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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 For n = 1, `E_(1)=-122.4eV`, for n = 3, `E_(3)=-13.6eV`. Thus the energy required to TRANSFER an electron from `E_(1)` level to `E_(3)` 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)=114.25Å` |
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