Question

2. A proton moves with a speed of 0.800c through a particle accelerator In the accelerator frame of reference, calculate (a) the total and (D) the kineic ene the proton in megaelectron volts. ME[ans: (a) 157 × 10 MeV;(b)626×10?ges0f

Answer 1

(a)

Since , Etotal =  m*c^2/sqrt (1 - v^2/c^2)

here, m = mass of proton = 1.67*10^-27 kg

c = velocity of light

v = velocity of proton = 0.800c

So,

Etotal = [(1.67*10^-27)*(3*10^8)^2]/sqrt (1 - 0.800c^2/c^2)

Etotal = 2.505*10^-10 J

Since 1 J = 6.242*10^18 eV

So,

Etotal = (2.505*10^-10)*(6.242*10^18) eV

Etotal = 1.564*10^3 MeV

(b)

Since,

Etotal = Ekinetic + E_rest

here, Erest = m*c^2

Erest = (1.67*10^-27)*(3*10^8)^2 = 1.503*10^-10 J

Erest = (1.503*10^-10)*(6.242*10^18) eV

Erest = 9.38*10^2 Mev

So,

Ekinetic = Etotal - Erest

Ekinetic = (1564 - 938) MeV

Ekinetic = 6.26*10^2 MeV

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