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Proton beam radiotherapy for cancer takes advantage of the fact that the rate of energy deposition increases as the protons slow down, as the inverse of the velocity. A typical treatment may address a cancer that lies under 150 mm of healthy tissue. 160 MeV protons will penetrate the healthy tissue depositing little energy and doing little damage. Then they come to a halt in the next 20 mm depositing the bulk of their energy there and doing enough damage to kill the cancer. If the protons lose half their kinetic energy in the cancer, and a sufficient dose is three Grays, calculate: i) The required power incident on the cancer to perform the treatment of a 20 mm cancer in five minutes

Proton beam radiotherapy for cancer takes advantage of the fact that the rate of energy deposition increases as the protons slow down, as the inverse of the velocity. A typical treatment may address a cancer that lies under 150 mm of healthy tissue. 160 MeV protons will penetrate the healthy tissue depositing little energy and doing little damage. Then they come to a halt in the next 20 mm depositing the bulk of their energy there and doing enough damage to kill the cancer. If the protons lose half their kinetic energy in the cancer, and a sufficient dose is three Grays, calculate: i) The required power incident on the cancer to perform the treatment of a 20 mm cancer in five minutes

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Proton beam radiotherapy for cancer takes advantage of the fact that the rate of energy
deposition increases as the protons slow down, as the inverse of the velocity. A typical
treatment may address a cancer that lies under 150 mm of healthy tissue. 160 MeV protons
will penetrate the healthy tissue depositing little energy and doing little damage. Then they
come to a halt in the next 20 mm depositing the bulk of their energy there and doing enough
damage to kill the cancer. If the protons lose half their kinetic energy in the cancer, and a
sufficient dose is three Grays, calculate:
i) The required power incident on the cancer to perform the treatment of a 20 mm cancer
in five minutes

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