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Ionisation energy of hydrogen atom is 13.6 eV. Then the energy required to ionize a hydrogen atom in which the electron is in the second orbit from the nucleus is (eV)

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10.2

3.4

-13.6

13.6

answer is B.

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Detailed Solution

${(I . E)}_{H} = 13.6 eV$

$n = 2$

$E = \frac{13.6}{2^{2}} Z^{2}$

$= \frac{13.6}{2^{2}} {(1)}^{2} = \frac{13.6}{4} = 3.4 eV$

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