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In 1932 Ernest Walton and John Cockcroft verified experimentally Einstein's equation that relates mass and energy - Leaving Cert Physics - Question 12 - 2022

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In 1932 Ernest Walton and John Cockcroft verified experimentally Einstein's equation that relates mass and energy. They accelerated protons through a potential diffe... show full transcript

Worked Solution & Example Answer:In 1932 Ernest Walton and John Cockcroft verified experimentally Einstein's equation that relates mass and energy - Leaving Cert Physics - Question 12 - 2022

Step 1

(i) Draw a labelled diagram of their apparatus.

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Answer

The labelled diagram should include:

  • Hydrogen discharge tube
  • Linear accelerator with voltage applied correctly
  • Target set at 45°
  • Screen with scintillations/microscope attached

Step 2

(ii) Write a nuclear equation for the interaction between a proton and a nucleus of lithium-7.

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The nuclear reaction can be represented as: p+Li-7Be-4+He-4\text{p} + \text{Li-}_7 \rightarrow \text{Be-}_4 + \text{He-}_4

Step 3

(iii) Convert this mass to kg. (Give your answer to six decimal places.)

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To convert the mass:

  • Given mass of 1H nuclide = 1.007825 u
  • Using the conversion factor, where 1 u = 1.66053906660 × 10⁻²⁷ kg,
  • The calculation is: 1.007825×1.66053906660×10271.673534×1027 kg1.007825 \times 1.66053906660 \times 10^{-27} \approx 1.673534 \times 10^{-27} \text{ kg}

Step 4

(iv) Explain the discrepancy between the value you have calculated and the value given for the mass of the proton on page 47 of the Formulae and Tables booklet.

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The discrepancy arises because the mass of the 1H nuclide includes the mass of the electron. The mass of the proton is a separate value:

  • The proton mass does not account for the electron, hence the value listed in the tables is greater than the value computed from the nuclide mass.

Step 5

(v) the kinetic energy of the proton as it collided with the metal.

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The kinetic energy (E) can be calculated using: E=qVE = qV, where q is the charge of the proton and V is the potential difference. Therefore,

  • Using the known values: E=(1.602176634×1019)×(70000)=1.12153257×1014 JE = (1.602176634 \times 10^{-19}) \times (70000) = 1.12153257 \times 10^{-14} \text{ J}

Step 6

(vi) the mass lost (in kg) during the interaction.

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The mass lost can be calculated using: Δm=Ec2\Delta m = \frac{E}{c^2}, where c is the speed of light ( c=3×108m/sc = 3 \times 10^8 m/s):

  • Therefore, the mass lost: Δm=1.12153257×1014(3×108)2=1.24376×1032 kg\Delta m = \frac{1.12153257 \times 10^{-14}}{(3 \times 10^8)^2} = 1.24376 \times 10^{-32} \text{ kg}

Step 7

(vii) the energy produced (in J) during the interaction.

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The energy produced equals the energy released due to the mass lost: Eproduced=Δmc2E_{produced} = \Delta m \cdot c^2

  • This can be calculated as: Eproduced=1.24376×1032×(3×108)2=1.120733×1014 JE_{produced} = 1.24376 \times 10^{-32} \times (3 \times 10^8)^2 = 1.120733 \times 10^{-14} \text{ J}

Step 8

(viii) the speed of the alpha particles formed during the interaction.

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Answer

Using conservation of momentum and energy: v=Emassαv = \frac{E}{\text{mass}_\alpha}, where mass of alpha particle approximately equals 4u=6.64466×1027 kg4u = 6.64466 \times 10^{-27} \text{ kg}. So,

  • Rearranging gives: v=1.12153257×10146.64466×10272.13×106m/sv = \frac{1.12153257 \times 10^{-14}}{6.64466 \times 10^{-27}} \approx 2.13 \times 10^6 m/s

Step 9

(ix) A proton may be classified as a hadron. Explain why.

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A proton is classified as a hadron because it is a composite particle made up of quarks, specifically two up quarks and one down quark. Hadrons are particles that interact via the strong force.

Step 10

(x) A proton may also be classified as a baryon. Explain why.

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Answer

A proton is classified as a baryon because it is a type of hadron formed from three quarks. Baryons are characterized by having half-integer spin and are subject to the Pauli exclusion principle, further indicating their composite quark structure.

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