`IE_1` for (H) and De are 13.6 eV 1 and 24.6 eV respectively . Thus enrgy liberated during the formation fo He by
` He^(2+) + 2e rarr H3,` is:

Options

A. ` 54.4 eV`
B. ` 49 . 2 eV`
C. ` 0. 27 , 4 eV`
D. ` 13. 6 lambda`

Answer

D

Explanation

`Hrarr H^+ + e , 13 . 6 eV`
` Herarr He ^+ + e, 24. 6 eV`
` IE_2" of "He = IE _1" of "HE^+`
` +E_(1H) xx Z^2 = 13 . 6 xx 4 = 54 . 4 eV`
` He^+ rarr He^(2+) , 54. 4 eV`
` :. Herarr He^(2+) + 2e, 24 . 6 + 54.4 = 79. 0 eV`.