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The enrgy of an electron in the first Borh orbit of H atom is `-13.6 eV` The possible energy values (s)of the excited state (s) for electron in bohr orbitsw of hydrogen is (are)A. `-3.4 eV`B. `-4.2 eV`C. `-6.8 eV`D. `+ 6.8 eV` |
Answer» Correct Answer - A The energy of an electron in Bohr orbit of hydrogen atom is given by the expeeression `E_(n) = - (Constant)/(n^(2))` When n takes only value for the first Bohr orbit `n = 1`and it is gigev that `E_(1) = -13.6 eV` Hence `E_(n) = (13.6 eV)/(n^(2))` Of the given value of energy only `-13.6 eV` can be obtained by substating `n = 2` in the above expression |
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