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The history of anti-matter begins in 1928 when a young English physicist named Paul Dirac predicted an anti-particle for the electron - Leaving Cert Physics - Question 10 - 2010

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The history of anti-matter begins in 1928 when a young English physicist named Paul Dirac predicted an anti-particle for the electron. (i) What is anti-matter? An ... show full transcript

Worked Solution & Example Answer:The history of anti-matter begins in 1928 when a young English physicist named Paul Dirac predicted an anti-particle for the electron - Leaving Cert Physics - Question 10 - 2010

Step 1

What is anti-matter?

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Answer

Anti-matter refers to a type of matter that has particles with the same mass as particles of ordinary matter but with opposite charge. The first anti-matter particle discovered was the positron, which is the anti-particle of the electron, denoted as e+e^+. When a particle meets its anti-particle, they annihilate each other, resulting in the conversion of their entire mass into energy, often manifesting as photons.

Step 2

What is meant by pair production?

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Answer

Pair production is a phenomenon where energy is converted into matter, specifically a particle and its anti-particle, due to interaction with a high-energy photon (often a gamma ray). This process conserves both energy and momentum.

Step 3

Calculate the kinetic energy of one of the particles produced.

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Answer

To find the kinetic energy of one particle produced from pair production, we use the energy-frequency relation:

E=hfE = h f

where h=6.626×1034 J sh = 6.626 \times 10^{-34} \text{ J s} is the Planck constant and f=3.6×1018 Hzf = 3.6 \times 10^{18} \text{ Hz}. Thus,

E=6.626×1034 J s×3.6×1018 Hz=2.376×1015 JE = 6.626 \times 10^{-34} \text{ J s} \times 3.6 \times 10^{18} \text{ Hz} = 2.376 \times 10^{-15} \text{ J}

Now, using the relation between energy and mass,

E=mc2E = mc^2

we substitute the rest mass (m=9.1×1031 kgm = 9.1 \times 10^{-31} \text{ kg} and c=3.0×108 m/sc = 3.0 \times 10^8 \text{ m/s}) to find the total energy:

E=(9.1×1031 kg)×(3.0×108 m/s)2=8.19×1014 JE = (9.1 \times 10^{-31} \text{ kg}) \times (3.0 \times 10^8 \text{ m/s})^2 = 8.19 \times 10^{-14} \text{ J}

The kinetic energy of one particle (after accounting for rest mass energy) is given by

Ek=EErest=2.376×1015 J8.19×1014 J=3.69×1014 JE_k = E - E_{rest} = 2.376 \times 10^{-15} \text{ J} - 8.19 \times 10^{-14} \text{ J} = 3.69 \times 10^{-14} \text{ J}.

Step 4

Construct the possible combinations.

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Answer

The meson family consisting of up and down quarks and their anti-particles results in the following combinations:

  • Combination 1: uu (up) + dd (down) [\text{Charge: } +1/3 + -1/3 = 0 ]
  • Combination 2: uu (up) + dˉ\bar{d} (anti-down) [\text{Charge: } +1/3 + +1/3 = +2/3 ]
  • Combination 3: uˉ\bar{u} (anti-up) + dd (down) [\text{Charge: } -1/3 + -1/3 = -2/3 ]
  • Combination 4: uˉ\bar{u} (anti-up) + dˉ\bar{d} (anti-down) [\text{Charge: } -1/3 + +1/3 = 0 ]

Step 5

What famous Irish writer first thought up the name 'quark'?

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Answer

The name 'quark' was first introduced by the famous Irish writer James Joyce.

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