As energy of `24.6 eV` is required to remove one of the required to remove both the electrons from a nuture brfore alon is
A
`38.2`
B
`94.2`
C
`51.8`
D
`79.0`
Text Solution
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The correct Answer is:
D
When one `e^(bar)` is removed from neutral helium atom , it becames a one `e^(bar)` speciel . For one `e^(bar)`speciel .we know `E_(n) = (- 13.62^(2))/(n^(2)) ev//atom` For helium ion ,`Z = 2` and for first arbit `n = 1` `:. E_(1) = (-13.6)/((1)^(2)) xx 2^(2) = - 54.4 eV` `:. ` Energy required to remove this `e^(bar) = 54.4 + 24.6 = 79 eV`
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