If one proton is travelling through space at 0.3c, what is it's kinetic energy in MeV?
We know the formula for kinetic energy is
KE = (1/2)mv2.
If we know that c, the speed of light, is 3x108ms-1 (you will be given this value in your Data & Formula books in the exam) and the proton is traveling at 0.3c, it has a velocity of
0.3x3x108 = 9x107ms-1.
Therefore v2 will be (9x107)2 = 8.1x1015.
We know that the mass of a proton is 1.67x10-27kg (again, you will be given this value in the exam).
Therefore kinetic energy in Joules (the SI unit, because we have currently only used SI units) is (1/2) x 1.67x10-27 x 8.1x1015 = 6.76x10-12J
It is known that 1eV (electron volt) is equal to 1.6x10-19J. Therefore dividing our energy in Joules by this value will give us our energy in eV.
(6.76x10-12) / (1.6x10-19) = 4.23x107eV.
As 1 MeV = 1x106eV, we can find our value in MeV by dividing it by 1x106;
(4.23x107) / (1x106) = 42.3MeV.
This is a standard unit used in particle energies, and is quite a common question in any A Level physics exam.
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