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The first ionisation potential of Na is ...

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

A

`-2.55 eV`

B

`-5.1 eV`

C

`-10.2 eV`

D

`+2.55 eV`

Text Solution

Verified by Experts

The correct Answer is:
B
`NararrNa^(+)+e^(-)" "1^(st)I.E`
`Na^(+)+e^(-)rarrNa` Electron gain enthalpy of `Na^(+)` Because reaction is reverse so then.
`DeltaH_(eg)=-5.1ev`.
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