Question

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 2.8×10−11J, has a current of 72 nA. The protons in the beam mostly come to rest within the tumor. The radiologist has ordered a total dose corresponding to 3.8×10−3J of energy to be deposited in the tumor.

Part A How many protons strike the tumor each second? Express your answer as a number of protons.

Part B: How long should the beam be on in order to deliver the required dose?

Answer 1

Let the charge of one proton is e = 1.602 * 10^-19 C

Let the energy of each proton is E = 2.8 * 10^-11 J

Total current in the proton beam is I = 72 nA = 72 n C/s = 72 * 10^-9 C/s

So , we can observe that a charge of Q = 72 n C is incident on the tumor per second

A ) If we take the total number of protons in the beam as n , then we can write

Number of protons * charge of each proton = total charge

n * e = Q

n = Q / e

n = ( 72 * 10^-9 ) / ( 1.602 * 10^-19 )

n = 45 * 10¹⁰

So the number of protons incident on the tumor per second is n = 45 * 10¹⁰

B ) Total energy of the proton beam is given by

E' = number of protons * energy of one proton

E' = 45 * 10¹⁰ * 2.8 * 10^-11

= 126 * 10^-1

= 12.6 J

So , the tumor receives an energy of 12.6 J per second

Total dose required to break the tumor is E'' = 3.8 * 10^-3 J

So the time in which an energy E'' is deposited is

T = total energy dose / energy received per second

= E'' / E'

= 3.8 * 10^-3 / 12.6

= 0.30 * 10^-3 s

= 0.3 ms

So the tumor must be exposed to proton beam for just 0.3 millisecond in order to kill it.