1. The problem statement, all variables and given/known data In an old TV cathode ray tube, electrons are accelerated by a potential of 15 kV between the gun and screen. Find: a) Kinetic energy of the electrons when they reach the screen b) Its respective wavelength c) Wavelength of the photons with maximum energy that are emitted from the screen. 2. Relevant equations E = hf (Energy of a photon with frequency f) f = c/[tex]\lambda[/tex] ([tex]\lambda[/tex] is the wavelength of respective photons) [tex]\lambda[/tex] = h/p (wavelength of electron depending on p, its momentum) 3. The attempt at a solution a) E = e*V = 15000 eV = 15000 * 1.6 x 10-19 = 2.4 x 10-15 J b) [tex]\lambda[/tex] = h/p But p = mv (where m = 9.1 x 10-31 kg) We also know that 1/2 * mv2 = E So, v = (2/m * E)1/2 => v = (2/(9.1 x 10-31)*2.4 x 10-15)1/2 = 0.73 x 108 m/s (not close enough to c, so we can use our electromagnetism approximation) Therefore, [tex]\lambda[/tex] = h/p = 6.626 * 10-34 / (9.1 x 10-31 * 0.73 x 108) = 9.97 x 10-12 m c) Let us suppose that there is 100% efficient exchange of energy between electron and photon. E = hf => f = E/h => f = (15000 * 1.6 x 10-19)/(6.626 x 10-34) = 3.62 x 1018 Hz But [tex]\lambda[/tex] = c/f => [tex]\lambda[/tex] = 8.28 x 10-11 m To me, I am using the formulas correctly... but the wavelength of light on part C is not in the visual spectrem. Since we are talking about a TV here... that certainly does not make sense! What am I doing wrong?? Thanks before hand for all the help. Tufts..