1.

If the ionization energy of `He^(+)` is `19.6xx10^(-18) J` per atom then the energy of `Be^(3+)` ion in the second stationary state is `:`A. `-4.9xx10^(-18)J`B. `-44.1xx10^(-18)J`C. `-11.025xx10^(-18)J`D. `-19.4xx10^(-18)J`

Answer» Correct Answer - 4
`E_(H)` in first orbit `=(-19.6xx10^(-18))/(4)J`
`E_(Be^(3+))` in second orbit `=-((19.4xx10^(-18))/(4))xx(16)/(4)=-19.4xx10^(-18)J`


Discussion

No Comment Found

Related InterviewSolutions