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If n=1 is taken to be the reference of the potential energy, what will be the kinetic energy (in eV) of an electron in second excited state in Li²+?
As its in question that second exited state so N=3 amd from formula total energy = -13.6*Z Square by N square answer is -13.6 and now we have formula total energy = -K.E-13.6 = -K.E SO, K.E is 13.6 eV OPTION 1 OR A
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