The first ionisation potential of `Na` is `5.1eV`. The value of eectrons gain enthalpy of `Na^(+)` will be

Options

A. `-2.55 eV`
B. `-5.1` eV
C. `-10.2 eV`
D. 2.55 eV

Answer

2

Explanation

`Na to Na^(+)+e^(-), I.P. =+5.1 eV`
`Na^(+) + e to Na to DeltaH_(eg) =-5.1 eV`