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In 1932 Cockcroft and Walton succeeded in splitting lithium nuclei by bombarding them with artificially accelerated protons using a linear accelerator - Leaving Cert Physics - Question 10(a) - 2009

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Question 10(a)

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In 1932 Cockcroft and Walton succeeded in splitting lithium nuclei by bombarding them with artificially accelerated protons using a linear accelerator. Each time a l... show full transcript

Worked Solution & Example Answer:In 1932 Cockcroft and Walton succeeded in splitting lithium nuclei by bombarding them with artificially accelerated protons using a linear accelerator - Leaving Cert Physics - Question 10(a) - 2009

Step 1

How were the protons accelerated? How were the alpha particles detected?

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Answer

The protons were accelerated using a high voltage power supply that creates an electric field in the linear accelerator. This electric field propels the protons toward the target, gaining sufficient kinetic energy to overcome the coulomb barrier of the lithium nucleus. The alpha particles produced from the lithium bombardment were detected using scintillation screens made of zinc sulphide or phosphor, which emit flashes of light upon impact with the alpha particles. Photomultiplier tubes are employed to convert this light into an electrical signal for detection.

Step 2

Write a nuclear equation to represent the splitting of a lithium nucleus by a proton.

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The nuclear equation can be represented as follows:

 36Li+ 11p2 24He\text{ }_3^6\text{Li} + \text{ }_1^1\text{p} \rightarrow 2 \text{ }_2^4\text{He}

This indicates that a lithium-6 nucleus reacts with a proton to produce two alpha particles.

Step 3

Calculate the energy released in this reaction.

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To calculate the energy released, we first need to determine the loss in mass:

Loss in mass = mass of lithium nucleus + mass of proton - 2(mass of alpha particle)

Substituting the values: Loss in mass=(1.6726×1026 kg)+(1.00728×1027 kg)2(6.6447×1027 kg)\text{Loss in mass} = (1.6726 \times 10^{-26} \text{ kg}) + (1.00728 \times 10^{-27} \text{ kg}) - 2(6.6447 \times 10^{-27} \text{ kg})

=1.32894×1029 kg= 1.32894 \times 10^{-29} \text{ kg}

Using Einstein's equation, energy released (E) can be calculated as: E=Δmc2E = \Delta m c^2

Where:

  • c=2.9979×108 m/sc = 2.9979 \times 10^8 \text{ m/s}
  • E=(1.32894×1029 kg)(2.9979×108 m/s)22.6×1012 JE = (1.32894 \times 10^{-29} \text{ kg})(2.9979 \times 10^8 \text{ m/s})^2 \approx 2.6 \times 10^{-12} \text{ J}

Step 4

Most of the accelerated protons did not split a lithium nucleus. Explain why.

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Answer

Most of the accelerated protons do not split a lithium nucleus primarily due to the empty space within the atomic structure. The protons may pass through the lithium nucleus without any interaction. This is because the majority of the volume of atoms consists of empty space between the nucleus and the electrons, resulting in a low probability of collision. Additionally, some protons simply do not have enough energy to overcome the repulsive forces from the nucleus.

Step 5

Explain why new particles are formed.

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New particles are formed during the collisions in the Large Hadron Collider due to the conversion of kinetic energy into mass, according to Einstein's mass-energy equivalence principle, E=mc2E=mc^2. When protons collide at high energies, the energy available in the interaction can create heavy particles which were not present before the collision, thus changing kinetic energy into mass.

Step 6

What is the maximum net mass of the new particles created per collision?

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The maximum net mass of the new particles created per collision can be calculated using the energy available from the kinetic energy of protons. Assuming each proton has a kinetic energy of 2.0 GeV:

1 GeV = 1.6×1010 J1.6 \times 10^{-10} \text{ J}

Total kinetic energy per proton: E=2.0 GeV=2.0×1.6×1010 J=3.2×1010 JE = 2.0 \text{ GeV} = 2.0 \times 1.6 \times 10^{-10} \text{ J} = 3.2 \times 10^{-10} \text{ J}

Using E=mc2E=mc^2, we find: m=Ec2=3.2×1010(3×108)23.56×1030 kgm = \frac{E}{c^2} = \frac{3.2 \times 10^{-10}}{(3 \times 10^8)^2} \approx 3.56 \times 10^{-30} \text{ kg}

This is the approximate maximum mass of new particles generated per collision.

Step 7

What is the advantage of using circular particle accelerators in particle physics?

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The primary advantage of using circular particle accelerators is that they can accumulate energy more efficiently over time. By continuously accelerating particles within a circular path, they gain energy with each pass, allowing for higher energy collisions with reduced space requirements. This design increases the frequency of particle collisions without necessitating long straight paths, facilitating numerous interaction events. Additionally, circular accelerators can provide more precise control over beam stability and focusing.

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