The lowest energy transition of a molecule with a HOMO-LUMO gap of 3 eV corresponds to the energy required to excite an electron from the highest occupied molecular orbital HOMO to the lowest unoccupied molecular orbital LUMO . To find the corresponding wavelength of this transition, we can use the equation:E = h * c / where E is the energy in joules , h is the Planck's constant 6.626 x 10^-34 Js , c is the speed of light 3.00 x 10^8 m/s , and is the wavelength in meters .First, we need to convert the energy from electron volts eV to joules J . The conversion factor is 1 eV = 1.602 x 10^-19 J. So, 3 eV = 3 * 1.602 x 10^-19 J = 4.806 x 10^-19 J.Now, we can solve for the wavelength: = h * c / E = 6.626 x 10^-34 Js * 3.00 x 10^8 m/s / 4.806 x 10^-19 J = 4.135 x 10^-7 mTo express the wavelength in nanometers nm , we can convert meters to nanometers by multiplying by 10^9: = 4.135 x 10^-7 m * 10^9 nm/m = 413.5 nmSo, the lowest energy transition of a molecule with a HOMO-LUMO gap of 3 eV corresponds to a wavelength of approximately 413.5 nm.