Nucleon number A and proton number Z; nuclide notation and isotopes

Large-angle alpha-particle scattering as evidence for a small, massive, positive nucleus, and how models of the atom changed

Thermionic emission and accelerating electrons through a p.d.: eV = ½ mv²

The roles of electric and magnetic fields in linear accelerators, cyclotrons and detectors

Deriving and using r = p/BQ

Conservation of charge, energy and momentum in interactions; reading particle tracks

Why high energies are needed to investigate the structure of nucleons

1. Alpha-particle scattering showed that an atom is mostly empty space with its positive charge and almost all its mass in a tiny nucleus, and accelerators use electric fields to speed charged particles up and magnetic fields to steer them.

2. ᴬ_ZX: A = nucleon number (protons + neutrons), Z = proton number; neutrons N = A - Z.

3. A heated filament releases electrons (thermionic emission); a p.d. V gives each one kinetic energy eV = ½ mv².

4. Electric fields accelerate charged particles (they do work); magnetic fields bend their paths (force perpendicular to velocity, no work done).

5. In a uniform magnetic field a charged particle moves in a circle of radius r = p/BQ: bigger momentum, bigger circle.

6. Charge, energy and momentum are conserved in every interaction; neutral particles leave no track.

7. To see detail of size d, a probe needs a de Broglie wavelength λ= h/p no bigger than d.

For a nucleon (≈ 10⁻¹⁵ m) that means energies of about 1 GeV.

A nuclide is written ᴬ_ZX, for example ²⁷₁₃Al: 13 protons and 27 − 13 = 14 neutrons. The number of neutrons is N = A - Z, and in a neutral atom the number of electrons equals Z.

Z fixes the element and its chemistry. Isotopes of an element have the same Z but different numbers of neutrons, so different A. The charge on a nucleus is +Ze, with e = 1.60×10⁻¹⁹ C. Protons and neutrons have almost the same mass (1.67 × 10⁻²⁷ kg), about 1800 times the mass of an electron, so A is a measure of the mass of the atom.

WORKED EXAMPLE · QUESTION

A nucleus of a radioactive isotope of strontium is written ⁹⁰₃₈Sr. (a) State the number of protons and neutrons in the nucleus. (b) Calculate the charge on the nucleus. (c) A stable isotope of strontium has two fewer neutrons. Write its symbol.

SOLUTION

(a) Z = 38 and N = A - Z: 38 protons, 90 − 38 = 52 neutrons

(b) Only the protons are charged: Q = 38 ×1.60×10⁻¹⁹ = 6.08 × 10⁻¹⁸ C

(c) Same element, so same Z; two fewer neutrons lowers A by 2: ⁸⁸₃₈Sr

A narrow beam of alpha particles (helium nuclei, charge +2e), all with the same energy, is aimed at a very thin sheet of gold foil, only a few hundred atoms thick. The apparatus is in a vacuum, so the alpha particles are not stopped or scattered by air. A detector, originally a zinc sulfide screen that gives a tiny flash for each alpha particle, watched through a microscope, is moved round the foil to count the alpha particles arriving at each angle.

Electrons cannot cause the large deflections: an alpha particle is about 7300 times as massive as an electron and pushes electrons aside, as a bowling ball would a table-tennis ball.

A head-on alpha particle slows down, stops for an instant and returns along its path. At the distance of closest approach all of its kinetic energy has become electric potential energy.

WORKED EXAMPLE · QUESTION

An alpha particle with kinetic energy 4.8 MeV travels directly towards a silver nucleus (Z = 47). Calculate the distance of closest approach.

SOLUTION

Convert to joules: Eₖ = 4.8×10⁶ ×1.60×10⁻¹⁹ = 7.68×10⁻¹³ J.

At closest approach all of it is electric potential energy:

Rearrange for r: r = 2.8 × 10⁻¹⁴ m (2.82 × 10⁻¹⁴ m)

This is about 10⁴ times smaller than an atom (≈ 10⁻¹⁰ m); the nucleus itself must be smaller still.

How models of the atom changed

Each model was kept until an experiment gave a result it could not explain.

That is how scientific models develop: they are tested by prediction and replaced or refined by evidence.

