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Relativistic speed of an electron

  1. Jul 1, 2013 #1
    1. The problem statement, all variables and given/known data

    Electrons are accelerated by a potential difference of 0.10 MV. Determine:

    a) The mass an accelerated electron

    b) The velocity of an accelerated electron


    The mass of the electron has been successful determined, which I give here. I was doing wrong when I was counting the speed based on the rest mass of the electron. Here, I would like to be sure that qU = 0.5v^2 is correct based on the relativistic mass. I have problems in this course, because I get questions that I've never seen before.


    2. Relevant equations

    qU = mc^2 – m0c^2

    qU = 0.5mv^2

    3. The attempt at a solution


    a)

    100'000 V = 1.602*10^-14 Joule

    m0c^2 = 8.18751672*10^-14 Joule

    1.602*10^-14 = (8.18751672*10^-14) - mc^2

    mc^2 = 9.78951672*10^-14

    m = (9.78951672*10^-14) / c^2

    m = 1.08917583*10^-30 kg

    This far I know the relativistic mass is correct, easy task.


    b)

    I try to use the relativistic mass in this formula:

    qU = 0.5mv^2

    1.602*10^-14 = 0.5(1.08917583*10^-30)v^2

    v^2 = 2.941673798*10^16

    v = 171’513’084 m/s

    Here, I’m not so sure that this is the right way. I only do believe so. I’m thankful for answers.
     
  2. jcsd
  3. Jul 1, 2013 #2

    Curious3141

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    Kinetic energy ##E_k = \frac{1}{2}mv^2## is *not* a relativistic formula. It's a Newtonian formula for kinetic energy, and has no place here. True, at speeds ##v<<c##, the relativistic formula closely approximates this, but this is only an approximation. You have no need of it here.

    Why are you equating "voltage" with "energy"? However, if you put eV as the unit on the LHS, that's correct because electron-volt is a unit of energy.

    Correct. This is the "rest energy" of the electron.

    Makes no sense. ##mc^2##, where ##m## represents "relativistic mass" of the electron, is the larger quantity, so the RHS should be reversed.

    But you did the rest of the calculation correctly. One caveat: you provided too many significant figures. In fact, because of rounding errors in your intermediate steps, many of your figures are incorrect. You should work purely in symbols until the very last step, then round off correctly.

    As I said, this is not a relativistic formula.

    Try instead:

    $$m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$

    and solve for ##v## (velocity). You've already calculated ##m## in the previous step.
     
  4. Jul 1, 2013 #3
    Thank you. I was suspecting this attempt being wrong.

    Ek = E – E0 = (mc^2) / ((1-v^2/c^2)^1/2) – mc^2

    (Ek + mc^2)*((1-v^2/c^2)^1/2) = mc^2

    1 – v^2/c^2 = 0.6994906196

    v^2 / c^2 = 0.3005093804

    v^2 = 2.700979513*10^16

    v = 164’346’570.2 m/s

    This is also 0.5481873588 c.

    I have not been writing all the steps correctly, I did it fast. But I do believe that this velocity is correct. Honestly, I had never seen this formula before, because I have so less experience of relativistic physics. The pit here was that these electrons were accelerated by a very high voltage, and therefore it was relativistiv velocity. All my tasks before has been classic, at voltage of maybe 500 V.

    What is the limit in voltage when it is a relativistic formula for accelerating electrons?
     
  5. Jul 1, 2013 #4
    Now, the thought hit med that maybe I was supposed to base the counting on tha relativistic mass I counted in previous task?
     
  6. Jul 1, 2013 #5
    After som trying to plug in the relativistic mass in the formula

    Ek = (mc^2) / ((1-v^2/c^2)^1/2) – mc^2

    I only came to the result 1, and does not work. I understand it as this formula does in itself count the relativistiv mass from rest mass automatically. So, I do believe that the velocity of
    164’346’570.2 m/s is correct.
     
  7. Jul 1, 2013 #6

    Curious3141

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    I couldn't understand what you're saying in post #5. But your answer for the speed of the electron in post #3 is in the correct ballpark.

    You should be careful about mixing up ##m## and ##m_0##. The former is used (in this context) to refer to relativistic mass, while the latter is simply the rest mass of the electron.

    I hope you understand that in your formulae in posts #3 and #5, you used ##m## when you should've used ##m_0## for consistency.

    Yes, it's easiest to just use the relativistic mass calculated in part a) to work out the velocity, as I suggested in my post. It is probably what the question setter intended you to do, since they guided you with a two-part question.

    BTW, the concept of "relativistic mass" has more or less fallen out of favour in modern physics (a lot of physicists simply consider "rest" or "invariant" mass to be the "only" true mass). But sometimes problems like this one still ask you to think in those terms.
     
  8. Jul 1, 2013 #7

    Curious3141

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    NO. Here the mass you should be "plugging in" is the *rest* mass, not the relativistic mass. Properly, you should have used the symbol ##m_0## instead of ##m##.
     
  9. Jul 1, 2013 #8
    I understand, there is a clear difference between m and m0. In my book, I was copying the formula

    Ek = (mc^2) / ((1-v^2/c^2)^1/2) – mc^2

    And this is exactly how it was written. From Ek = E – E0 I did suppose that this is the same relationship as Ek = mc^2 – m0c^2. I would therefore like to write the formula as following:


    Ek = (m0c^2) / ((1-v^2/c^2)^1/2) – m0c^2


    I suppose this is what you mean? Excuse me, but my book is misleading me.
     
  10. Jul 1, 2013 #9

    Curious3141

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    I don't think I can agree that the book is misleading you unless they use inconsistent symbols. It's fine if they defined rest mass to be represented by ##m##. As I said, in most modern treatments, "rest mass" is the only mass to be considered. Relativistic mass is not emphasised, and instead other properties such as kinetic energy and momentum are used to cover the concept.

    e.g. if I call ##m## the (rest) mass, then I define momentum, total energy and kinetic respectively as:

    $$p = {\gamma}mv$$

    and

    $$E = {\gamma}mc^2$$

    and

    $$E_k = (\gamma-1)mc^2$$

    where

    $$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$

    then everything remains consistent.

    However, this particular question does ask for "relativistic mass" in part a) (otherwise the question makes little sense). Nothing wrong with this per se, it's just an older-fashioned way of looking at things.
     
    Last edited: Jul 1, 2013
  11. Jul 1, 2013 #10
    Thank you for sharing your knowledge. I'm beginner in relativistic physics, and my teacher explain bad. In my E-mail contact, the teacher tells me "use this relativistic mass in the equation for velocity". I was confused.

    In found it therefore correct to use 9.1093826*19^-31 kg as the only value of m in both places in this equation: Ek = (mc^2) / ((1-v^2/c^2)^1/2) – mc^2
     
  12. Jul 1, 2013 #11

    Curious3141

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    And that's fine, because the ##m## there is the rest mass. The "relativistic mass" is ##{\gamma}m##, where ##\gamma## was defined in my last post. It's equivalent to the first term in that difference (ignoring the factor of ##c^2##).

    Basically, what that equation is saying is that the kinetic energy is the difference between the relativistic mass*##c^2## - the rest mass*##c^2##. Makes sense considering that the particle started out with an energy of ##mc^2## at rest, and ended up with an energy of ##{\gamma}mc^2## in motion. The difference in those energies has to be the energy of motion, also known as kinetic energy.
     
  13. Jul 1, 2013 #12
    Yes, was pretty sure about that in your latest post, when it was γ. I was more asking idiot safe, to be absolutely sure. Thank you very much.
     
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