InterviewSolution
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InterviewSolution
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In H-atoms, the energy of electron in the nth orbit is given as `E_(n) = - (13.6)/(n^(2)) eV` Show that `E_((n +1)) - E_(n) = (13.6 xx 2)/(n^(3)) eV` for large value of n. |
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Answer» `E_((n +1)) - E_(n) = - (13.6)/((n +1)^(2)) - (-(13.6)/(n^(2))) = 13.6 [(1)/(n^(2)) - (1)/((n +1)^(2))] = 13.6 xx ((n +1)^(2)- n^(2))/(n^(2) (n +1)^(2))` `= (13.6 xx (2n +1))/(n^(2) (n +1)^(2)) = (13.6 xx 2n)/(n^(2) (n +1)^(2)) + (13.6)/(n(n + 1)^(2)) = (13.6 xx 2)/(n(n +1)^(2)) + (13.6)/(n^(2) (n + 1)^(2))` As n is large, `n + 1 ~=n`. Hence, `E_((n +1)) - E_(n) ~= (13.6 xx 2)/(n xx n^(2)) + (13.6)/(n^(2) xx n^(2)) = (13.6 xx 2)/(n^(3))` (Neglecting second term because `n^(4)` is very large) |
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