Calculate the two highest wavelength of the radiation emitted when hydrogen atoms make transition from higher state to `n = 2`

Explanation

The higest wavelength curresponds to the lowest energy of transition .This will be the case for the transition `n = 3 to n= 2` The second higest wavelength corresponds to the transition` n = 4 to n= 2`
The energy of the state n is `E_(0) = (E_(1))/(n^(2))`
`Thus E_(2) = -(13.6eV)/(4) = -3.4eV`
E_(3) = -(13.6eV)/(9) = -1.5eV`
E_(4) = -(13.6eV)/(16) = -0.85eV`
The higest wavelength is `lambda_91)= (hc)/(Delta E)`
`= (1242evnm)/((3.4eV- 1.5eV)) = 654nm`
The second higest wavelength is
`lambda_(2) = (1242eVnm)/((3.4eV- 0.85ev)) =487nm`