From `N` atoms of an element A when half the atoms transfer on electron to the another atom `405 kJ m01^(-1)` of energy was found to be consumed. An additional energy of `745 kJ mo1^(-1)` was further required to convert all the `A^(Theta)` ions to `A^(o+)`. Calculate the ionisation energy and the electron affinity of atom A in eV .

Explanation

`N//2 (A rarr A^(o+)), IE (+),`
`N//2 (A rarr A^(Theta)), EA (-)`
`rArr (405 xx 10^(3)eV)/(1.5xx10^(-19)) = ((IE)/(2)xx6xx10^(23) -(EA)/(2)xx6xx10^(23)) ..(i)`
Now, `N//2 (A^(Theta) rarr Ararr A^(o+))`
`rArr (745xx10^(3)eV)/(1.6xx10^(-19)) = ((IE)/(2)xx 6 xx 10^(23) -(EA)/(2) xx 6xx10^(23)) ...(ii)`
Solve for `IE` and `EA` using, equations (i) and (ii), we get `IE = 11.93eV` and `EA = 3.52 eV`
Second method:
Let `x = IE, y = EA`
`{:[{:( :.(1)/(2)(x-y)=405kJ),(),( :.(1)/(2)(x+y)=745kJ):}]{:(..(1)),(),(..(2)):}{:("Solve from equations"),(),((i)and(ii)),(x=1150kJ=11.91eV),(),(y=340kJ=3.52eV):}:}`