Calculate the wavelength of the spectral line when an electron jumps from n=7 to n=4 level in an atom of hydrogen.

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`