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Find total energy, `PE & KE` of electron in : (A) `2^(nd)` orbit of `He^(+)` ion. (B) `1^(st)` excited state of `Be^(+3)` ion. |
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Answer» `E_(n) =- 13.6 xx(Z^(2))/(n^(2)) (eV)/("atom"),E_(n) =- KE = (PE)/(2)` `(A) E_(2)=-13.6xx(2^(2))/(2^(2))=-13.6" "(B) E_(3)=-13.6xx(4^(2))/(2^(2))=-54.4eV` `KE = 13.6eV, PE = - 27.2 eV" "KE = 54.4 eV, PE=-108.8eV` |
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