Thermionic emission: a metal filament is heated by a current, and some free electrons near the surface gain enough energy to escape from the metal. In an electron gun the hot filament (cathode) sits near a positive anode with a small hole in it. The p.d. V between them sets up an electric field that accelerates the electrons towards the anode, and a narrow beam passes through the hole. Beyond the anode there is no field, so the electrons move at constant velocity. The tube is evacuated so that electrons do not lose energy in collisions with gas molecules.

The electronvolt: 1 eV is the energy gained by an electron accelerated through 1 V, so 1 eV = 1.60 × 10⁻¹⁹ J. An electron through 3.2 kV gains 3.2 keV; an alpha particle (charge 2e) through the same p.d. gains 6.4 keV.

A charge Q moving at speed v at right angles to a uniform magnetic field of flux density B feels a force F = BQv (4.5), with direction from Fleming's left-hand rule; for a negative particle, point the "current" finger opposite to its velocity. The force is always perpendicular to the velocity, so it does no work: the speed and kinetic energy stay constant and only the direction changes. A force of constant size at right angles to the motion is exactly what circular motion needs, so the particle moves in a circle.

WORKED EXAMPLE · QUESTION

Electrons are released by thermionic emission and accelerated from rest through a p.d. of 3.2 kV. The beam then enters a uniform magnetic field of flux density 2.0 mT at right angles to the field. (a) Show that the speed of the electrons is about 3.4 × 10⁷ m s⁻¹. (b) Calculate the radius of their circular path.

SOLUTION

(a) Kinetic energy gained = eV = 1.60×10⁻¹⁹ ×3200 = 5.12×10⁻¹⁶ J. Rearranging eV = ½ mv²: v = √(2 ×5.12×10⁻¹⁶ ÷9.11×10⁻³¹) = 3.35 × 10⁷ m s⁻¹

(b) p = 9.11×10⁻³¹ ×3.35×10⁷ = 3.05×10⁻²³ kg m s⁻¹, so r = p/BQ = 3.05×10⁻²³ ÷(2.0×10⁻³ ×1.60×10⁻¹⁹) = 0.095 m (9.5 cm)

The speed is about 0.11c, so the non-relativistic equation is a good approximation here.

The linear accelerator (linac)

A linac is a line of hollow metal drift tubes in an evacuated pipe, connected alternately to the two terminals of a high-frequency alternating supply. Inside a tube there is no electric field, so the particle drifts at constant speed. It is accelerated only in the gaps, where neighbouring tubes are at different potentials. While the particle is inside a tube the p.d. reverses, so when it reaches the next gap the field there again points the right way to accelerate it.

The particle must spend the same time, half a period of the supply, in every tube.

As it speeds up it covers more distance in that time, so the tubes get longer: L = v/2f. Once the particles approach the speed of light their speed hardly changes, so the tubes become almost equal in length.

WORKED EXAMPLE · QUESTION

Protons enter a linac with negligible energy and gain 100 keV at each gap. The alternating supply has a frequency of 25 MHz. (a) Calculate the speed of a proton in tube 1. (b) Calculate the length of tube 1. (c) Explain why tube 4 is twice as long as tube 1.

SOLUTION

(a) Kinetic energy after one gap = 100 keV = 1.0 × 10⁵ × 1.60 × 10⁻¹⁹ J = 1.60 × 10⁻¹⁴ J: v = √(2 ×1.60×10⁻¹⁴ ÷1.67×10⁻²⁷) = 4.38 × 10⁶ m s⁻¹

(b) Time in a tube = half a period = 1 ÷ (2 × 25 × 10⁶) = 20 ns, so L = 4.38×10⁶ ×2.0×10⁻⁸ = 0.0875 m (8.75 cm)

(c) In tube 4 the kinetic energy is 4 × 100 keV. v ∝√(Eₖ), so the speed is √4 = 2 times larger; the time is the same, so the tube is twice as long (17.5 cm).

The cyclotron

A cyclotron has two hollow D-shaped electrodes (dees) with a narrow gap between them, in a vacuum, between the poles of a large magnet giving a uniform field B perpendicular to the dees. Inside a dee there is no electric field; the magnetic force makes the particle travel a semicircle at constant speed. At the gap an alternating p.d. accelerates the particle each time it crosses, twice per revolution. After each crossing the particle is faster, so the next semicircle has a larger radius (r ∝v): the path spirals outwards until the particle leaves at the edge.

