Guest

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+?

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+? 

Question Image
Grade:12

1 Answers

SAKSHI
16 Points
6 years ago
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

Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free