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the energy of each photon is E=h*f compaire frequances of ultraviolet and infrared
Einstein was awarded the 1921 Nobel Prize for his explanation of the photoelectric effect. Experimentalists had found that '... [sic] when a metallic surface is exposed to electromagnetic radiation above a certain threshold frequency (typically visible light), the light is absorbed and electrons are emitted. In 1902, Philipp Eduard Anton von Lenard observed that the energy of individual emitted electrons increased with the frequency, or color, of the light. This was at odds with James Clerk Maxwell's wave theory of light, which predicted that the electron energy would be proportional to the intensity of the radiation. ...(Wikipedia)' Einstein's theory, published in 1905, accounted for the emission of electrons from a surface, when it is illuminated by light, by describing light a a 'photon' or quantised packet of energy. The energy of the photon was described as being directly proportional to the frequency of the light. With the constant of proportionality, within the energy 'E' / frequency 'f' relationship, being Planck's constant 'h'. Thus, a photon of light has a 'quantised' energy given by the expression: - E = hf Hence, if 'light' may be found in the decreasing wavelength range of: infra-red, red, orange, yellow, green, blue, indigo, violet, and ultra-violet, then because of the relationship between frequency 'f' and wavelenght 'λ' and the speed of light 'c' (from the wave model of light): - c = λ x f for decreasing electromagnetic (or light) wavelength the frequency is increasing. Thus, as the frequency of the light increases so too must its energy (from Einstein's equation (above)). Hence, ultra-violet light has more energy than infra-red radiation because it has a shorter wavelength and, thus, a much higher frequency. Source(s):
sure that ultra violet has highest energy
um ultraviolet has the lowest wavelength therefore it has a higher frequency :)
so it will have more energy :)