A) \[\frac{13.6}{11}eV\]
B) \[\frac{13.6}{112}eV\]
C) \[13.6\times {{(11)}^{2}}\,eV\]
D) \[13.6\text{ }eV\]
Correct Answer: C
Solution :
The energy of nth orbit of hydrogen like atom is, \[E_{n}^{2}=-13.6\frac{{{Z}^{2}}}{{{n}^{2}}}\] Here, Z =11 for Na atom.\[10{{e}^{-l}}\]are removed already. For the last electron to be removed \[n=1\] Hence, \[E_{n}^{2}=-13.6\times \frac{{{(11)}^{2}}}{{{(1)}^{2}}}\] \[=-13.6\times {{(11)}^{2}}eV\]
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