Periodicity-2

X(g) --> X⁺(g) ΔH (for 1/2 mole)

so ionisation energy = 2ΔH as ionisation energy is calculated per mole

X(g) --> X⁺(g) + X^-(g) ΔH₂

now ΔH₂ = ΔH + EA/2 (as only half mole is ionised to X-)

=> EA = 2(ΔH₂ - ΔH)

Wrong answer !!!!!
check it again ..there is a silly mistake .

Let the I.E of X be I and electron affinity of X be E kj/mol .
X(g)→X⁺ (g)+e
[1/(N_{A V})×N₀/2]=∆H
I =[(2∆H×N_AV)/N₀] kj/mol
(where N_AV is the Avogadro”s no)
Similarly
X(g)→X⁺ (g)+e
X(g)+e→X^- (g)
∆H+[E/(N_{A V})×N₀/2]=∆H₂
E = ((2(∆H₂-∆H))/N₀ ) kj/mol
Got the answer !!
see where u did mistake !!

I thought that N₀ is Avogadro's no. we use N_A or N_o as avogadro..