Question

In proton-beam therapy, a high-energy beam of protons is fired at a tumor. The protons come to rest in the tumor,depositing their kinetic energy and breaking apart the tumor’s DNA, thus killing its cells. The protons gain their energy by being discharged through a type of capacitor with a potential difference of 20MV and have a starting speed of 1.88 x 10^8 m/s. Using what you know of the behavior of a point charge in an electrostatic field and the conservation of energy solve the following: A: What is the change in kinetic energy of a proton after moving through the capacitor. B: What is the final kinetic energy of the proton after moving through the capacitor. C: What is the speed of the proton after moving through the capacitor

Answer 1

(A) change in KE = loss in electric PE

delta(KE) = q deltaV = (1e)(20 MV)

= 20 MeV

(B) v = 1.88 x 10^8 = (1.88 x 10^8)/(3 x 10^8) c

v = 0.627c

Y = 1 / sqrt[1 - (v/c)^2] = 1.283

inital KE = (Y - 1)m0 c^2

= (1.283- 1) (938.27 MeV/c^2)(c^2)

= 265.74 MeV  

KEf = 265.74 + 20 MeV = 285.74 MeV .....Ans

(B) KE = (Y - 1) m0 c^2

Y - 1 = 285.74 / 938.27 = 0.305

Y = 1.305

and Y = 1 / sqrt[1 - (v/c)^2]

1 - (v/c)^2 = (1/1.305)^2

v = 0.642c

v = 1.93 x 10^8 m/s

answered by: anonymous