How many Cl atoms can you ionize in the process Cl→Cl++e− by the energy liberated from the process Cl+e−→Cl− for Avogadro number of atoms?
Given, IP=13.0eV and EA=4.29eV.
1×1023
1.5×1023
2×1023
2.5×1023
A
1×1023
B
1.5×1023
C
2.5×1023
D
2×1023
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Solution
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Energy liberated =4.29×6.02×1023
Now, by 13 eV energy number of atoms that get ionized = 1
So, by 4.29×6.02×1023eV, number of atoms ionize =113×4.29×6.02×1023≈2×1023
Hence, the correct option is C
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