Unformatted text preview: . 2a) The units of a photon’s energy in this particular problem is Joules per second squared. This is shown here: Joulessecond x meterssecondmeters The meters cancel out, leaving you with Joulesseconds2 . 2b) = E hcλ h = 6.6256 x 10-34 | c = 299,792,458 m/s | λ = 5.9 x 10-6 meters = .-( , , )( . - ) E 6 6256 x 10 34 299 792 458 5 9 x 10 6 = 3.367 x 10-20 Js2 2c) The input in this program would only be wavelength, since that is the only value that keeps changing. H and c would remain constant, since they are predetermined measurements or values. The output would be E . In my algorithm, I would have the user input the wavelength by making the wavelength a float....
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This note was uploaded on 04/14/2010 for the course COMM 2064 taught by Professor Jimkuypers during the Spring '08 term at Virginia Tech.
- Spring '08