A) \[-30.6\text{ }eV\]
B) \[~-13.6\text{ }eV\]
C) \[-24.6\text{ }eV\]
D) \[~-28.6\text{ }eV\]
Correct Answer: A
Solution :
Key Idea: Energy of electron \[=\frac{-13.6{{Z}^{2}}}{{{n}^{2}}}eV\] where Z = atomic number of element, for Li = 3 \[n=\]number of orbit, for last \[{{e}^{-}}\]of \[Li,n=2\] \[\therefore \]Energy of last electron of Li \[=\frac{-13.6\times {{(3)}^{2}}}{{{(2)}^{2}}}=\frac{-13\times 9}{4}\] \[=-30.6\,eV\]
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