The first ionization energy of $H$ and $H e$ are $13.6 e V$ and $24.6 e V$ respectively. If the energy which would be given out during the formation of ground state of $H e$ atom from $H^{+}$ is $X e v$ then $- X$ is :

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Solution

Given:
$H ⟶ H^{+} + e; I E_{1} = 13.6 e V$
$H e ⟶ H^{+} + e; I E_{1} = 24.6 e V$

We have to determine the values of
${H e}^{+} + e ⟶ {H e}^{+}; Δ H = a$
${H e}^{+} + e ⟶ {H e}; Δ H = b$
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$H^{2 +} + 2 e ⟶ {H e}; Δ H = (a + b)$

The $I E_{1}$ of $H^{+} = I E_{1 H} \times 2^{2} = 13.6 \times 4 = 54.4 e V$

$∴ a = - 54.4 e V$

Also $H^{+} + e ⟶ {H e}; I E_{1} = 24.6 e V$

$H^{+} + e ⟶ {H e}; I E_{1} = 24.6 e V$

$b = - 24.6 e V$

Thus, total energy given out $= a + b = - 79 e V$

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