E=mc^2 Conversion to eV=u: What Happens to the Distance Units?

  • Thread starter LouFerrigno
  • Start date
  • Tags Tags
    E=mc^2 Units
4 replies · 5K views · last reply
LouFerrigno
Messages
3
Reaction score
0
Hi,
I'm new to the site and not sure if I'm posting in the right place as this is not exactly a homework problem, but just a problem in general. I'm looking at E=mc2 and how the units can be broken down into eV=u (I think?). I don't know what happens to the distance units, though, if we keep seconds as the time units.

Homework Statement


[/B]
What happens to the distance units when E=mc2 is converted from joules=(kg)(m/s)2 into eV=u? Where eV is electron volts and u is atomic mass units.

Homework Equations


[/B]
1u= 1.66x10-27 kg
1eV=1.6x10-19 J
c2=9.315x108 eV/u

3. Attempt at Solution

I really don't know what to do here. Can it just stay meters?
Any help is appreciated.
 
Work out the required scale factor... i.e. if your mass were in 4-kilogram lots, then the distance would need to be in units of 2 meters to keep energy in joules. But if you needed the energy is units of a quarter joule, what do you need to do to the meter?
 
LouFerrigno said:
and how the units can be broken down into eV=u (I think?)
You can convert them into each other using this equation, but they are not equal.

Plugging in m = 1u = 1.66*10-27 kg gives an energy of mc^2 = 1.66*10-27 kg * (3*108 m/s)2 = 1.5*10-10 J = 1.5*10-10 J * (1eV / (1.6*10-19 J)) = 0.94 * 109 eV or about 940 MeV.

The multiplication by 1eV / (1.6*10-19 J) works because this fraction is equal to 1, and 1 kg m2/s2 = 1 J.
 
Simon Bridge said:
Work out the required scale factor... i.e. if your mass were in 4-kilogram lots, then the distance would need to be in units of 2 meters to keep energy in joules. But if you needed the energy is units of a quarter joule, what do you need to do to the meter?

Oh, ok I get it. I guess I was over complicating it in my head. Thanks a lot.
 
mfb said:
You can convert them into each other using this equation, but they are not equal.

Plugging in m = 1u = 1.66*10-27 kg gives an energy of mc^2 = 1.66*10-27 kg * (3*108 m/s)2 = 1.5*10-10 J = 1.5*10-10 J * (1eV / (1.6*10-19 J)) = 0.94 * 109 eV or about 940 MeV.

The multiplication by 1eV / (1.6*10-19 J) works because this fraction is equal to 1, and 1 kg m2/s2 = 1 J.

Thanks. This helps.