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A doubly ionized lithium atom is hydrogen like with atomic number Z = 3 . Find the wavelength of the radiation required to excite the electron in Li^(2+) from the first to the third Bohr orbit. Given the ionization energy of hydrogen atom as 13.6 eV. |
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Answer» Solution :The energy of `n^(th)` orbit of a hydrogen-like ATOM is given as `E_(n) = - (13.6Z^(2))/(n^(2))` Thus for `Li^(2+)` atom, as Z= 3 the electron energies of the first and third Bohr orbits are Forn = 1 , `E_(1) ` = - 122.4eV, For n = 3 , `E_(3)`= - 13.6eV. Thus the LEVEL is, E = `E_(3) - E_(1) `= - 13.6 - (-122.4) = 108.8eV Therefore , the radiation needed to CAUSE this transition should have photons of this energy. hv = 108.8 eV The WAVELENGTH of this radiation is or `lambda = (hc)/(108.8eV) = 114.25`Å |
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