Calculate the ionisation energy of hydrogen atom as well as energy needed to promote its electron from first energy level to third energy level

Explanation

Ionization enthalpy n=5
for ionization n`=prop =n_(f )`
in initial electron is in n=5 so `n_(i) =5 ` hydrogen
z=1
`Delta E = 2.18 xx 10^(18) ((1)/(n_(1)^(2) ) - (1)/(n_(f )^(2)))= IE_(5)`
`=2.18 xx 10^(18) ((1)/(5^(2)) - (1)/(prop^(2)))`
`=(2.18 xx 10^(18))/( 25) J`
ti balance ionisation enthalpy of `IE_(5) ` and `IE_(1)` :
`(1E_(5))/( 1E_(1)) = (8.72 xx 10^(20)J )/( 2.18 xx 10^(18) J) = 4.-036 xx 10^(2) = 0.04`
ionization enthalpy of n=1 compare to n=5 is 0.04
for `n_(5)` less and `n_(1)` more ionization enthalpy
`IE(5) : IE (1)`
`=0.04 :1`
`4: 100`
`1: 25`
So IE (1) is approx 25 time more than 1E(5 )`