A) 3.6 V
B) 10.2 V
C) 13.6V
D) 3.4V
Correct Answer: B
Solution :
From the relation \[\Delta E=13.6\,eV-\frac{13.6\,eV}{{{n}^{2}}}\] \[=13.6\,eV-\frac{13.6\,eV}{{{(2)}^{2}}}=10.2\,eV\] Therefore, excitation potential \[=\frac{10.2}{e}eV=10.2\,V\]
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