Cosmic rays are high-energy particles that come from space - AQA - A-Level Physics - Question 1 - 2021 - Paper 1

To find the speed of nucleus X, we use the kinetic energy (KE) formula:

$KE = \frac{1}{2} mv^2$

We can rearrange this to solve for speed (v):

$v = \sqrt{\frac{2 \times KE}{m}}$

Substituting the values:

$KE = 215 \text{ MeV} = 215 \times 1.6 \times 10^{-13} \text{ J} \approx 3.44 \times 10^{-11} \text{ J}$

$v = \sqrt{\frac{2 \times 3.44 \times 10^{-11} \text{ J}}{8.02 \times 10^{-26} \text{ kg}}} \approx \sqrt{8.59 \times 10^{14}} \approx 2.93 \times 10^7 \text{ m/s}$

In the decay of the pion (π-) which produces a positron (e+) and an electron neutrino (νe), we need to check the conservation of charge and lepton number:

• The initial charge of the pion is -1 (as π- has a charge of -1).

  • After decay: Positron: +1, Electron neutrino: 0, Total charge: +1 + 0 = 0.
  • Thus, charge is conserved.

• For lepton number: The initial lepton number is 1 (the pion is a meson and does not contribute):

  • After decay: Lepton number from the positron: +1 and from the neutrino: +1.
  • Therefore total lepton number = 1 = initial lepton number, hence conserved.

In this decay of the K+ into an anti-muon (μ-) and a muon neutrino (νμ), we can analyze strangeness:

• Strangeness is a quantum number that represents the presence of strange quarks. K mesons contain strange quarks.
• The K+ has a strangeness of +1 because it contains a strange quark.
• The decay products (anti-muon and muon neutrino) do not contain any strange quarks, leading to a strangeness of 0 in the final state.
• Since the strangeness must be conserved in weak decays, and because we started with a strangeness of +1, the decay indicates a change in strangeness is allowed, which is unique to weak interactions.

An example equation for a K+ decay that involves only hadrons is:

$K^+ \rightarrow \pi^+ + \pi^0$

In this process, a kaon decays to produce a positive pion and a neutral pion, both of which are hadrons.