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Nuclear and Quantum Physics - Structure of Matter

Grade 11IBPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Rutherford-Geiger-Marsden experiment provided evidence for the nuclear model of the atom, showing that the nucleus is small, dense, and positively charged, as most α\alpha-particles passed through gold foil while a few were deflected at large angles.

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Nuclide notation is written as ZAX^A_Z X, where AA is the nucleon (mass) number and ZZ is the atomic (proton) number. Isotopes are atoms of the same element with the same ZZ but different AA.

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Hadrons are particles made of quarks. They are subdivided into Baryons (made of 3 quarks, e.g., protons uuduud and neutrons uddudd) and Mesons (made of one quark and one anti-quark).

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Leptons are fundamental particles that do not experience the strong nuclear force. Examples include the electron (e−e^-), muon (μ\mu), tau (τ\tau), and their associated neutrinos (νe,νμ,ντ\nu_e, \nu_{\mu}, \nu_{\tau}).

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Quarks carry fractional charges: up (uu), charm (cc), and top (tt) have a charge of +23e+\frac{2}{3}e; down (dd), strange (ss), and bottom (bb) have a charge of −13e-\frac{1}{3}e.

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The four fundamental forces and their exchange particles (gauge bosons) are: Electromagnetic (photons γ\gamma), Strong (gluons gg), Weak (W+,W−,Z0W^+, W^-, Z^0 bosons), and Gravitational (graviton).

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Conservation laws: In any nuclear reaction, charge (QQ), baryon number (BB), lepton number (LL), and mass-energy must be conserved. Strangeness (SS) is conserved in strong and electromagnetic interactions but can change by ±1\pm 1 in weak interactions.

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Confinement: Quarks cannot exist in isolation because the force between them increases as they are pulled apart. The energy required to separate them creates a new quark-antiquark pair (pair production).

📐Formulae

E=hfE = hf

c=fλc = f\lambda

λ=hp\lambda = \frac{h}{p}

ΔE=Δmc2\Delta E = \Delta m c^2

R=R0A1/3R = R_0 A^{1/3}

💡Examples

Problem 1:

During β−\beta^- decay, a neutron decays into a proton, an electron, and an antineutrino. Represent this using quark notation and verify the conservation of charge.

Solution:

udd→uud+e−+νˉeudd \rightarrow uud + e^- + \bar{\nu}_e

Explanation:

A down quark (dd) changes into an up quark (uu) via the weak interaction. The initial charge of the neutron (uddudd) is 00 (+23−13−13=0+\frac{2}{3} - \frac{1}{3} - \frac{1}{3} = 0). The final charges are: proton (uuduud) =+1= +1, electron (e−e^-) =−1= -1, and antineutrino (νˉe\bar{\nu}_e) =0= 0. Total final charge =+1−1+0=0= +1 - 1 + 0 = 0. Charge is conserved.

Problem 2:

A meson is composed of an up quark and an anti-strange quark (usˉu\bar{s}). Determine the charge and the strangeness of this particle.

Solution:

Charge =+1e= +1e, Strangeness =+1= +1

Explanation:

The charge of an up quark (uu) is +23e+\frac{2}{3}e. The charge of a strange quark (ss) is −13e-\frac{1}{3}e, so the charge of an anti-strange quark (sˉ\bar{s}) is +13e+\frac{1}{3}e. Total charge =+23+13=+1e= +\frac{2}{3} + \frac{1}{3} = +1e. The strangeness of an ss quark is −1-1, so the strangeness of an sˉ\bar{s} quark is +1+1.

Problem 3:

Calculate the energy of a photon with a frequency of 5.0×1014 Hz5.0 \times 10^{14} \text{ Hz}. (Use h=6.63×10−34 J sh = 6.63 \times 10^{-34} \text{ J s})

Solution:

E=(6.63×10−34)×(5.0×1014)=3.315×10−19 JE = (6.63 \times 10^{-34}) \times (5.0 \times 10^{14}) = 3.315 \times 10^{-19} \text{ J}

Explanation:

Energy is calculated using the formula E=hfE = hf. Substituting the values for Planck's constant and the given frequency yields the energy in Joules.

Structure of Matter Grade 11 Notes & Examples | IB Physics