A) 13.6 eV
B) \[13.6\times 11\ eV\]
C) \[\frac{13.6}{11}eV\]
D) \[13.6\times {{(11)}^{2}}eV\]
Correct Answer: D
Solution :
(Eion)Na\[={{Z}^{2}}{{({{E}_{ion}})}_{H}}={{(11)}^{2}}13.6\ eV\]
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