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Calculate the energy of `Li^(+2)` atom for `2^(nd)` excited state. |
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Answer» `E =-13.6xx(Z^(2))/(n^(2))` `because Z=3` and `e^(-)` exist in `2^(nd)` excited state, means `e^(-)` present in `3^(rd)` shell i.e. `n=3` `therefore E=-13.6 xx((3)^(2))/((3)^(2))=-13.6 eV//"atom"` |
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