do the tubes get longer along the line?
The particle speeds up at every gap but must spend the same half-period in each tube, so each tube must be longer.
Cyclotron path shape
A spiral of semicircles with growing radius; each gap crossing increases the speed and so the radius.
Track curving one way with B into the page vs the opposite way
Opposite curvature means opposite sign of charge. With B into the page, a positive particle curves anticlockwise.
A track that spirals inwards
The particle is losing kinetic energy (momentum) by ionising; it travels from the large-radius end to the small-radius end.
A 'V' of two tracks starting from a point with nothing visible entering
A neutral particle (no track) decayed there into two oppositely charged particles.
Alpha scattering: three observations and conclusions
Most pass through → mostly empty space. Some deflected → concentrated positive charge. Very few back → tiny nucleus with most of the mass.
Why can't electrons in the foil cause large alpha deflections?
An alpha particle is about 7300 times as massive as an electron and pushes electrons aside.
Why did alpha scattering rule out the plum-pudding model?
Spread-out positive charge gives only a weak field everywhere, so it could never turn an alpha particle back.
Role of the electric field in an accelerator
It does work on the particle and increases its speed (in the gaps of a linac or cyclotron).
Role of the magnetic field in an accelerator or detector
Its force is perpendicular to the velocity, so it changes direction only (no work): it bends the path into a curve.
How does a detector use ionisation?
A charged particle ionises atoms along its path; droplets, bubbles or electrical pulses form on the ions and mark the track.
Why are high energies needed to study the inside of a nucleon?
Detail is resolved only if λ ≲ its size (~10⁻¹⁵ m). λ = h/p, so a very large momentum — energies of order GeV — is needed.
Conservation laws used to analyse particle interactions
Charge, momentum (as a vector) and total energy are conserved in every interaction.
Electron accelerated from rest through 2.0 kV. Speed?
v = √(2 × 1.60 × 10⁻¹⁹ × 2000 / 9.11 × 10⁻³¹) = 2.7 × 10⁷ m s⁻¹. (No — derive it from V = W/Q and Ek = ½mv²)
Proton at 3.0 × 10⁶ m s⁻¹ perpendicular to a 0.50 T field. Radius?
r = mv/BQ = 1.67 × 10⁻²⁷ × 3.0 × 10⁶ / (0.50 × 1.60 × 10⁻¹⁹) = 0.063 m. (Yes)
Cyclotron for protons with B = 1.5 T. Frequency of the p.d.?
f = BQ/2πm = 1.5 × 1.60 × 10⁻¹⁹ / (2π × 1.67 × 10⁻²⁷) = 2.3 × 10⁷ Hz. (No — derive it)
6.0 MeV alpha head-on to a tin nucleus (Z = 50). Closest approach?
r = k(2e)(50e)/Ek = 8.99 × 10⁹ × 3.2 × 10⁻¹⁹ × 8.0 × 10⁻¹⁸ / 9.6 × 10⁻¹³ = 2.4 × 10⁻¹⁴ m. (No — derive it)
Track of radius 0.25 m for a particle of charge e in a 0.80 T field. Momentum?
p = BQr = 0.80 × 1.60 × 10⁻¹⁹ × 0.25 = 3.2 × 10⁻²⁰ kg m s⁻¹. (Yes)
Momentum needed for a de Broglie wavelength of 2.0 × 10⁻¹⁵ m?
p = h/λ = 6.63 × 10⁻³⁴ / 2.0 × 10⁻¹⁵ = 3.3 × 10⁻¹⁹ kg m s⁻¹. (Yes)
Charge of an alpha particle in r = p/BQ or closest-approach problems
2e = 3.20 × 10⁻¹⁹ C, not e (and not 4e).
Energies given in keV, MeV or GeV
Convert to joules (× 1.60 × 10⁻¹⁹ J per eV) before using ½mv² or p²/2m.
'The magnetic field accelerates the particles'
Wrong: the magnetic force does no work. Only the electric field changes the speed.
Deducing the sign of a charge from a track
State field direction, velocity and force (towards the centre of the curve), then apply Fleming's left-hand rule; reverse the current for negative charges.
Deriving r = p/BQ
Start with 'the magnetic force provides the centripetal force': BQv = mv²/r, so r = mv/BQ = p/BQ.

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