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If in the hydrogen atom Potential Energy at $\infty$ is chosen to be $13.6$eV then the ratio of Total energy and Kinetic Energy(with the sign) for $1$st Bohr Orbit is?

My Attempt
I just simply gave the answer given the relation that, $$\text{T.E}:\text{K.E}:\text{P.E} = -1:1:-2$$

Because I think we don't consider the reference frame while speaking of this relation.

But the answer given is 0. Any help would be appreciated.

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  • $\begingroup$ What would be the potential energy of the 1st orbit in this reference frame? Then what would be the total energy? $\endgroup$ – Ivan Neretin Aug 15 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 at 8:57
  • $\begingroup$ 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). $\endgroup$ – Ivan Neretin Aug 15 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 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? $\endgroup$ – Ivan Neretin Aug 15 at 9:53

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