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Unformatted text preview: Test # 3 Solutions 1(a) The ground state of the Hydrogen atom ( n = 1) is E 1 = 13 . 6eV. Thus, we require exactly 13 . 6eV to ionize the Hydrogen atom in its ground state. 1(b) With E initial = 0 eV and E final = 3 . 4 eV, we have E = hf = E initial E final = 0 ( 3 . 4) = 3 . 4 eV. Then, using the fact that h = 4 . 14 10 15 eV s, and that f = c/ where c = 3 . 10 8 m s 1 is the speed of light, we get c = E initial E final = 3 . 4 eV = hc E = 4 . 14 10 15 eV s 3 . 10 8 m s 1 3 . 4eV = 3 . 65 10 7 m Thus, the emitted photon has a wavelength of nearly 3 . 65 10 7 m. We may also use the alternative approach, where we use the formula 1 = R ( 1 n 2 1 m 2 ) In the above expression, R = 1 . 097 10 7 m 1 . However, we now need to be careful, since we are not starting off with the Hydrogen atom. Instead, we are starting out with a single proton, to which an electron is brought so that they together form a Hydrogen atom in the first excited state (...
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This note was uploaded on 10/20/2011 for the course PHY 309L taught by Professor Sai during the Spring '08 term at University of Texas at Austin.
 Spring '08
 SAI
 Physics

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