1.

The ratio of the energy ofthe electron in the ground state of H to the electron in the first excited state of Be^(3+)is ,

Answer»

`1:4`
`1:8`
`1:16`
`16:1`

Solution :`E_(n)=-(1311.8Z^(2))/(n^(2))kJ"MOL"^(-1)`
`E_(1)(H)=-1311.8`
first excitedstate of `Be^(3+)` is n=2
`E_(2)(Be^(+))=(-1311.8xx16)/4(n=2,z=4)`
`(E_(1)(H))/(E_(2)(Be^(3+)))=1/4`


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