1. Tardigrade
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  3. Chemistry
  4. The ionisation potential of K would be numerically equal to

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Q. The ionisation potential of K would be numerically equal to

Solutions

Solution:

$K ( g )+ IE \longrightarrow K ^{+}( g )+ e ^{-}$
$K ^{+}( g )+ e ^{-} \rightarrow K ( g )+ EA$
Second equation is reverse of the first. The energy released in the second process would be equal to the energy absorbed in the first.