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The ratio of the energy of the electron in ground state of hydrogen to the electron in first excited state of `Be^(3+)` is :A. `1 : 4`B. `1 : 8`C. `1 : 16`D. `16 : 1` |
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Answer» Correct Answer - A (a) `E_n = - (13.6 Z^2)/(n^2) eV`, where `Z = "atomic number" = 4` `N = "number of orbits"` `Be^(3 +) (n = 2)` (first excited state) `E_1 = -13.6 eV, E_2 - 13.6 xx 16//4`. |
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