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Grade: 12

                        

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+?

2 years ago

Answers : (1)

SAKSHI
16 Points
							
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
2 years ago
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