WORKED EXAMPLE · QUESTION

A cyclotron accelerates protons in a uniform magnetic field of 0.80 T. The protons leave at a radius of 0.45 m. The p.d. across the gap is 20 kV. (a) Calculate the frequency of the alternating p.d. (b) Calculate the maximum kinetic energy of the protons, in MeV. (c) Estimate the number of complete revolutions each proton makes.

SOLUTION

(a) One revolution of the proton = one cycle of the supply: f = BQ/2πm = (0.80 ×1.60×10⁻¹⁹) ÷(2π×1.67×10⁻²⁷) = 1.22 × 10⁷ Hz (12.2 MHz)

(b) Maximum momentum from r = p/BQ with r = 0.45 m: p = 0.80 ×1.60×10⁻¹⁹ ×0.45 = 5.76×10⁻²⁰ kg m s⁻¹. Then Eₖ = p²/2m = (5.76×10⁻²⁰)² ÷(2 ×1.67×10⁻²⁷) = 9.93×10⁻¹³ J = 6.2 MeV (6.21 MeV)

(c) Each revolution includes two gap crossings, each adding 20 keV: 6.21 MeV ÷ (2 × 20 keV) ≈ 155 revolutions

Check: v = p/m = 3.45×10⁷ m s⁻¹ ≈ 0.11c, so the fixed frequency stays in step.

Ionisation

A fast charged particle knocks electrons out of the atoms it passes, leaving a trail of ions. In a cloud chamber droplets form on the ions; in a bubble chamber small bubbles form; in electronic detectors the ions are collected as a pulse of current. The trail is the track. A heavily ionising particle (large charge, slow, for example an alpha particle) leaves a short, thick track; a fast electron leaves a long, thin one. Neutral particles (neutrons, photons, neutral hadrons) do not ionise directly and leave no track; their presence is inferred from gaps and from conservation laws.

Deflection

A magnetic field across the detector curves each track. The direction of curvature gives the sign of the charge (Fleming's left-hand rule); the radius gives the momentum, p = BQr. A particle losing energy by ionisation slows down, so its radius shrinks and its track is a spiral inwards, which also shows which way it was travelling.

Two particles with charges of equal size but opposite sign, and the same momentum, curve with the same radius in opposite directions.

Conservation laws in interactions

Charge: total charge before = total charge after. A neutral particle must decay into products whose charges add to zero.

Momentum (a vector): add components along two perpendicular directions. A particle at rest that decays into two particles sends them off in opposite directions with equal-sized momenta.

Energy: total energy is conserved (including rest-mass energy, 4.8). Kinetic energy alone need not be.

WORKED EXAMPLE · QUESTION

A bubble chamber has a uniform magnetic field of 1.5 T into the page. A neutral particle leaves no track, then decays into two particles with charges of magnitude e. Track 1 starts at 20° above the original direction and has radius 0.40 m; track 2 starts at 35° below it and has radius 0.24 m (figure above). (a) Explain what conservation of charge tells you about the two particles. (b) Calculate the momentum of each particle just after the decay. (c) Show that momentum is conserved at right angles to the original direction. (d) Calculate the momentum of the neutral particle.

SOLUTION

(a) Total charge before = 0, so the charges after must add to zero: one particle is +e and the other −e, and they curve in opposite directions. With B into the page, Fleming's left-hand rule gives an anticlockwise curve for a positive charge, so track 1 is the positive particle.

(b) p = BQr: p₁ = 1.5 ×1.60×10⁻¹⁹ ×0.40 = 9.6 × 10⁻²⁰ kg m s⁻¹; p₂ = 1.5 ×1.60×10⁻¹⁹ ×0.24 = 5.76 × 10⁻²⁰ kg m s⁻¹

(c) The neutral particle had no sideways momentum, so the sideways components must cancel: 9.6×10⁻²⁰sin 20° = 3.28×10⁻²⁰ and 5.76×10⁻²⁰sin 35° = 3.30×10⁻²⁰ kg m s⁻¹, equal and opposite within 1%.

(d) Add the components along the original direction: p = 9.6×10⁻²⁰cos 20° + 5.76×10⁻²⁰cos 35° = 9.02×10⁻²⁰ + 4.72×10⁻²⁰ = 1.37 × 10⁻¹⁹ kg m s⁻¹

To reveal detail, a probe must be diffracted or scattered by it, and that only happens strongly when its wavelength is similar to or smaller than the size of the detail (Unit 2, electron diffraction). A moving particle has a de Broglie wavelength λ= h/p, so a shorter wavelength needs a larger momentum, and so a larger energy.

