There is a simple relationship between the energy for a photon of light ephoton in units of ev (electron volts) and its wavelength λ in nm (nanometers) given by: ephoton (ev) = (1240 ev nm) / (λ nm). for example, if light has a wavelength of 620 nm, then its energy ephoton = (1240 ev nm) / (620 nm) = 2 ev. visible light has photon energies of ~1.7 to 3 ev and wavelengths of ~400 to 750 nm (λblue = 475 nm, λgreen = 510 nm, λyellow = 570 nm, and λred = 650 nm).

what is the wavelength λ in nm for light with a photon energy ephoton = 2 ev?

what is the photon energy for light with a wavelength λ = 630 nm?

in the electromagnetic spectrum, x-rays have much higher photon energies (~100 to 100,000 ev) and shorter wavelengths (~0.01 to 10 nm) than visible light. what is the photon energy for an x-ray with a wavelength λ = 6 nm?