Calculate the wavelength and wave number of the spectrum, line when an electron of H atom falls from a higher energy state `n = 3` in a state `n = 2` also determine the energy of a photonm to ionic this atom by removing the electron from the second orbit .Compare it with the energy of photon required to ionic the atom by removing the electron from the ground state

Explanation

First calculate the energy difference `(Delta E)` between the white `n = 3 and n = 2` using
`Delta E = 2.18 xx 10^(-18)Z^(2)((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))J`
`Delta E_((3 rarr2)) = 2.18 xx 10^(-18)(1)^(2)((1)/(2^(2))- (1)/(3^(2))) J(Z = 1)`
`= 3.03 xx 10^(-19) J`
Now this energy difference in the energy of the photon emitted
`E_("photon") = hv = (hc)/(lambda) = hc bar v `
`rArr (hc)/(lambda) = 3.03 xx 10^(-19) rArr lambda = 6560.3 Å`
and `bar v = (1)/(lambda) = 1.52 xx 10^(-6)m^(-1)`

To ionic the atom from `n = 2` the transition will be `n = 2 rarr n = oo`
`Delta E_(( 2 rarr oo)) = 2.18 xx 10^(-19) xx 1^(2) xx ((1)/(2^(2)) - (1)/(oo^(2))) J`
`= 5.45 xx 10^(-19) J`
TO ionic the atom from the ground state `(n = 1)` and transition is `1 rarr oo`
`Delta E = 2.18 xx 10^(-19) xx 1^(2) xx ((1)/(1^(2)) - (2)/(oo^(2)))`
`2.18 xx 10^(-18) J`