What is the value of electron affinity of $N a^{+}$ if $I E_{1}$ of Na = 5.1 eV ?

0 eV

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-3.52 eV

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-5.1 eV

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-2.84 eV

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Solution

The correct option is C -5.1 eV
$e^{-}$ affinity is opposite to $I E$ , i.e., $e^{-}$ affinity is energy released/required during addition of $e^{-}$ whereas $I E$ is the energy required/released during removal of $e^{-}$ .

Thus, $E A = - I E$

$∴ E A (N a^{+}) = - 5.1 e V$

Hence, the correct option is $C$

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