Particle accelerators for proton beam therapy
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Particle accelerators for proton beam therapy
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[ID] => 559976
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[post_date] => 2025-01-09 22:03:50
[post_date_gmt] => 2025-01-10 03:03:50
[post_content] => Practice Passage (Question 1-5)
*This passage is the property of Khan Academy and has been reformatted into an AAMC-style interface in their entirety by MedLife Mastery. MedLife Mastery does not endorse and is not an affiliate of Khan Academy.
Proton beam therapy (PBT) is used to treat cancers in sensitive areas such as the eyes or brain. High energy protons are produced by a particle accelerator and steered toward the patient’s tumor by a series of magnets. Once in the patient’s body, 80% of the protons only interact with electrons, not nuclei. Protons lose energy by ionizing atoms in those interactions. Those protons will slow slightly in the skin and other “shallow” tissues without significantly deviating from a straight path. Once slowed, protons rapidly lose their remaining energy over a short distance as they capture an electron and decelerate to thermal speeds. This is sometimes called a “depth charge” effect. How deep the proton penetrates before releasing this burst of energy depends on its initial speed and the composition of the tissue. By adjusting how much energy protons have when they enter the patient, physicians can target where the proton will most rapidly ionize tissue atoms. This gives doctors some control over how much energy gets deposited in the shallower tissue (the “skin dose”), and all but eliminates irradiation of deeper tissues (the “shadow dose”). This pattern of energy loss is illustrated by the proton’s narrow “Bragg peak” (Figure 1). Combining beams with different energies allows the new beam to deposit energy with a plateau-like profile that can damage a tumor across its full depth. This combined dose forms a Spread Out Bragg Peak (SOBP; see Figure 1’s combined dose curve).
Figure 1. Percent therapeutic energy dose (dose necessary to kill tumor cells) deposited in tissue by proton beams by depth. Curve for a single proton energy is a normal Bragg Peak. The Combined dose is (offset from the 250MeV dose at the right end for illustration only) is an SOBP.
Adapted from Aamiller, Creative Commons. CC by SA 3.0
[post_title] => Particle accelerators for proton beam therapy
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[quiz_unique_key] => 578908434
[question] => What is the most likely fate of a PBT-beam proton entering tissue?
[value] => Array
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[answer] => 4
[description] => Reason for the Correct Answer:
A deflection (rather than reflection or emission) implies that the muons retain some of their original momentum as they pass through the material.
Most protons only interact with electrons, not nuclei.
Most protons give off energy, slowing down and then capturing an electron.
A proton with an electron is a hydrogen atom.
)
[answers] => Array
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[0] => Array
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[each_answer] => A. The proton decays into a neutron and an electron.
)
[1] => Array
(
[each_answer] => B. It is absorbed by the nucleus of an atom in the tissue.
)
[2] => Array
(
[each_answer] => C. The proton continues its path and exits the body.
)
[3] => Array
(
[each_answer] => D. It becomes a hydrogen atom.
)
)
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[1] => Array
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[quiz_unique_key] => 3873426850
[question] => If a 2 T steering magnet acts on a proton moving at 2 x 106 m/s, approximately how much work does the magnet do on the moving charge along a 0.1 m long path?
[value] => Array
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[answer] => 1
[description] => Reason for the Correct Answer:
Work = Force x Distance cos(θ).
FB = qv x B, where FB, v, and B are vectors.
In other words, Force, velocity, and magnetic field are all perpendicular to each other on a plane. The “right hand rule” to find the direction of the magnetic force utilizes this same principle. This means that FB and v are perpendicular, so θ = 90°
If = 90°, cos(θ) = 0, so work = 0 J.
)
[answers] => Array
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[0] => Array
(
[each_answer] => A. 0 J
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[1] => Array
(
[each_answer] => B. 6 x 10-14 J
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[2] => Array
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[each_answer] => C. 4 x 105 eV
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[3] => Array
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[each_answer] => D. More information is needed.
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[2] => Array
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[quiz_unique_key] => 83407773
[question] => If a physician wanted to treat a tumor that 18 and 24 cm in depth, protons of which energy is she most likely to consider adding to the SOBP in Figure 1?
[value] => Array
(
[answer] => 3
[description] => Reason for the Correct Answer:
The SOBP in Figure 1 is the combination of 225 MeV and 250 MeV proton beams.
Protons are constantly slowing down by giving off energy as they travel through the tissue, so higher energy protons have their “depth charge” at greater depth than lower energy protons.
Look at Figure 1. The new proton energy should have its peak between the 225 MeV and 250 MeV beams to raise the plateau closer to the 100% therapeutic dose, which is required to kill tumor cells.
Only the 240 MeV falls between 225 MeV and 250 MeV.
)
[answers] => Array
(
[0] => Array
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[each_answer] => A. 200 MeV
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[1] => Array
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[each_answer] => B. 210 MeV
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[2] => Array
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[each_answer] => C. 240 MeV
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[3] => Array
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[each_answer] => D. 270 MeV
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[3] => Array
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[quiz_unique_key] => 2377279144
[question] => Holding all other variables constant, a 10% increase in which parameter would most increase the power used to produce a beam with a circular cross section if all of the energy used went into creating the beam itself?
[value] => Array
(
[answer] => 3
[description] => Reason for the Correct Answer:
Power = total energy/time [J/s].
The total amount of energy needed to produce a beam is the total number of particles x their energy, T = N x E [n x J/n = J], where “n” is the number of protons.
The total number of particles per unit time = total area x flux = A x Φ [m2 x n/s]
Total area, 
A 10% increase in E or Φ is like multiplying the power needed by 1.10 (=1+10%). A 10% increase in r is like multiplying by 1.21 (=1.102).
Increasing the radius by 10% increases power requirements by more than 20%, but a 10% increase to the flux or proton energy would only increase power requirements by 10%, so increasing the beam radius would increase power requirements the most.
)
[answers] => Array
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[0] => Array
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[each_answer] => A. Particle flux rate, Φ
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[1] => Array
(
[each_answer] => B. The energy of each proton, E
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[2] => Array
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[each_answer] => C. The radius of the beam, r
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[3] => Array
(
[each_answer] => D. All of the above would change power consumption equally
)
)
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[4] => Array
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[quiz_unique_key] => 2261298308
[question] => If a proton were to ionize a ground-state carbon atom in the patient’s body by ejecting an electron with a final kinetic energy of 50eV, which orbital’s electron would slow the proton the most?
[value] => Array
(
[answer] => 1
[description] => Reason for the Correct Answer:
Slowing the proton means forcing it to lose energy. The more energy it loses, the more it slows down.
Ionization is the removal or addition of an electron.
The most tightly bound electrons will take the most energy to remove from the atom.
The most tightly bound electrons are those closest to the nucleus.
The 1s electrons are closest to the nucleus, so they are the most tightly bound. The proton will slow the most when it gives up the energy to knock them away from the carbon atom.
)
[answers] => Array
(
[0] => Array
(
[each_answer] => A. 1s
)
[1] => Array
(
[each_answer] => B. 2s
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[2] => Array
(
[each_answer] => C. 2p
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[3] => Array
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[each_answer] => D. 3s
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[559976|1] => D
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[559976|3] => C
[559976|4] => C
[559976|5] => A
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