Calculate the energy emitted when electrons in one gm. atom of hydrogen undergo transition giving the spectral lines of lowest energy in the visible region of its atomic spectra.

For visible line spectrum i.e., Balmer series and for minimum energy transition :We know that n1 = 2 and n2 = 3                                    [∵  1st line has minimum energy]                           1λ=RH1n12−1n22×Z21λ=1.097×107m−1122−132×12or                       1λ=1.097×107m−114−19                           =1.097×10−7m−1×536                          =0.1524×107m−1=1.524×106m−1∴                  λ=6.56×10−7m     Now                    E=hcλ=6.62×10−34Js×3×108ms−16.56×10−7m                              =3.0274×10−19J  Energy corresponding to 1gm atom of Hydrogen                        E=3.0274×10−19×6.02×1023J   =18.22×104J=182.2×103J   =182.2kJ.