Ratio of Total energy to Kinetic energy in hydrogen atom when the reference is changed

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.

• What would be the potential energy of the 1st orbit in this reference frame? Then what would be the total energy? – Ivan Neretin Aug 15 at 7:46
• P.E = $-13.6$eV in this frame. But would the K.E be still the same? – Tony Aug 15 at 8:57
• 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). – Ivan Neretin Aug 15 at 9:11
• Potential energy would be $-27.2$eV in the normal case. In this reference, it would be $-13.6$eV. – Tony Aug 15 at 9:50
• 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? – Ivan Neretin Aug 15 at 9:53