If in the hydrogen atom potential energy at $\infty$ is chosen to be $\pu{13.6 eV},$ then what is the ratio of total energy and kinetic energy (with the sign) for the first Bohr orbit?

I think we don't consider the reference frame, and I answered with the relation

$$E_\mathrm{tot} : E_\mathrm{k} : E_\mathrm{p} = -1 : 1 : -2,$$

but the answer given is zero. Why?

  • $\begingroup$ What would be the potential energy of the 1st orbit in this reference frame? Then what would be the total energy? $\endgroup$ Aug 15, 2019 at 7:46
  • $\begingroup$ P.E = $-13.6$eV in this frame. But would the K.E be still the same? $\endgroup$
    – Tony
    Aug 15, 2019 at 8:57
  • 1
    $\begingroup$ Of course K.E. will be still the same, but P.E. will not. In particular, it won't be -13.6 (it was like that when we had $0$ at $\infty$, which is not the case anymore). $\endgroup$ Aug 15, 2019 at 9:11
  • $\begingroup$ Potential energy would be $-27.2$eV in the normal case. In this reference, it would be $-13.6$eV. $\endgroup$
    – Tony
    Aug 15, 2019 at 9:50
  • $\begingroup$ Sure. My fault, I confused it with the total energy. Well, anyway, K.E. stays the same as P.E. changes; what will happen to the total energy and the ratio? $\endgroup$ Aug 15, 2019 at 9:53


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