`H^(+)` ions are in excited state from where maximum 6 types of photons can be emitted. If E is the separation energy of electron in this excited state of `H^(+)` ion, then how many times of energy E is required to ionize Hydrogen atom ?

Answer

4

Explanation

`6=(n(n-1))/(2)" "n=4`
Separation energy, `E=+13.6 xx(z^(2))/(n^(2))=+13.6xx((2)^(2))/((4)^(2))=3.4 eV`
l.E. of Hydrogen atom `=x xxE=13.6 eV`
`x=(13.6)/(E)=4`