Question

IS Halliday, Fundamentals of Physics, 11e PRINTER VERSION RACK NEXE Chapter 26, Problem 022 Your answer is partially correct. Try again In a nuclear experiment a proton with kinetic energy 0.77 MeV moves in a circular path in a uniform magnetic field. What energy must (a) an alpha partide (+2e, m = 4.0 u) and (b) deuteron (a = te, m - 2.0 u) have if they are to circulate in the same circular path?

Answer 1

Radius Kinetic energy q proton, Icp= 1 mp op = 0.77 Mev of comlar path is given by Ramprle - mp up Up B same crudar penth so reding is constant eB R SIDLO of alpha particle is moving in same circular path, then. R= me u 4mp & น B 2e B from and B equation Ч гр 14 mp Up d = hep 2e B RB Kinetic energy of x-particle IG= £m let Ka = 2 Lumpy (1 = £ mpref = kp &=0.77 MeV For deuteron ma = 2mp , q=te R=malla WB 2 mp la e B 2 mp led - molle ☺ eB eR - up = 1 -- Kinehe energy a deuteron, Kot = ma vp = { izmol uning? Ka- mely q 0.77 MeV 4 2 Kal=0.385 Mev 키