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

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Homework Help Overview

The discussion revolves around the conversion of energy units from joules to electron volts (eV) and atomic mass units (u) in the context of the equation E=mc². Participants are exploring how distance units are affected during this conversion process, particularly when maintaining seconds as the time unit.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning how to handle distance units when converting energy units from joules to eV and u. There is a discussion about the implications of using different mass scales and how that affects the distance measurement.

Discussion Status

Some participants have provided insights into the conversion process and the necessary scale factors involved. There appears to be an ongoing exploration of the relationships between the units, but no consensus has been reached on the specific handling of distance units.

Contextual Notes

Participants are considering the implications of using different mass units (e.g., kilograms versus atomic mass units) and how this affects the energy calculations and unit conversions. There is also mention of potential confusion regarding the relationships between the units involved.

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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