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

Options

A. `1:4`
B. `1:8`
C. `1:16`
D. `16:1`

Answer

A

Explanation

`E_(n)=-(1311.8Z^(2))/(n^(2))kJ"mol"^(-1)`
`E_(1)(H)=-1311.8`
first excited state 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`