A) \[\frac{\text{13}\text{.6}}{\text{11}}\text{eV}\]
B) \[\frac{\text{13}\text{.6}}{\text{112}}\text{eV}\]
C) \[\text{13}\text{.6}\times {{\text{(11)}}^{2}}\,\text{eV}\]
D) \[\text{13}\text{.6}\,\text{eV}\]
Correct Answer: A
Solution :
The energy of \[nth\]orbit of hydrogen like atom is, \[{{E}_{n}}=-13.6\frac{{{Z}^{2}}}{{{n}^{2}}}\] Here, \[Z=1\]for Na-atom. 10 electrons are removed already. For the last electron to be removed \[n=1.\] \[\therefore \] \[{{E}_{n}}=\frac{-13.6\times {{(11)}^{2}}}{{{(1)}^{2}}}eV\] \[=-13.6\times {{(11)}^{2}}eV\]
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