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

Explanation

`E_(n)=-(21.8xx10^(-19))/n^(2)" J atom"^(-1)`
For ionization from 5th orbit, `n_(1)=5, n_(2)=oo`
`:. Delta E=E_(2)-E_(1)=-21.8xx10^(-19) (1/n_(2)^(2)-1/n_(1)^(2))=21.8xx10^(-19) (1/n_(1)^(2)-1/n_(2)^(2))=21.8xx10^(-19) (1/5^(2) -1/oo)`
`=8.72xx10^(-20) J`
For ionization from 1st orbit, `n_(1)=1, n_(2)=oo`
`Delta E'=21.8xx10^(-19) (1/1^(2)-1/oo)=21.8xx10^(-19) J`
`(Delta E')/(Delta E)=(21.8xx10^(-19))/(8.72xx10^(-20))=25`
Thus, the energy required to remove electron from 1st orbit is 25 times than that required to remove electron from 5th orbit.