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I Electronvolt unit

  1. Mar 14, 2016 #1
    I'm doing some numerics on a system, where the relevant length scale is nanometers and energy scale is electronvolts. But are these two scales not affected by the choice of each other? I'm kind of confused. What should I choose as units for a system like mine.
     
  2. jcsd
  3. Mar 14, 2016 #2

    davenn

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    they measure 2 totally different things
    nanometres is a length measurement ( eg a physical distance, a wavelength etc)

    electron Volt is a measurement of the energy of an electron ...
    from wiki
    In physics, the electronvolt[1][2] (symbol eV; also written electron volt) is a unit of energy equal to approximately 160 zeptojoules (symbol zJ) or 6981160000000000000♠1.6×10−19 joules (symbol J).


    Dave
     
  4. Mar 15, 2016 #3

    Drakkith

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    Whoa, where did all those extra numbers and the spade come from!? When I copy the same line I get the following:

    In physics, the electronvolt[1][2] (symbol eV; also written electron volt) is a unit of energy equal to approximately 160 zeptojoules (symbol zJ) or 1.6×10−19joules (symbol J).

    No. An electronvolt is an amount of energy equal to moving a unit of charge equal to an elementary charge (the charge of an electron or proton) across a potential difference of 1 volt. Since the energy required to move an electron or proton through 1 volt does not depend on distance, the two scales do not affect one another.
     
  5. Mar 15, 2016 #4

    davenn

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    dunno LOL ... it's just the way it pasted for me in PF
    I tried a couple of times and it wouldn't change so gave up and left as is :rolleyes::rolleyes:
     
  6. Mar 15, 2016 #5

    DrClaude

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    You have to be careful that you use the correct value for the fundamental constants. For instance, ħ would be in eV⋅s. If you write the equations with all the proper units, it should all be clear.
     
  7. Mar 15, 2016 #6

    davenn

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    who was that directed to ? :smile:
     
  8. Mar 15, 2016 #7
    But SI the units of hbar are J*s = N*m*s
    So in terms isn't hbar in units of nm equal to hbar(nm) = 10^9 * hbar(SI)
    maybe I am just confusing myself.
     
  9. Mar 15, 2016 #8

    DrClaude

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    The OP.


    Yes. Considering that ħ = 6.582119×10-16 eV⋅s, in a system of units where SI units are used everywhere except for length, which is in nm, and energy, which is in eV, ħ has a numerical value of 6.582119×10-7.

    Edit: my comment that SI units are used everywhere except length and energy might not hold. I have to do a few checks first.
     
  10. Mar 15, 2016 #9

    DrClaude

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    Indeed, since E = ML2T-2, if E is in eV and L in nm, you need to modify the unit of mass or of time to get a consistent system.
     
  11. Mar 15, 2016 #10

    Drakkith

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    So is my above post (#3) incorrect then?
     
  12. Mar 15, 2016 #11

    DrClaude

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    You answered no to the question "But are these two scales not affected by the choice of each other?", which was correct, but your comment
    is actually irrelevant. Choosing one Drakkith as the unit of energy would also work, however that is defined.

    One can choose L = nm and E = eV and derive a consistent set of units. One can still choose for instance a unit of time or of mass, but not both, as otherwise the system of units will be inconsistent.
     
  13. Mar 15, 2016 #12
    So will I be in good shape if I just use:
    hbar = 10^9* hbar(in eV)
    I am very confused, but I guess in the above I have redefined s=10^9 * s, i.e. a new unit of time.
     
  14. Mar 15, 2016 #13

    DrClaude

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    No.

    What you have is E = M L2 T-2, so T = (M L2 E-1)1/2, which gives a unit of time (lets call it ##\tau##)
    $$
    \tau = \left( \frac{\mathrm{kg}\ 10^{-18}\ \mathrm{m}^2}{1.602177 \times 10^{-19}\ \mathrm{J}} \right)^{1/2} = 2.498301\ \mathrm{s}
    $$
    So ħ = 6.582119×10-16 eV s becomes
    $$
    \hbar = 6.582119\times10^{-16}\ \mathrm{eV}\,\mathrm{s} \times \frac{\tau}{2.498301\ \mathrm{s}} = 2.634638\times10^{-16}\ \mathrm{eV}\,\tau
    $$

    If you decide to redefine mass instead of time, then ħ doesn't change: it is still ħ = 6.582119×10-16 eV s.
     
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