Problem 4 (20 points The particle accelerator at Stanford University is three ki

Question

Problem 4 (20 points The particle accelerator at Stanford University is three kilometers long and accelerates electrons (mron 9.1110kg) to a speed of 0.999 999 999 7 c, which is very nearly equal to the speed of light. (a) what is the rest energy (in MeV) of the electrons? (1 eV = 1.602x10s Joules) (b) Find the magnitude of the relativistic momentum of the electrons. Comparing it with the nonrelativistic value, is it bigger, smaller? By what factor? c) Compute the electron's total energy (in GeV). -31 19

Explanation / Answer

Given

mass of electron is m_e = 9.11*10^-31 kg ,

rest energy of the electron is E= m0*C^2

E = 9.11*10^-31*(3*10^8)^2 J = 8.199*10^-14 J

E = 8.199*10^-14 /(1.6*10^-19) eV = 512437.5 eV

E = 0.512437 MeV

b) relativistic momentum is  

P = m0*v/sqrt(1-v^2/c^2)

P = (9.11*10^-31*0.9999999997*3*10^8)/(sqrt(1- (0.9999999997*3*10^8)^2/(3*10^8)^2)) kg m/s

P = 1.11574*10^-17 kg m/s

c) total energy of the electron is  

E = m0c^2/(sqrt(1-v^2/c^2))

E = 9.11*10^-31*(3*10^8)^2/(sqrt(1-(0.9999999997*3*10^8)^2/(3*10^8)^2) J

E = (3.3472277337643*10^-9)/(1.6*10^-19) eV

E = 20.92 G eV

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