A) 13.6 eV
B) 27.2 eV
C) 54.4 eV
D) 100 eV
Correct Answer: C
Solution :
Ionic potential\[=\frac{{{Z}^{2}}\times 13.6}{{{(1)}^{2}}}\] For He, \[Z=2\] \[\therefore \]Potential\[=\frac{{{(2)}^{2}}\times 13.6}{{{(1)}^{2}}}\] \[=54.4eV\]
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