I have been tasked with using the one dimensional particle in a box model to calculate the zero point energy of an hydrogen atom with different sizes. Using the formula $E_1=h^2/(8mL^2)$ for a box with length 0.3nm and 1cm I receive results that are very huge, which made me question the validity of my approach.

Is the one dimensional rigid box model really valid for atoms?

  • $\begingroup$ This is a classic type of question to ask in semi-advanced QM courses, typically followed by using perturbation theory to improve the initial treatment with a particle in a box. Without more information, it is hard to tell for sure if you've gone wrong, but my memory is that using a particle in a box is a pretty bad approximation of the hydrogen atom. This makes sense as the hydrogen atom energy levels scale as $1/n^2$ while the PIB ones scale as $n$, so they are qualitatively dissimilar. Also, I think the PIB should give too large an energy as you impose an artificial barrier. $\endgroup$ – jheindel Sep 13 '19 at 21:20
  • $\begingroup$ Very thankful for your answer. What I've done is just plug in the mass for an hydrogen atom together with the length of by box(0.3nm and 1cm). I was to answer in Joules and eV but receive answers in the magnitude of 10^21 eV and 10^27 eV. Until now we haven't covered perturbation theory, so I'm not sure if this is what the teacher will request of us later on. $\endgroup$ – Per Arne Sep 13 '19 at 21:28
  • $\begingroup$ That looks to me like a unit error. Make sure you have a teeny-tiny mass for the hydrogen atoms if you are working in SI units. $\endgroup$ – jheindel Sep 13 '19 at 22:15
  • 2
    $\begingroup$ After reviewing my calculations I found a mistake. Now for the box of length 0.3nm I get the result ~2.227 meV. This seems more correct to me, but I'm not sure. Thanks for your help. $\endgroup$ – Per Arne Sep 13 '19 at 22:38
  • 1
    $\begingroup$ @PerArne Sounds like you might have an answer to submit! $\endgroup$ – LordStryker Sep 13 '19 at 23:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.