Assume that the electron in Li2+ ion is in third orbit. Calculate (5) i) the radius of the orbit, and ii) the total energy of the electron
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radius=0.529 (n^2/z)A
where n is the no of orbit in which electron move
z is atomic no of element
so, radius =0.529 (9/3)= 1.587A
ii) total energy =-13.6 (z^2/n^2)
=-13.6 (9/9)= -13.6 eV/atom
where n is the no of orbit in which electron move
z is atomic no of element
so, radius =0.529 (9/3)= 1.587A
ii) total energy =-13.6 (z^2/n^2)
=-13.6 (9/9)= -13.6 eV/atom
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