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Question

How many chlorine atoms can you ionize in the process ClCl++e, by the energy liberated from the following process? Cl+eCl for 6 ×1023 atoms. Given electron affinity of Cl = 3.61 eV/atom, and IP of Cl= 17.422 eV/atom


A

1.24×1023atoms

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B

9.82×1020atoms

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C

2.02×1015atoms

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D

None of these

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Solution

The correct option is A

1.24×1023atoms


Energy released in conversion of 6 ×1023 atoms of Cl ions = 6 ×1023× electron affinity

= 6×1023×3.61=2.166×1024 eV

Let x Cl atoms be converted to Cl+ ion.

Energy absorbed = x × ionization energy

x×17.422=2.166×1024

x=1.243×1023atoms


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