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In proton beam therapy, a beam of high-energy protons is used to kill cancerous cells in a tumor. In one system, the beam, which consists of protons with an energy of $2.8 \times 10^{-11} \mathrm{J},$ has a current of 80 nA. The protons in the beam mostly come to rest within the tumor. The radiologist has ordered a total dose corresponding to $3.6 \times 10^{-3} \mathrm{J}$ of energy to be deposited in the tumor.

a. How many protons strike the tumor each second?

b. How long should the beam be on in order to deliver the required dose?

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here for the solution for the party since I called to any bitey where Isaac all to 80 multiplied by 10 to the power minus nine ampere equal to minus 1.6, multiplied by 10 departments 19 column and take equal to one second now rearranging migration for and we get an equal to it by e. So by substituting the values in the equation, we get an equal to 80. Multiply by 10 10 to the power minus 19. Multiply by one divided by 1.6, multiplied by 10 to the power minus 19. From here we get an equal to to five multiply by 10 to the power 11 so the number of proteins strikes the turmoil. Per second is why multiplied by 10 to the power 11 for the party. Let the proton BMP on the human for is debt five multiplied by 10 to the power 11 multiplied by two pointed multiplied by 10 to the power minus 11 multiplied by t called to 3.6 multiplied by 10 to the power minus three As the energy deposited by the proton beam per second is five multiplied by 10 to the power 11 multiplied by seven to pointed, multiplied by 20. Power minus 11. Which is this? No, here to solve 40. We get equal to 3.6, multiplied by 24 minus three, divided by five. Multiplied by 2.8. From here, we get equal to 25 points. Seven multiplied by 10 to depart, minus five seconds. So this is the solution. Step by step. Please go through this.

DR APJ ABDUL KALAM TECHNICAL UNIVERSITY

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