Or in other words which forces keep electrons in orbitals and prevent it from flying away or crashing into the nucleus according to modern understanding?

  • thebestaquaman@lemmy.world
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    3 months ago

    There are a lot of good answers here already, but I’ll try to attack the question from a new angle.

    Firstly, yes: they experience an attractive force from the nucleus, and would in principle have their lowest possible potential energy if they were located exactly in the nucleus. An equilibrium state is the state with lowest energy, so why aren’t they exactly in the nucleus?

    Consider that an electrons position and speed cannot be exactly defined at the same time (uncertainty principle). So an electron with an exact position could have any speed. If you compute the expectation value of a particles kinetic energy, when the particle can have any speed, you’ll find that it’s divergent (goes to infinity).

    So: Because an electron with an exactly defined position must have infinite kinetic energy, the equilibrium state cannot be an electron with an exactly defined position, and so cannot be an electron exactly in the nucleus. So what do we do?

    We have to make the electrons position “diffuse”. Of course, that means it is no longer exactly inside the nucleus, so it gains some potential energy, but on the other hand it can move more slowly and has lower kinetic energy.

    The equilibrium state is the state we find where the trade off between kinetic and potential energy gives us the lowest total energy, which is described as a 1s orbital. The electron is “diffuse” enough to have a relatively low kinetic energy, and “localised” enough to have a relatively low potential energy, giving as low total energy as possible.

    Once you start adding more electrons you need to start taking Pauli exclusion into account, so I won’t go there, but the same manner of thinking still essentially holds up.

    • JWBananas@lemmy.world
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      3 months ago

      So: Because an electron with an exactly defined position must have infinite kinetic energy

      There are an infinite number of velocities I can use to get up off the couch right now.

      That does not mean that I will get up off the couch at infinite velocity.

      • thebestaquaman@lemmy.world
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        3 months ago

        Yes, there are an infinite number of velocities you can use, but if you look at their distribution, you’ll find that it quickly goes to zero somewhere around 1-2 m/s, so the expectation value of the velocity is convergent.

        If you have an object with a velocity taken from a distribution that doesn’t approach zero sufficiently fast as the velocity goes to infinity, the expectation value diverges. A simple example would be a person that would be half as likely to get up at a velocity of 2 m/s as 1 m/s, and half as likely to get up at 4 m/s as 2 m/s, etc.

        The more mathematical version of the same argument is to compute the kinetic energy of a particle whose wavefunction is a delta pulse (i.e. a particle whose position is exactly defined), and you’ll find that the particle has infinite energy.

      • SmoothOperator@lemmy.world
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        3 months ago

        You choose a velocity from an infinite number of options, but the electron exists in a superposition of all those options.