A) \[-30.6\,\,eV\]
B) \[-13.6\,\,eV\]
C) \[-24.6\,\,eV\]
D) \[-28.6\,\,eV\]
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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