1.

As energy of `24.6 eV` is required to remove one of the required to remove both the electrons from a nuture brfore alon isA. `38.2`B. `94.2`C. `51.8`D. `79.0`

Answer» Correct Answer - 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`


Discussion

No Comment Found

Related InterviewSolutions