How much energy is required to ionise a H - atom if the electron occupies ` n=5 ` orbit ? Compare your answer with the ionization energy of H-atom (energy required to remove the electron from ` nth` orbit ).

Explanation

The expression of energy is given by,
` E_(n)= -(2.18xx10^(-18)Z^(2))/(n^(2))`
where ,S
Z = atomic number of the atom
n = principal quantum number
For ionization from ` n_(1) = 5 "to" n_(2) =oo`
`DeltaE = E_(oo)-E_(5)`
`[{-((2.18xx10^(-18)J)(1^(2)))/((oo)^(2))}-{((2.18xx10^(-18))(1)^(2))/((5)^(2))}]`
` ( 2. 18 xx 10^(-18)J)(1/((5)^(2)) " " ("since "1/oo = 0 )`
`= 0.0872xx 10 ^(-18) J `
`DeltaE= 8.72 xx 10^(-20) J `
hence , the energy required to ionization from n = 5 to n =
` oo is 8.72 xx 10^(-20)) J `
energy required for `n_(1) = 1 "to" n = `
`oo`
`DeltaE' = E_(oo)-E_(1)`
`[{-((2.18xx10^(-18))(1^(2)))/((oo)^(2))}-{((2.18xx10^(-18))(1)^(2))/((1)^(2))}]`
` (2.18xx10^(-18))[ 1-0]`
` 2.18 xx 10^(-18)`
hence m less enengy is required to ionize an electron in the 5th obital of hydrogen atom as compared to that in the ground state.