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What is the translational kinetic energy, in MeV, of a proton moving at 0.54 of the speed of light ?

User Radhika
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We would apply the relativistic kinetic energy formula which is expressed as

KE = (y - 1)mc^2 = [1/√(1 - v^2/c^2 - 1]mc^2

where

m is the mass of proton

c is the speed of light

v is the speed of the proton

From the information given,

m = 1.67 x 10^-27 kg

c = 3 x 10^8 m/s

v = 0.54c

Thus,

KE = 1.67 x 10^-27 x (3 x 10^8)^2[1/√(1 - (0.54c/c)^2 - 1]

KE = 28.27 x 10^-12 J

We would convert from joules to mev

1 J = 6,241,509,343,260 MeV

28.27 x 10^-12 = 28.27 x 10^-12 x 6,241,509,343,260

KE = 176.48 Mev

the translational kinetic energy is 176.48 Mev

User Steve Grossi
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