Units, Relativistic Momentum

jk4

I was working a problem in a Modern Physics book:
Find the momentum (in MeV/c) of an electron whose speed is 0.600c.

My first approach was:
mass of electron = 9.1E-31 kg

[tex]\sqrt{1 - \frac{(0.600c)^{2}}{c^{2}}} = 0.800[/tex]

[tex]p = \frac{9.1E-31 * 0.600c}{0.800} = 2.04E-22[/tex] (ignoring units)

then I needed to convert to MeV/c so with some messing around I ended up dividing by 1,000,000 and then multiplying by c to get the exact answer in the book. But this bothered me because I thought I had MeV then multiplying by c to get the answer in the book in MeV/c which doesn't make sense....

I then realized if I first convert the mass into [tex]\frac{MeV}{c^{2}}[/tex] then the units work out perfectly. But I'm still curious why I get the same answer doing it the first way, could someone please help me understand why it works out?

JesseM

According to this, 1 MeV = 1.60217653 * 10^-13 J (and 1 J = 1 kg * m^2 / s^2). So, 1 MeV/c = (1.60217653 * 10^-13 / 299792458) kg * m / s = 5.34428565 * 10^-22 kg * m / s. And if the mass of an electron is 9.11 * 10^-31 kg, then its momentum at 0.6c is (9.11 * 10^-31 * 0.6 * 299792458 / 0.8) kg * m / s = 2.05 * 10^-22 kg * m / s. This is 0.384 the size of 5.34428565 * 10^-22 kg * m / s which I found above for the value of 1 MeV/c, so the momentum of the electron at 0.6c should be 0.384 MeV/c...is this different from what you got?

jk4

I got 0.383MeV/c so ya that's what I got. I was just confused about how the units worked themselves out in my first method of computing the answer, but thanks to your post I see it.

JesseM

So is your question from the original post resolved? I was confused about what you meant when you said you divided by 1,000,000 and multiplied by c to get the correct answer, since 2.04*10^-22 * 299792458 / 1,000,000 = 6.12 * 10^-20, which isn't even close to 0.384.