Question

`"A"_0/2` atoms of X(g) are converted into X⁺(g) by absorbing energy E₁. `"A"_0/2` ions of X⁺(g) are converted into X⁻(g) with release of energy E₂. Hence ionization energy and electron affinity of X(g) are ______.

विकल्प

• `(2"E"_1)/("A"_0), (2("E"_1 - "E"_2))/("A"_0)`

• `(2"E"_1)/("A"_0), (2("E"_2 - "E"_1))/("A"_0)`

• `(("E"_1 - "E"_2))/("A"_0), (2"E"_2)/("A"_0)`

• None of these

Answer 1

`"A"_0/2` atoms of X(g) are converted into X⁺(g) by absorbing energy E₁. `"A"_0/2` ions of X⁺(g) are converted into X⁻(g) with release of energy E₂. Hence ionization energy and electron affinity of X(g) are `bbunderline((2"E"_1)/("A"_0), (2("E"_2 - "E"_1))/("A"_0))`.

Explanation:

\[\ce{X(g) -> X^+(g)}\]; I.E.

∴ For `"A"_0/2` absorbed energy

= `"A"_0/2` I.E. = E₁

⇒ I.E. = `(2"E"_1)/"A"_0`

\[\ce{\underset{↓}{X^+(g)} -> X(g) -> \underset{↓}{X^-(g)}}\]

released I.E released E.A.

For `"A"_0/2` ion, released energy

= `"I.E." xx "A"_0/2 + ("E.A.")"A"_0/2` = E₂

⇒ E.A. = `(2("E"_2 - "E"_1))/("A"_0)`