How many chlorine atoms can you ionize in the process `Cl ->Cl^+ + e^–`, by the energy liberated from the following process ?
`Cl + e^– ->Cl^–` for `6×10^(23)` atoms Given electron affinity of Cl =3.61 eV, and IP of Cl = 17.422 eV.
| A. |
`1.24×10^23` atoms |
 |
| B. |
`9.82×10^20` atoms |
| C. |
`2.02×10^15` atoms |
| D. |
None of these |
Energy released in conversion of `6×10^23` atoms of `Cl^–` ions = `6×10^23` electron affinity
= `6 × 10^23×3.61`
= `2.166×10^(24)` eV
Let x Cl atoms be converted to `Cl^+` ion. Energy absorbed
= x × ionization energy
`=>x × 17.422 = 2.166×10^24 `
`=>x =1.243 ×10^23` atoms