Topic 7 Grand Challenge: The Particle Discovery Mission

Topic 7 Grand Challenge: The Particle Discovery Mission

To cover the entirety of Topic 7: Nuclear and Particle Physics, this multi-stage challenge integrates the historical methods of probing matter (7A), the engineering of accelerators and detectors (7B), and the classification of the subatomic "zoo" via the Standard Model (7C).

Context: A research team is investigating the internal structure of nucleons by colliding high-energy particles. To do this, they must produce a probe, accelerate it to a wavelength small enough to "see" the nucleus, and analyze the products of the resulting collision.

Part 1: Probing Matter (7A)

Question 1: Electron Creation

Electrons are released from a heated metal filament in an evacuated chamber.

• Name the process responsible for this release.
• Explain why a vacuum is necessary for the experiment.

Worked Solution:

1. Name of Process:

The process is called thermionic emission.

2. Necessity of a Vacuum:

An evacuated chamber is vital to ensure that the freed electrons do not collide with air molecules. Such background collisions would cause the electrons to scatter uncontrollably, lose kinetic energy, or get absorbed before hitting the target material.

Question 2: Wavelength Calculations

The electrons are accelerated through a potential difference ($V$) of $2500\text{ V}$.

• Calculate the de Broglie wavelength ($\lambda$) of these electrons.
• Discuss why these specific electrons might be suitable for probing the spacing of atoms in a crystal, but would require significantly more energy to investigate the structure of an individual proton.

Worked Solution:

1. Constants Required (A-level Specification):

• Mass of electron ($m_e$) = $9.11\times {10}^{-31}\text{ kg}$
• Charge of electron ($e$) = $1.60\times {10}^{-19}\text{ C}$
• Planck constant ($h$) = $6.63\times {10}^{-34}\text{ J s}$

2. Relate Kinetic Energy ($E_k$) to Potential Difference:

$$E_k=eV=(1.60\times {10}^{-19})\times 2500=4.00\times {10}^{-16}\text{ J}$$

3. Derive Momentum ($p$):

$$E_k=\frac{p^2}{2m_e}⟹p=\sqrt{2m_e{E}_k}$$

$$p=\sqrt{2\times (9.11\times {10}^{-31})\times (4.00\times {10}^{-16})}\approx 2.70\times {10}^{-23}\text{ kg m s}{-}^1$$

4. Compute de Broglie Wavelength ($\lambda$):

$$\lambda =\frac{h}{p}=\frac{6.63\times {10}^{-34}}{2.70\times {10}^{-23}}\approx 2.46\times {10}^{-11}\text{ m}$$

de Broglie Wavelength (λ) = 2.5 × 10⁻¹¹ m (or 0.025 nm)

5. Structural Scale Discussion:

To produce a clear diffraction pattern when probing structural systems, the de Broglie wavelength of the particle must be comparable to or smaller than the spacing of the structures being examined ($\lambda \le d$).

• Atomic Spacing: Atomic gaps in a typical crystal lattice are on the order of ${10}^{-10}\text{ m}$. Since $2.46\times {10}^{-11}\text{ m}<{10}^{-10}\text{ m}$, these electrons are perfectly suited for crystal probing.
• Proton Structure: An individual proton has a structural scale of roughly ${10}^{-15}\text{ m}$ (1 femtometer). The current wavelength is far too large to resolve something that small. Since $\lambda \propto \frac{1}{\sqrt{V}}$, resolving inside a proton requires a massive increase in acceleration potential to shrink the wavelength to the femtometer scale.

Part 2: Accelerators and Detectors (7B)

Question 3: Cyclotron Field Dynamics

To reach the required higher energies, protons are instead introduced into a cyclotron with a magnetic field strength of $0.60\text{ T}$.

• Explain the roles of the alternating potential difference and the uniform magnetic field in this machine.
• Derive the expression for the cyclotron frequency ($f$) and show that the time taken for a particle to complete a semicircle is independent of its speed.

Worked Solution:

1. Core Component Functions:

• Uniform Magnetic Field: Acts at right angles to the motion of the protons within the hollow D-shaped electrodes ("Dees"). It applies a constant centripetal magnetic force ($F=Bqv$) that forces the charged particles to move in circular arcs, keeping them confined within the accelerator.
• Alternating Potential Difference (A.C. Voltage): Is applied across the narrow gap between the two Dees. It creates an electric field that accelerates the protons every time they cross the gap, increasing their kinetic energy and orbital radius. The polarity alternates periodically to match the transit time of the protons.

2. Derivation of Cyclotron Frequency Formula:

The centripetal force holding a particle of mass $m$ and charge $q$ in an arc of radius $r$ is provided entirely by the magnetic Lorenz force:

$$$$Bqv=mv2r⟹v=Bqrm

Substitute linear velocity into the angular period cycle formulation ($v=2\pi rf$):

$$2\pi rf=\frac{Bqr}{m}$$

Canceling out the radius variable ($r$) on both sides yields the final frequency relation:

$$f=\frac{Bq}{2\pi }$$

3. Proof of Velocity Independence:

The time $t$ required to complete one exact semicircle inside a single Dee is exactly half of the total cycle period ($T=\frac{1}{f}$):

$$t=\frac{T}{2}=\frac{1}{2f}=\frac{\pi m}{Bq}$$

Because the velocity parameter ($v$) does not appear in this final expression, the time taken to complete a semicircle depends solely on the fixed invariants of mass, charge, and field strength. As a proton gains speed, its orbital path length scales up proportionally to match, keeping the duration constant.

