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

Answer

`6561 Å, 4863 Å` (Approx)

Explanation

`lambda_(3 to 2)=12400/(DeltaE_(3 to 2))=12400/1.89=6561 Å, lambda_(4 to 2)=12400/(DeltaE_(4 to 2))=12400/2.55=4863 Å`