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