Which state of the triply ionized `Be^(+++)` has the same orbital radius as that of the ground state of hydrogen? Compare the energies of two states.

Answer

`2;4`

Explanation

Suppose orbital radius of groung state of hydrogen (n=1) is `r_(1)`
Radius of nth orbit of hydrogen like atom is
`r_(n)=(n^(2))/Z r_(1)`
for `Be^(+++), Z=4`
As `r_(n)=r_(1)`, therefore, `(n^(2))/4=1 or n=2`
Let energy of electrons in ground state of hydrogen `=E_(1)`. Energy of electron in nth state of hydrogen like atom
`E=(Z^(2))/(n^(2)) E_(1)=(4^(2))/(2^(2)) E_(1)=4E_(1)`
`:. (E_(2))/(E_(1))=4`