Find the binding energy of the electron in the ground state of hydrogen-like ions in whose spectrum the third line of the Balmer series is equal to `108.5nm`.

Explanation

For the third line of Balmer series
`omega=RZ^(2)((1)/(2^(2))-(1)/(5^(2)))=(21)/(100)RZ^(2)`
Hence `lambda=(2pi c)/(omega)=(200pi c)/(21RZ^(2))`
`Z=sqrt((200pi c)/(21R lambda))`
Subsitution gives `Z=2`. Hence the binding energy of the electron in the ground state of this ion is
`E_(b)= 4E_(H)= 4xx13.65= 54.6 eV`
The ion is `He^(+)`.