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Deriving the ground state energy of a hydrogen atom?

  1. Sep 24, 2014 #1
    1. The problem statement, all variables and given/known data
    Hello! I am trying to derive the ground state enegry of a hydrogen atom, and have come to

    [itex]U=\frac{-mk_{0}^{2}Ze^{4}}{n^{2}\hbar^{2}}[/itex]

    Problem is, I know there should e another factor of 2 in the denomenator because I get the ground state energy of hydrogen as being 27.145eV rather than 13.6eV

    2. Relevant equations
    Centripetal force is [itex]F_{c}=\frac{mv^{2}}{r}[/itex]

    Coulomb's law is [itex]F=\frac{k_{0}q_{1}q_{2}}{r^{2}}[/itex]

    Potential is [itex]U=\frac{k_{0}q_{1}q_{2}}{r}[/itex]

    And using Bohr's idea that the angular momentum can only be multiplies of [itex]\hbar[/itex], so

    [itex]mvr=n\hbar[/itex]

    so [itex]v=\frac{n\hbar}{mr}[/itex]

    [itex]v^{2}=\frac{n^{2}\hbar^{2}}{m^{2}r^{2}}[/itex]

    3. The attempt at a solution
    For a system with a single electron orbiting a nucleus with Z protons, and equating the centripetal force with Coulomb force,

    [itex]\frac{mv^{2}}{r}=\frac{k_{0}Ze^{2}}{r^{2}}[/itex]

    [itex]mv^{2}=\frac{k_{0}Ze^{2}}{r}[/itex]

    Now substituting in for [itex]v^{2}[/itex],

    [itex]\frac{n^{2}\hbar^{2}}{m^{2}r^{2}}=\frac{k_{0}Ze^{2}}{mr}[/itex]

    So [itex]r=\frac{n^{2}\hbar^{2}}{mke^{2}}[/itex]

    Substituting this is for the potential,

    [itex]U=\frac{-mk_{0}^{2}Ze^{4}}{n^{2}\hbar^{2}}[/itex]

    Please tell me where have gone wrong!
    Thank you in advance :)
     
  2. jcsd
  3. Sep 24, 2014 #2
    13.6eV is the total energy of that system...so in addition to potential energy , you'll have to take kinetic energy into consideration , which you apparently have not
     
  4. Sep 27, 2014 #3
    Hi throneoo, thanks for your reply!

    How would I take into account the kinetic energy to derive the 13.6eV? Also, I'm not too sure that kinetic energy is the problem here because the answer I got was exactly twice what it should be, so I think it may be more likely that I missed a factor of a half somewhere. Though I am not too sure...
     
  5. Sep 27, 2014 #4

    Simon Bridge

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    You find the kinetic energy and add it to the potential energy.
    You did make a miscalculation though ... what sign should the potential energy be?
     
  6. Sep 27, 2014 #5

    Orodruin

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    Does ##mv^2## remind you of something that you would typically see in an expression for kinetic energy? This would be the hand-waving approach but would be similar in rigour to applying the Bohr model. Alternatively you could actually compute the ground state wave function and its corresponding kinetic energy term - or apply the virial theorem.
     
  7. Sep 27, 2014 #6
    you are very close to deriving the expression for the kinetic energy in terms of the potential energy....just derive it and plug it into the expressionfor the total energy . that should you a ground state energy of -13.6eV
     
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