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In 1932 J.D - Leaving Cert Physics - Question 10 - 2013

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In 1932 J.D. Cockcroft and E.T.S. Walton accelerated protons to energies of up to 700 keV and used them to bombard a lithium target. They observed the production of ... show full transcript

Worked Solution & Example Answer:In 1932 J.D - Leaving Cert Physics - Question 10 - 2013

Step 1

How did Cockcroft and Walton accelerate the protons?

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Answer

Cockcroft and Walton utilized high voltage and a large electric field to accelerate protons. They employed a voltage multiplier circuit that produced high electric potentials, allowing protons to gain significant energy as they were directed towards the lithium target.

Step 2

How did they detect the alpha-particles?

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Answer

Detection of the alpha-particles was performed using scintillators that emitted flashes of light when struck by the particles. This light was further captured with a zinc sulfide screen, which aided in visualizing the alpha-particles.

Step 3

Write the nuclear equation for the reaction that occurred and indicate the historical significance of their observation.

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The nuclear reaction can be represented as:

36Li+11p24He+13H\mathrm{^6_3Li + ^1_1p \rightarrow ^4_2He + ^3_1H}

The historical significance lies in it being the first experimental verification of the equation E=mc2E=mc^2 and the first artificial splitting of the nucleus, leading to the discovery of transmutation using artificially accelerated particles, which later garnered them the Nobel Prize.

Step 4

Calculate the speed of a proton that has a kinetic energy of 700 keV.

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The kinetic energy of the proton is given as 700 keV, which can be converted to joules:

700 keV=700×1.6×1019 J=1.12×1013 J700 \text{ keV} = 700 \times 1.6 \times 10^{-19} \text{ J} = 1.12 \times 10^{-13} \text{ J}

Using the formula for kinetic energy, where:

E=12mv2,E = \frac{1}{2} mv^2,

Rearranging gives:

v=2Emv = \sqrt{\frac{2E}{m}}

Taking the proton mass as m=1.67×1027m = 1.67 \times 10^{-27} kg, we can calculate:

v=2×1.12×10131.67×10271.16×107 m/sv = \sqrt{\frac{2 \times 1.12 \times 10^{-13}}{1.67 \times 10^{-27}}} \approx 1.16 \times 10^7 \text{ m/s}

Step 5

Why is the tube evacuated?

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Answer

The tube is evacuated to create a vacuum, which prevents particles from colliding with gas molecules. This increases the mean free path of the accelerated particles and allows them to travel without interference, ensuring higher energy transfers and more efficient collisions with the target.

Step 6

What is the purpose of accelerating the particles to high velocities?

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Particles are accelerated to high velocities to increase their kinetic energy, allowing them to overcome the Coulomb barrier that exists between positively charged nuclei. Higher velocities ensure sufficient energy for the particles to collide and induce nuclear reactions.

Step 7

What is the purpose of the magnets?

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Magnets are used in circular accelerators to bend the paths of charged particles. This bending allows particles to maintain circular orbits within the accelerator, enabling them to recirculate and gain more energy with each pass through the acceleration segments.

Step 8

Give an advantage of a circular accelerator over a linear accelerator?

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A significant advantage of a circular accelerator is its ability to accelerate particles to higher energies. As particles traverse the same path multiple times, they can be accelerated repeatedly, making it possible to achieve a greater final energy than a linear accelerator would allow.

Step 9

Can an accelerator of this design be used to accelerate neutrinos? Explain your answer.

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An accelerator of this design cannot effectively accelerate neutrinos because neutrinos are electrically neutral and interact very weakly with matter. Unlike charged particles that can be manipulated with electric and magnetic fields in accelerators, neutrinos would pass through most structures without interaction, making it impractical to accelerate them using conventional methods.

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