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During a nuclear interaction an antiproton collides with a proton - Leaving Cert Physics - Question 10 - 2006

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During a nuclear interaction an antiproton collides with a proton. Pair annihilation takes place and two gamma ray photons of the same frequency are produced. What ... show full transcript

Worked Solution & Example Answer:During a nuclear interaction an antiproton collides with a proton - Leaving Cert Physics - Question 10 - 2006

Step 1

What is a photon? Calculate the frequency of a photon produced during the interaction.

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Answer

A photon is a quantum of electromagnetic radiation, which exhibits wave-particle duality. It has no mass and travels at the speed of light, which is approximately ( 2.998 \times 10^8 , \text{m s}^{-1} ).

To calculate the frequency of the photon produced, we can use the energy-mass equivalence principle:

E=mc2E = mc^2

For the pair annihilation, the total energy available is calculated using the mass of the proton and antiproton:

[ m = 2 \times (1.673 \times 10^{-27} , \text{kg}) = 3.346 \times 10^{-27} , \text{kg} ]

The energy produced during annihilation is:

[ E = (3.346 \times 10^{-27} , \text{kg}) \times (3 \times 10^8 , \text{m s}^{-1})^2 ]

This yields the energy in Joules. The frequency ( f ) of the photon is then given by:

[ f = \frac{E}{h} ]

where ( h = 6.626 \times 10^{-34} , \text{Js} ). After substituting the values, we can find ( f ).

Step 2

Why are two photons produced? Describe the motion of the photons after the interaction.

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Answer

Two photons are produced to conserve momentum. In a nuclear interaction, when an antiproton collides with a proton, the annihilation event creates two photons that travel in opposite directions. This ensures that the total momentum before and after the collision remains the same, as momentum must be conserved in isolated systems. After the interaction, the photons move away from the annihilation point at the speed of light. Since they travel in opposite directions, the momentum is balanced and conserved.

Step 3

How is charge conserved during this interaction?

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Charge conservation is a fundamental principle in physics that states that the total electric charge in an isolated system never changes. In this interaction, before annihilation, we have one proton (charge +1) and one antiproton (charge -1), leading to a total charge of zero. After the annihilation occurs, two photons are produced, and photons are electrically neutral (charge = 0). Thus, the total charge before (0) equals the total charge after (0), demonstrating that charge is conserved.

Step 4

After the annihilation, pairs of negative and positive pions are produced. Explain why.

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Answer

Pions are produced as a result of the energy released during the electron-positron annihilation. According to Einstein's mass-energy equivalence principle ( E = mc^2 ), the energy from the annihilation event converts into matter, creating pairs of positive and negative pions to conserve both baryon number and charge. The pair production must also conserve other quantities such as energy and momentum, which is satisfied by producing a particle-antiparticle pair.

Step 5

Pions are mesons that consist of up and down quarks and their antiquarks. Give the quark composition of (i) a positive pion, (ii) a negative pion.

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Answer

(i) A positive pion (( \pi^+ )) consists of an up quark and a down antiquark: ( u\bar{d} ).

(ii) A negative pion (( \pi^- )) consists of a down quark and an up antiquark: ( d\bar{u} ).

Step 6

List the fundamental forces of nature that pions experience.

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Pions experience four fundamental forces of nature:

  1. Electromagnetic force.
  2. Strong nuclear force.
  3. Weak nuclear force.
  4. Gravitational force (although very weak compared to the others).

Step 7

A neutral pion is unstable with a decay constant of 2.5 × 10^−13 s^−1. What is the half-life of a neutral pion?

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The half-life ( T_{1/2} ) of a particle is related to its decay constant ( \lambda ) by the formula:

T1/2=ln2λT_{1/2} = \frac{\ln{2}}{\lambda}

Substituting in the decay constant:

[ T_{1/2} = \frac{0.693}{2.5 \times 10^{13} , ext{s}^{-1}} = 2.772 \times 10^{-13} , \text{s} = 2.8 \times 10^{-13} , \text{s} ]

Thus, the half-life of a neutral pion is approximately ( 2.8 \times 10^{-13} , \text{s} ).

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