E=mc^2 units

  • #1

Main Question or Discussion Point

When solving with E=mc^2 what units should I use in the answer??
For example: (if I have an object with a mass of 43kg)
E=mc^2
E=(43kg)(299792458m/s)^2
E=3.86464727 × 10^18
Now here's my problem, do I include units of J/kg (joules per kilogram) or eV (electron volts)??
If someone could help me it would be very helpful.

Thanks,
Al
 

Answers and Replies

  • #2
28,902
5,160
What is the SI unit of energy?
 
  • #3
126
0
Joules are kg*m2/s2. Since the units you chose for mass and c are kg, m and s, Joules are what you get.
 
  • #4
Pengwuino
Gold Member
4,989
15
Electron volts would require the mass to be given in [itex]{{eV}\over{c^2}}[/itex], which is common in high energy physics or when you're talking about atomic scale stuff. For example, the rest mass of a proton is [itex]938 \times 10^8 {{eV}\over{c^2}}[/itex]. By itself, eV is not an SI unit.
 
  • #5
Oh ok so the unit would be J/kg, no?
Or would it just be J, because I read that it might be J/kg, so which one??
 
  • #6
jtbell
Mentor
15,505
3,300
E=(43kg)(299792458m/s)^2
E=3.86464727 × 10^18
1. What do you get with you multiply/divide the units together as indicated by your calculation? (Don't do any conversions!)

2. What is a joule (J) in terms of kg, m, and s?
 
  • #7
1,506
17
The SI unit of energy is Joule.
eV is a convenient unit of energy in some areas of physics. The relationship between eV and Joule is
1eV = 1.6 x 10^-19 J
 
  • #8
Oh alright so the unit would be Joule.

Thanks all,
Al
 
Last edited:
  • #9
Khashishi
Science Advisor
2,815
493
Either is correct, since joules and electron volts are both units of energy. If you are working with a macroscopic sized system, you probably want to use joules. If you are working with a microscopic system (i.e. single particles) then electron volts are probably more convenient.
 
  • #10
Either is correct, since joules and electron volts are both units of energy. If you are working with a macroscopic sized system, you probably want to use joules. If you are working with a microscopic system (i.e. single particles) then electron volts are probably more convenient.
Ah ok now that explanation makes a lot of sense thanks.

Al
 

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