Problem 28.46 In proton-beam therapy , a high-energy beam of protons is fired at

Question

Problem 28.46

In proton-beam therapy, a high-energy beam of protons is fired at a tumor. As the protons stop in the tumor, their kinetic energy breaks apart the tumor's DNA, thus killing the tumor cells. For one patient, it is desired to deposit 8.0

In proton-beam therapy, a high-energy beam of protons is fired at a tumor. As the protons stop in the tumor, their kinetic energy breaks apart the tumor's DNA, thus killing the tumor cells. For one patient, it is desired to deposit 8.0??10^2J kV potential difference. Part A What is the total charge of the protons that must be fired at the tumor? ^j of proton energy in the tumor. To create the proton beam, protons are accelerated from rest through a 11000kV

Explanation / Answer

from conservation of energy,increase in potential energy of protons equals to decrease in kinetic energy.

hence we get (delta K =-delta U)

Given that proton energy is 0.12j

and potential difference is 1.2*10^7 v

we know that the equation for the charge is

Q=(-delta K)/delta v

=0.12/(1.2*10^7)

=10^-8c

=10 nc

hence total charge of proton is 10nc

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