1.

the amount of energy released when 10^(6) atoms of iodine in vapour state are converted to I^(-) ions is 4.8 xx 10^(-13) J. What is the electron affinity of iodine in ev/atom.

Answer»

`2.0`
`2.5`
`2.75`
`3.06`

Solution :`because` Energy released for `10^(6)` ATOMS `=4.9xx10^(-13)J`
`THEREFORE` Energy released for 1 ATOM `=(4.9xx10^(-13))/(10^(6))J`
`=(4.9xx10^(-13)xx10^(-6))/(1.6xx10^(-19))`
`=3.06eV`.


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