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What is the change in energy of an atom if the wavelength of the photon absorbed by the atom is equal to 4.64 x 10-7 meters? (Planck’s constant is 6.626 x 10-34 joule seconds; the speed of light is 2.998 x 108 m/s)? 1.428 x 10-41 joules second 3.074 x 10-26 joules 3.440 x 10-19 joules/second 4.281 x 10-19 joules 10.255 x 10-49 joule second
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Energy of photon = hv where h = Plancks constant and v = frequency. Increase in energy of atom = hv = 6.626*10^-34*v Using velocity = frequency*wavelength 2.998*10^8 = v*4.64*10^-7 v = (2.998/4.64)*10^15 Hz = 6.461*10^14Hz Therefor increase in energy of atom is: E = 6.626*10^-34*6.461*10^14 J E = 4.281*10^-19 J
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