In this problem, it is necessary to introduce the atomic number Z into the equation.Â
E=(2.18×10^-18 J)(Z^2 )|1/(ni^2 )-1/(nf^2 )| E=(2.18×10^-18 J)(2^2 )|1/(6 ^2 )-1/(4 ^2 )|=3.02798×10^-19 JÂ
Plug the E value calculated into the equation: (wavelength is always positive)Â
E=hc/λ 3.02798×10^-19 J=hc/λ λ=6.56×10^-7 mÂ
656 nm is in the visible spectrum