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SOLUTIONS TO EXERCISE 16
hc 6.63 * 10^(-34) J sec * 3 * 10^17 nm/s (a) Energy = ------ = -------------------------------------- Lambda 350 nm = 5.68 * 10^-19 Joule (b) The wavelength of a photon with 2.1 * 10^(-18) Joule is given by hc 6.63 * 10^(-34) J sec * 3 * 10^17 nm/s Lambda = ------ = -------------------------------------- = 94.7 nm Energy 2.1 * 10^(-18) J From the equation, you can see that the larger the wavelength gets, the smaller the energy gets. Thus, the maximum wavelength of a hydrogen ionizing photon here is 94.7 nm. Also from the equation, the smaller the wavelength, the larger the energy; since you only need a photon with more energy than 2.1 * 10^(-18) J, the minimum wavelength of a hydrogen ionizing photon is zero. You can also reason that the minimum wavelength would the minimum wavelength allowed by nuclear processes; this is not necessary, but just for reference, a photon with a wavelength shorter than about 0.001 nm will often turn into matter and antimatter spontaneously. (c) Using Wien's Law, we know that Lambda * Temperature = 2.9 * 10^6 nm K. So for Lambda = 94.7 nm, T = 2.9 * 10^6 / 94.7 = 30600 K. The sun's surface temperature is about 5800 K, so such an object would have a surface temperature over five times that of the Sun. |