When hydrogen atom excites from ground state what happens to its kinetic and potential energies?

Well, as the electron in the hydrogen atom gets excited, energy put into it that causes such an excitation (e.g. a laser!) transforms into kinetic energy #K# that allows the electron to overcome some of the potential energy of attraction #V# that the nucleus enforces.

The higher the electron gets in energy, the more these energies balance out, i.e.

#overbrace(K)^"Used to excite" + overbrace(V)^"Holds electron in atom" >= V#

where #V = -e^2/(4piepsilon_0r) < 0# is the coulombic potential energy enforced by the hydrogen atom nucleus interacting with the electron. Here, #K# can be greater than #V#.

Now suppose you add energy until you ionize the atom.

The electron starts as the #color(red)("red")# wave shown above.

The more energy you put into exciting the atom, the larger #K# is, until it reaches #K_max = V# (the #color(lightgreen)("green")# wave is almost there).

When #K > V# (as in the #color(blue)("blue")# wave), the electron escapes with some excess kinetic energy #K_"xs"# (in that case, define #K = K_max + K_"xs"#), while the required kinetic energy to escape #K_max = DeltaH_"IE"# is known as the ionization energy:

#K_"xs" = overbrace(K)^"Used to ionize atom" + overbrace(V)^"Holds electron in atom" >= 0#

And in that case, this resembles what is known from the photoelectric effect:

#K_"xs" = hnu - phi >= 0#

where #hnu# is the incoming light energy and #phi# is the threshold energy #hnu_0#, or the work function.

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