## anonymous 4 years ago In what region of the electromagnetic spectrum would you expect to find radiation having an energy per photon 100 times that associated with 988nm radiation?

Well, if we use the Planck-Einstein equation...$E=\frac{hc}{\lambda}$Now, let's plug in a few numbers, and see where that leaves us...$E=\frac{(6.626*10^{-34}J*s)(3.00*10^8\frac{m}{s})}{988nm}$Regardless of what the unit conversions end up being, if we increase the left side of the equation by a factor of 100 (make the energy 100 times that which it is), we need to do the same thing to the other side. Let's go ahead and see what happens when we do that...$(100)E=\frac{6.626*10^{-34}J*s)(3.00*10^8\frac{m}{s})(100)}{988nm}$Cancelling out the 100 on the top of the RHS with the 988nm makes it...$(100)E=\frac{(6.626*10^{-34}J*s)(3.00*10^8\frac{m}{s})}{9.88nm}$This means that, for the energy to be equal, the wavelength of the photon needs to have a wavelength of 9.88nm. According to Wolfram Alpha ( http://www.wolframalpha.com/input/?i=9.8nm ), this falls into the "extreme ultraviolet" to "soft x-ray" part of the spectrum. Depending on the specifics of your course, you may need to report it in a different way.