A nucleon is about 10⁻¹⁵ m across. Electrons from an ordinary electron gun have wavelengths around 10⁻¹¹ m: fine for the spacing of atoms in a crystal, far too long to see inside a nucleon. The momenta needed are so large that the particles travel at very nearly the speed of light, and the energies are of the order of GeV. Only large accelerators reach them. Scattering electrons of this energy off protons and neutrons gave the first evidence that nucleons contain smaller particles, quarks (4.8).

WORKED EXAMPLE · QUESTION

To study structure inside a proton, electrons with a de Broglie wavelength of 1.0 × 10⁻¹⁵ m are needed. (a) Calculate the momentum of such an electron. (b) Show that the electron cannot be treated as moving slowly compared with light. (c) For a particle moving at very nearly c, energy ≈ pc. Estimate the energy of the electrons in GeV. (d) Compare with the electrons in Worked example 3.

SOLUTION

(a) p = h/λ= 6.63×10⁻³⁴ ÷1.0×10⁻¹⁵ = 6.63 × 10⁻¹⁹ kg m s⁻¹

(b) Try v = p/m = 6.63×10⁻¹⁹ ÷9.11×10⁻³¹ = 7.3×10¹¹ m s⁻¹, more than 2000 times c, which is impossible, so p = mv does not apply.

(c) pc = 6.63×10⁻¹⁹ ×3.00×10⁸ = 1.99×10⁻¹⁰ J; ÷ 1.60 × 10⁻¹⁹ gives ≈ 1.2 GeV

(d) The 3.2 kV electrons had p = 3.05×10⁻²³ kg m s⁻¹, so λ= 6.63×10⁻³⁴ ÷3.05×10⁻²³ = 2.17×10⁻¹¹ m: about 22 000 times too long.

You do not need relativistic equations in this unit; the point to explain is why the energy must be so high.

3 Electric fields and 4.5 Magnetic fields and forces

The closest-approach calculation uses V = Q/4πε₀ r; the circular paths come from F = BQv and Fleming's left-hand rule.

1 Momentum and 4.2 Circular motion

r = p/BQ is F = mv²/r in disguise; track problems use conservation of momentum in two dimensions and Eₖ = p²/2m.

Electron diffraction (Unit 2), 4.8 The standard model and nuclear physics (Unit 5)

λ= h/p links accelerator energy to the size of detail a beam can reveal. Energies in MeV and GeV, creating particles from energy and quarks inside nucleons continue on 4.8.

1. How many protons, neutrons and electrons are there in a neutral atom of radon-222 (Z = 86)?

2. Why did the alpha-scattering result rule out the "plum pudding" model?

3. An electron is accelerated from rest through 600 V. Calculate its kinetic energy and its speed.

4. A proton moves at 2.0 × 10⁶ m s⁻¹ at right angles to a uniform magnetic field of 0.25 T. Calculate the radius of its path.

5. Two tracks start from the same point, with equal radii, curving in opposite directions. Nothing visible enters the point. What can you deduce?

6. Explain why the drift tubes of a proton linac get longer along the accelerator.

1. 86 protons, 222 − 86 = 136 neutrons and 86 electrons.

2. In that model the positive charge is spread through the whole atom, so the electric field inside is weak everywhere and could only deflect an alpha particle by a tiny angle. Deflections of more than 90° need the positive charge and mass concentrated in a very small nucleus.

3. Eₖ = 1.60×10⁻¹⁹ ×600 = 9.6 × 10⁻¹⁷ J; v = √(2 ×9.6×10⁻¹⁷ ÷9.11×10⁻³¹) = 1.45 × 10⁷ m s⁻¹.

4. r = mv/BQ = (1.67×10⁻²⁷ ×2.0×10⁶) ÷(0.25 ×1.60×10⁻¹⁹) = 0.084 m.

5. A neutral particle (no track) decayed there. The products have opposite charges (charge conservation and opposite curvature). If their charges are equal in size, equal radii mean equal-sized momenta.

6. The proton gains energy at every gap, so it moves faster in each successive tube. The supply reverses every half-period, so the proton must spend the same time (half a period) in each tube; at a higher speed it travels further in that time, so each tube must be longer (L = v/2f).