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

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    E=mc^2 Units
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The discussion focuses on converting the equation E=mc² from joules to electron volts (eV) and atomic mass units (u), specifically addressing the implications for distance units when time remains in seconds. The conversion involves understanding that 1 u equals 1.66 x 10^-27 kg and 1 eV equals 1.6 x 10^-19 J, leading to a derived energy of approximately 940 MeV for 1 u mass. Participants clarify that while the units can be converted, they are not directly equal, and the appropriate scale factor must be considered based on mass and energy requirements. Ultimately, the discussion concludes that the conversion process is manageable and not as complicated as initially thought.
LouFerrigno
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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.
 
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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.
 
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