If the binding energy of the electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of Li⁺⁺ is

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122.4 eV

3.4 eV

30.6 eV

13.6 eV

answer is B.

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Detailed Solution

$E_{n} = \frac{13 .6}{n^{2}} \times Z^{2} . For first excited state n = 2 and for {Li}^{+ +}, z = 3 \Rightarrow E = \frac{13 .6}{4} \times 9 = 30 .6 eV$

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