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Why does an electron shell further away from nucleus has higher energy?

  1. May 16, 2016 #1
    Using electrical potential energy =1/4πεo Q1Q2/r , a particle further away from nucleus has lower magnitude of energy

    Using coulomb's law, a particle further away from nucleus experiences weaker attraction, hence less energy is needed to maintain orbit* around that e-shell compared to a electron shell closerr to nucleus, hence the one closer to nucleus supposedly should have higher energy.

    *i know in reality e- does not orbit around a atom, but its position exists as a probability density of radial probability function.
     
  2. jcsd
  3. May 16, 2016 #2
    [SORRY] Thread's name is truncated!!! Should've been "
    Why does an electron shell further away from nucleus has higher energy level "
     
  4. May 16, 2016 #3

    Drakkith

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    Staff: Mentor

    That's not how energy and force are related. In fact, it requires zero energy to keep something in a classical orbit (ignoring the inevitable loss of energy via EM radiation due to the electron being accelerated). What do you know about work?
     
  5. May 16, 2016 #4
    Alright, maybe not directly related. E=WD/t =(F×D)/t

    Would u elucidate abt my first point regarding energy,
    And the 2nd point regarding that e- further away having weaker attraction compared to a e- nearer to nucleus,
    Both points in which are contrary to the fact that electron shells of further distance away has higher energy level?
    Please, and Thank you !
     
  6. May 16, 2016 #5

    Drakkith

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    Staff: Mentor

    Time should not be in there. Including time would make this energy per unit of time, which is power, a different concept. Work is essentially defined as the product of a force times a distance. Specifically this product is actually a dot product, which means that the force has to act in a direction parallel to the direction of the displacement. For example, a planet in a perfectly circular orbit always experiences a force perpendicular to its direction of motion, so no work is performed on the planet by gravity and no energy is gained or lost from it at any time.

    I am attempting to.

    Now, consider what would happen if a photon was absorbed by an inner electron. The electron would jump up to a new energy level, using the energy provided by the photon. In other words, the photon performed work on the electron, which requires a transfer of energy from the photon to the electron. So, our electron has absorbed energy and moved away from the nucleus. This is the same thing that happens to rockets we shoot into interplanetary space. If we want to get it into an solar orbit outside of Earth's orbit, we have to perform work to move against the force of gravity. The rocket actually winds up with less speed and kinetic energy than it had before, but it ends up gaining potential energy, more than enough to offset the loss of kinetic.

    A similar principle happens inside of atoms. The electrons further away from the nucleus have more potential energy than the ones closer in. If you were to consider them as being in classical orbits around the nucleus, then you could think of them as being accelerated as they fall towards the nucleus during a transition, converting potential energy into kinetic energy, and then releasing part of that kinetic energy in the form of radiation or heat and ending up in a stable orbit closer to the nucleus than before. But since we are dealing with quantum rules here we can't have a nice pretty picture like that. There is no gradual acceleration, only an instantaneous transition and release of energy. Still, the end result would is the same. The electron drops into a lower energy level and releases energy in the form of radiation or heat.
     
  7. May 16, 2016 #6

    Dale

    Staff: Mentor

    This claim is only true if both particles are positively charged or both particles are negatively charged. If one particle is positive and the other is negative then the potential energy is lower the closer it is. Don't forget the signs!
     
  8. May 16, 2016 #7
    [SORRY TYPO] :
    It's position exists as a probability density OR* radial probability function

    is it possible to edit the contents of the post?
     
  9. May 16, 2016 #8

    Drakkith

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    Only for a short time after posting.
     
  10. May 16, 2016 #9

    I see. Yes, if a electron is closer to a proton, potential energy is more negative, hence less.
     
  11. May 16, 2016 #10
    However there's one aspect i don't understand. In chemistry, the electron shell of higher principal quantum number, has higher energy levels !
     
  12. May 16, 2016 #11
    Ahh yes ! Okay, i understand better now.
    Yes, no energy is needed for an electron to maintain an orbit as acceleration is perpendicular to direction of motion

    Removing e- from a atom or transitioning it to a higher energy level requires energy.

    I realised my misconception.
    What about the concept of why electron shells of higher principal quantum number have higher energy level ?
     
  13. May 16, 2016 #12

    Drakkith

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    Staff: Mentor

    That's beyond my expertise, sorry. I expect it has something to do with the way the math works, but I don't know for sure.
     
  14. May 16, 2016 #13
    I think by EPE = 1/4πε( Qproton Qe-) /r,
    As r increase, EPE becomes less -ve(i.e. more +ve) , indicating higher energy level
     
  15. May 16, 2016 #14

    Dale

    Staff: Mentor

    This is correct for a pair of classical point particles. For actual protons and electrons it is only part of the answer (the potential well). The rest requires the uncertainty principle and the kinetic energy. In the smaller orbitals the PE is lower but the KE is higher. This essentially prevents the electron from crashing down into the nucleus.
     
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