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Question

How many chlorine atoms can you ionize in the process Cl+eCl++e, by the energy liberated from the following process:
Cl+eClfor 6×1023 atoms
given electron affinity of chlorine is 3.61 eV and ionization energy of chlorine is 17.422 eV

A
1.243×1022 atoms
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B
1.243×1023 atoms
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C
2.243×1022 atoms
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D
2.243×1023 atoms
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Solution

The correct option is B 1.243×1023 atoms

Energy released in conversion of =6×1023 atoms to Cl ions.
=6×1023×electron affinity=6×1023×3.61=2.166×1024 eV
Let x, chlorine atoms are conversed to Cl+ ion.
Energy absorbed =N×ionization energy=N×17.422 eV
N×17.422 eV=2.166×1024
N=1.243×1023 atoms
hence, number of chlorine atom = 1.243×1023 atoms

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