Question 4: Bubble Chamber Observations

The accelerated protons collide with a target, and the resulting interaction is captured in a hydrogen bubble chamber.

• Explain how a bubble chamber allows physicists to track charged particles and why some particles (like neutrinos) do not leave tracks.
• A track is observed to spiral inwards. Describe two reasons why a particle’s path might behave this way.

Worked Solution:

1. Tracking Mechanism:

A bubble chamber contains a liquefied gas (often superheated liquid hydrogen) kept just below its boiling point. When a fast-moving, charged particle passes through, it ionizes the atoms along its path. The energy released by these localized ionizations causes the liquid to boil instantly at those points, forming tiny visible bubbles of vapor that chart the particle's trajectory.

• Unseen Neutral Particles: Particles like neutrinos or neutrons carry zero net electric charge ($q=0$). Because they have no charge, they do not exert electrostatic forces on atomic electrons and cause virtually no ionization, leaving the liquid completely undisturbed.

2. Explaining the Inward Spiral:

The orbital radius equation established previously is $r=\frac{mv}{Bq}$. An inward spiral means the tracking radius is steadily decreasing over time. This happens due to two primary physics factors:

• Loss of Velocity/Kinetic Energy: As the particle plows through the liquid medium, it continuously loses kinetic energy due to ongoing ionization collisions with the liquid atoms. As its velocity ($v$) drops, its path radius ($r$) shrinks proportionally.
• Relativistic Mass Adjustments: If the particle is traveling at near-light velocities and slowing down, its effective relativistic mass ($m$) decreases along with its momentum, further reducing the turning radius in the uniform magnetic field.

Part 3: The Particle Zoo (7C)

Question 5: Evaluating Interactions

A high-energy collision results in the following proposed reaction:

$$p+p\to p+n+{\pi }^{+}$$

Evaluate whether this reaction is possible by testing for the conservation of charge, baryon number, and lepton number.

Worked Solution:

We check the conservation of quantum numbers across the reaction by comparing the sum of the initial states (left) against the final products (right):

Quantum Number	Initial States ($p+p$)	Final States ($p+n+{\pi }^{+}$)	Status
Charge ($Q$)	$(+1)+(+1)=+2$	$(+1)+0+(+1)=+2$	Conserved
Baryon Number ($B$)	$(+1)+(+1)=+2$	$(+1)+(+1)+0=+2$	Conserved
Lepton Number ($L$)	$0+0=0$	$0+0+0=0$	Conserved

Conclusion: The reaction is fully allowed and physically possible.

Question 6: Hadron Classification

One of the products, the ${\pi }^{+}$ meson, is known to be a hadron.

• Define the difference between a baryon and a meson in terms of their quark composition.
• Deduce the quark structure of the ${\pi }^{+}$ meson, given that it has a charge of $+1$ and a baryon number of $0$.

Worked Solution:

1. Structural Classification Definitions:

• Baryons: Are composite hadrons containing exactly three quarks ($qqq$) bound together (or three antiquarks for antibaryons).
• Mesons: Are composite hadrons made of one quark and one antiquark ($q\overline{q}$) paired together.

2. Deduction of ${\pi }^{+}$ Quark Structure:

Since it is a meson, it must have a $q\overline{q}$ configuration. We select from the standard first-generation quarks:

• Up quark ($u$): Charge = $+\frac{2}{3}$, Baryon Number = $+\frac{1}{3}$
• Down quark ($d$): Charge = $-\frac{3}{3}$, Baryon Number = $+\frac{1}{3}$
• Anti-down quark ($\overline{d}$): Charge = $+\frac{1}{3}$, Baryon Number = $-\frac{1}{3}$

Let's test the combination $u\overline{d}$:

$$\text{Net Charge}=Q_u+Q_{\overline{d}}=(+\frac{2}{3})+(+\frac{1}{3})=+1$$

$$\text{Net Baryon Number}=B_u+B_{\overline{d}}=(+\frac{1}{3})+(-\frac{1}{3})=0$$

Quark Structure of π⁺ = ud

Question 7: Exchange Particles

The interaction that holds quarks together within these nucleons is the strong nuclear force.

Identify the exchange boson associated with this force and compare its range to that of the photon in the electromagnetic force.

Worked Solution:

1. Force Carrier Identification:

The exchange boson mediating the strong force between individual quarks is the gluon. (Note: The residual strong force holding nucleons together inside a nucleus can also be mediated by mesons like the pion).

2. Range Comparison Analysis:

• Gluon (Strong Force): Has a highly restricted, short operational range (approximately ${10}^{-15}\text{ m}$). This short range is due to a phenomenon called color confinement, meaning the strong force does not extend outside the boundaries of the hadron.
• Photon (Electromagnetic Force): Has a rest mass of zero. Because it is massless, the virtual photons mediating the electromagnetic force have an infinite range, though the force's strength drops off following an inverse-square law ($\propto \frac{1}{r^2}$).