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

In summary: So in summary, when converting E=mc2 from joules=(kg)(m/s)2 to eV=u, the distance units can be converted by multiplying by a scale factor of 1eV / (1.6*10-19 J) to keep the energy in the same units.
  • #1
LouFerrigno
3
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.
 
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  • #2
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?
 
  • #3
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.
 
  • #4
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.
 
  • #5
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.
 

1. What does E=mc^2 represent?

E=mc^2 is the famous equation developed by Albert Einstein that represents the relationship between mass and energy. It states that the energy (E) of an object is equal to its mass (m) multiplied by the speed of light (c) squared.

2. How is E=mc^2 related to eV=u?

E=mc^2 is related to eV=u through the conversion factor of 1 eV (electron volt) being equivalent to 1.602 x 10^-19 joules. This means that the energy (E) in joules can be converted to electron volts (eV) by dividing by 1.602 x 10^-19, and vice versa.

3. What are the units of E=mc^2?

The units of E=mc^2 are joules (J) for energy, kilograms (kg) for mass, and meters per second (m/s) for the speed of light.

4. How is E=mc^2 used in scientific research?

E=mc^2 is used in scientific research to understand the relationship between mass and energy, and how they can be converted into each other. It is also used in fields such as nuclear physics and astrophysics to calculate the energy released from nuclear reactions and the mass of particles.

5. Is E=mc^2 the full equation for mass-energy equivalence?

No, E=mc^2 is not the full equation for mass-energy equivalence. The full equation is E^2 = (mc^2)^2 + (pc)^2, where p represents momentum. However, E=mc^2 is a simplified version that is often used for practical calculations.

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