I am interested in knowing how much is one eV of energy. Everywhere I found are the technical definitions. Can anybody please tell me how much is this much energy. I need something which I can feel. I mean how much work I can do with 1 eV? Can I drive a 1000cc car for 1hour? Any of example in context of real life usage would be interesting.


1 Answer 1


An electronvolt is just the energy acquired when an electron falls through a potential of 1 volt, which means $$1\: {\rm eV} = e \times 1\:{\rm V} = 1.6 \times 10^{-19}\: {\rm J}$$

When you lift your $2.5\:{\rm kg}$ laptop (a 15-inch Apple MacBook Pro, for example) by a foot, you do work of approximately $2.5\: {\rm kg} \times 10\: {\rm m\,s^{-2}} \times 0.3 \: {\rm m} = 7.5 \:{\rm J}$ which is about $4.7 \times 10^{19}\:{\rm eV}$. So an $\rm eV$ is a really low energy scale by everyday standards.

One $\rm TeV$ (a teraelectronvolt) is about the energy of motion of a flying mosquito.

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    $\begingroup$ In CERN in running LHC project, the scientists say that they are going to do a collision of 8 TeV. So whats the big deal in that? I may be understanding something wrong. Please clear my concept.... $\endgroup$ Apr 5, 2012 at 13:11
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    $\begingroup$ A subatomic particle is much, much lighter than a mosquito. In a back-of-the-envelope calculation, let's say a mosquito is 1.7 grams and a proton is 1.7 $\times 10^{-27}$ kg. So to have the same kinetic energy as a flying mosquito, the proton must be travelling about $10^{12}$ times faster than the mosquito. That's a big deal. $\endgroup$
    – ptomato
    Apr 5, 2012 at 13:32
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    $\begingroup$ Also in LHC they are running around 100 trillion protons, each having 8 TeV. The 100 trillion protons weight about 1 nanogram in total (that is like nothing), but their net kinetic energy is huge: it is like a 300 tones train going at 100 km per hour. Imagine that something weighting 1 nanogram has an energy of a fast going train. That puts things into perspective. $\endgroup$
    – mpv
    Dec 5, 2013 at 0:04
  • $\begingroup$ @mpv - Yikes! That's one gnarly train wreck! Thanks for the above answers and comments. $\endgroup$
    – pr1268
    May 1, 2016 at 9:21

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