Do Electrons have MASS?

  1. Whirling around Atoms are the Electrons. I was wondering if they have MASS? If so, it would seem there tremendous rotational velocity would cause them to fly away from the nucleus but they don't, so I kinda think maybe they do not have mass, only some form of energy. Can you help answer this for me?
     
  2. jcsd
  3. Yes, electrons have mass. A single electron weighs about 9.109534 x 10[itex]^{-31}[/itex] kg.

    The reason they don't 'fly away': quantum mechanics.
     
  4. ZapperZ

    ZapperZ 30,165
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    There is a very simple experiment often done in college level intro physics labs that measure the e/m ratio of electrons. Here, the path of electrons coming off a hot cathode are bent by an external uniform magnetic field (typically generated by a helmholtz coil). As one increases the potential between the anode and the cathode (thus imparting more kinetic energy to the electrons), one sees that the path of the electrons are bent less and less. An object with a mass that is travelling faster will require a greater force to bend its path by the same amount, or given a constant bending force (as in this experiment), an object with mass that is moving faster will be bent less. This essentially is the same principle applied in a mass spectrometer!

    This experiment would have been useless, or give weird results, if electrons do not have a mass.

    Zz.
     
  5. GCT

    GCT 1,769
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    Yes electrons have mass. Most chemical phenomena are based on this fact. For your latter question, remember that E=mc^2.
     
  6. The great evil of the Bohr model strikes again!

    Electrons do not whirl around atoms. (or nuclei, as I think you meant to say) They exist as an uncertain "electron cloud".

    In general, massless particles travel at c. The electron does not travel at c, so it has rest mass.
     
  7. You are correct about the "uncertainty principle" in that we can not be certain where an electron will be at any particular cloud, but I believe that it is still up for debate as to wether it is a cloud or a cloud due to the whirling of the electrons around the nucleus.

    I guess this would depend on wether the electon was considered to be a particle or a wave????

    Nautica
     
  8. selfAdjoint

    selfAdjoint 8,147
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    Even if the electron is considered a particle, it does not have a well defined position, trajectory, or momentum, due to uncertainty. So the whirling concept brings in false assumptions - at least to my mind.
     
  9. I will agree with that. So are you saying the "cloud" has mass and there is not actually a point mass.

    Nautica
     
  10. The electron is a point mass, but when it's in a "cloud" it's uncertain where this mass is, and the more you know where it is the less you know when it was there. Why this is important is because it's not just that our instruments suck it’s that the information isn't even there to "know."
     
    Last edited: Jan 23, 2004
  11. The "clouds", or orbitals, are solutions to Schrodinger's equation (not directly an uncertainty principle issue). We're quite certain they (orbitals) exist, and that they represent a probability density of finding the electron. The notion of a "chunk" of matter moving around is classical and fall apart very quickly in the quantum regime.

    If you consider the shape of p or d orbitals, there's no way the electron can be "whirling". Furthermore (to kill the whirling idea altogether), ask yourself what happens if a charge accelerates.
     
  12. Nereid

    Nereid 4,014
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    orbitals - quintessential QM

    Very good GRQC, we should use the concept of orbitals - as in chemical bonds etc - more often in explanations of QM, the nature of the electron, etc.
     
    Last edited: Feb 7, 2004
  13. It's all clear today about the electrondynamics. however i can not stop to think that how did physisits or chemists measure the mass of an electron. and if this was a measurement what was the reference and how much error is acounted for?
     
  14. Astronuc

    Staff: Mentor

    Last edited: Mar 11, 2007
  15. jtbell

    Staff: Mentor

    I think you left out a few words there:

    "The electron's charge to mass ratio was known (Thomson)..."
     
  16. Astronuc

    Staff: Mentor

    Yeah - I left out some words while changing the order of the sentence. :rolleyes:

    Thank jt - I added the crucial words.
     

  17. This is true, but how would you describe the "motion" of a free electron? It would be fine to describe it as a particle with velocity as well as a wave which is a solution to the schrodinger eq with no potential
     
  18. Even if they behaved classically, they still wouldn't "fly away," as long as they had a low enough energy. After all, planets most certainly have mass and also orbit the sun.
     
  19. Cathode Rays

    Hey...i really dont get this part of dischage tubes....a potential difference is applied and the 2 electrodes are cathode and anode...now if electrons are produced and emitted at the cathode...shouldnt they always hit the anode...since they would need to hit the anode to complete the circuit?? :S Cuz for cathode rays in TV's they go through the anode...and i dont understand how this works :S....??? Any ideas anyone please??
     
  20. An electron in any orbital isn't going anywhere, because an orbital is a stationary state solution to the S.E.

    However, the probability density does 'swirl' in practice because the wavefunction is a linear combination over all orbitals. That allows for interference between the phases of differing orbitals creating time dependent regions of low and high probability.

    The 'swirls' are not much like planetary orbits, but there is movement.

    ------------------

    This applet should convince you that particle wavefunctions do swirl in a potential well.

    http://www.falstad.com/qm2dosc/

    Edit: This applet is even better- it's for the H-atom

    http://www.falstad.com/qmatom/

    You can easily combine an S orbital together with a P orbital and watch oscillations occur.
     
    Last edited: Apr 11, 2007
  21. it says so on the formula sheet ... (9.11 x 10^negative something ) :smile:
     
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