1.

The one electron species having ionization energy of 54.4 eV is

Answer»

H
`He^(+)`
`B^(4+)`
`Li^(2+)`

Solution :`I.E. = (13.6 Z^(2))/(N^(2)) eV`
`= 13.6 Z^(2)` for one-electron SPECIES
`:. 13.6 Z^(2) = 54.4 or Z^(2) = 4 or Z = 2, " i.e., " He^(+)`


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