Equation used? I know that the same Equation works: e=13.605 eV z^2/n^2 But I ha
ID: 989517 • Letter: E
Question
Equation used? I know that the same Equation works: e=13.605 eV z^2/n^2 But I have been told that the second problem there is no n in the equation and that saying n=1 to calculate the enegy to remove an electron is not correct and the equation in general does not include an n term.In my mind these are exactly the same equation and n just equals 1. But I have explained this and I am Still being told that only the ionization potential energy has involved. I know they give the same Answer but I feel like the n is included in the equation and it is just one.
If you cannot give me a exact detailed explaination please leave this unanswered.
Explanation / Answer
In both the problems the same equation is used that is :
E =-13.605eV Z2 /n2
where E = ionization energy, Z = no. of protons and n = number of orbital where the electron lies
In first case O7+
electronic configuration of O = 8 => 1s2,2s2,2p4
now O7+ = 1s1 that means the remaining electron lies in the 1s orbital hence value of n=1
Z = 8
n= 1 (that is 1s orbital)
therefore E = (-13.605eV x 82) /12 = 870.72eV
the square of 1 do not make any difference being in denominator so not mentioned in the equation of problem 1
whereas in problem 2
electronic configuration of H = 1 => 1s1orbital.............................ground state
in mentioned problem this only one electron of H atom is present in 4f orbital
therefore n= 4 (i.e.4f orbital)
Z = 1
E = (-13.605eV x 12) / 42 = 0.8504eV
Thus this difference in your observation is that n is the number of orbital where the valence electron lies not the number of electrons present.
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