Author Jonathan David
Author Jonathan David
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Particle Physics and Cosmology: Fundamental Particles—A History

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Particle Physics and Cosmology: Fundamental Particles—A History

Citation: The content below is based on the most recent edition of University Physics with Modern Physics.

Important Formulas

  • Energy-Mass Relation:

        \[E = mc^2\]

  • De Broglie Wavelength:

        \[\lambda = \frac{h}{p}\]

  • Quantum Energy Levels:

        \[E_n = -\frac{13.6}{n^2} \, \text{eV}\]

  • Relativistic Energy:

        \[E^2 = (pc)^2 + (m_0c^2)^2\]

Key Terms and Definitions

  • Atom: The smallest unit of matter retaining the properties of an element, initially thought to be indivisible.
  • Electron: The first discovered fundamental particle, negatively charged, identified by J.J. Thomson in 1897.
  • Proton: A positively charged particle found in the nucleus, discovered by Ernest Rutherford in 1917.
  • Neutron: A neutral particle in the nucleus, discovered by James Chadwick in 1932.
  • Quarks: Fundamental constituents of protons and neutrons, introduced by Murray Gell-Mann and George Zweig in 1964.
  • Higgs Boson: A particle discovered in 2012 that gives mass to other particles via the Higgs field.

Historical Timeline

  • 1897: Discovery of the electron by J.J. Thomson.
  • 1911: Ernest Rutherford proposes the nuclear model of the atom.
  • 1917: Discovery of the proton by Ernest Rutherford.
  • 1932: Discovery of the neutron by James Chadwick.
  • 1964: The quark model is proposed by Murray Gell-Mann and George Zweig.
  • 2012: Discovery of the Higgs boson at CERN’s Large Hadron Collider.

Example

If an electron is accelerated through a potential difference of V = 100 \, \text{V}, calculate its de Broglie wavelength.

Using the de Broglie wavelength formula:

    \[\lambda = \frac{h}{p}\]

The momentum (p) is given by:

    \[p = \sqrt{2 m e V}\]

Substituting values (h = 6.63 \times 10^{-34} \, \text{Js}, m = 9.11 \times 10^{-31} \, \text{kg}, e = 1.6 \times 10^{-19} \, \text{C}):

    \[p = \sqrt{2 (9.11 \times 10^{-31}) (1.6 \times 10^{-19}) (100)}\]

    \[p \approx 5.39 \times 10^{-24} \, \text{kg m/s}\]

Substituting into \lambda:

    \[\lambda = \frac{6.63 \times 10^{-34}}{5.39 \times 10^{-24}}\]

    \[\lambda \approx 1.23 \times 10^{-10} \, \text{m}\]

Result:

The de Broglie wavelength of the electron is approximately 0.123 \, \text{nm}.

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