Suppose that means were available for stripping `28` electrons from `_29 Cu` in vapour of this metel. a. Compute the first three energy level for the remaining electron. b. Find the wavelength of the spectral line of the series for which `n_(1) = 1, n_(1),3,4`. What is the ionization potential for the last electron?

Suppose that means were available for stripping `28` electrons from `_

1.	Suppose that means were available for stripping `28` electrons from `_29 Cu` in vapour of this metel. a. Compute the first three energy level for the remaining electron. b. Find the wavelength of the spectral line of the series for which `n_(1) = 1, n_(1),3,4`. What is the ionization potential for the last electron?
Answer» Since `E_(n) prop (Z^(2))/(n^(2))` in a hydrogen-like atom , the energy levels will be `29^(2) = 841` time the corresponding energies for hydrogen. `:. E_(n) = 841 xx ((-13.6)/n^(2)) or E_(1) = - 11.44 ke V` `E_(2) = - 2.86 ke V` `E_(3) = - 1.27 ke V` b. Making use of `Delta E = E_(n) - E_(1)` `implies lambda _(1) = 1.44 Å. lambda _(2) = 1.22 Å, and lambda _(3) = 1.15 Å`. Lonization potential is the potential corresponding to energy `\|E_(1)\|`. `V_(oo) = (\|E_(1)\|)/(e) = 11.44 kV`

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