1.

Theionizationenergyhydrogenatom is13.6ev .Theionizationenergyof He

Answer»

13.6 eV
54.4eV
122.4eV
ZERO

Solution :`E_(He = (+13.6z^(2) He^(+))/(n^(2) He^(+))`
`E_(H) = (13.6 Z^(2)H)/(n^(2)H)`


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