The ratio of the enrgy of the electron in groun state of hydrogen to the elecron in first excited state of `Be^(2+)` is :

Options

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

Answer

A

Explanation

`E_(2_((Be^(3+))) = E_(2(H)) xx Z^2` ,
Also ` E_(2 (H)) = E_(1(H0))/2^2`
` ,. E_(2(Be^(3+0)) = (E_(1(H)))/2^2 xx 4^2 = 4 xx E_(1(H))`.