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The `IP_(1) `of H is `13.6 eV` it is expoxed induced radiation .Find the wavelength of these ijnduced radiation

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`E_(1)` of H atom `= - 13.6 eV`1
Energy given to H atom
`= (6.625xx 10^(-34) xx 3.0 xx 10^(8))/(1028 xx 10^(-10))`
`= 1.933 xx 10^(-18) J = 12.07 eV`
Energy of H atom after excitation
`= - 13.6 + 1207 = - 1.53 eV`
`:. E_(n) = (E_(1))/(n^(2))`
`:. n^(2) = (-13.6)/(-1.53) = 9 :. n = 3`
The electron in H atom is excited to third shell
:. induced `lambda_(1) = (hc)/(E_(3) - E_(1))`
`E_(1) = -13.6 ev, E_(3) = - 1.53 eV`
`lambda_(1) = (6.626 xx 110^(-34) xx 3 xx 10^(6))/((-1.53 + 13.6) xx 1.602 xx 10^(-19)) = 1028 xx 10^(-10) m`
`:. lambda = 1028 Å`
`1` induced `lambda_(2) = (hc)/(E_(2) - E_(1)))`
`E_(1) = -13.6 eV, E_(2) = -(13.6)/(4) eV`
`:. lambda_(2)= (6.625 xx 10^(-34) xx 3 xx 10^(6))/((-(13.6)/(4) +13.6)xx1.602 xx 10^(-`19))`
`:. induced lambda_(3) = (hc)/(E_(3) - E_(2))`
`E_(1) = - 13.6 eV,E_(2) = - (13.6)/(4) eV`
`E_(3) = -(13.6)/(9) eV`
`:. lambda_(3) = (6.626 xx 10^(-34) xx 3 xx 10^(8))/((-(13.6)/(4) + (13.6)/(4)) xx1.602 xx 10^(-19))`
`= 6568 xx 10^(-10) m = 6568 Å`

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