# T 6562314 nm d 6562912 nm h 6564695 nm d t 00598 nm h

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T = 656.2314 nm; ? D = 656.2912 nm; ? H = 656.4695 nm ? D ? T = 0.0598 nm; ? H ? T = 0.2381 nm 78 ··· Suppose that the interaction between an electron and proton were of the form F =–Kr , where K is a constant, rather than 1 /r 2 . If the stationary state orbits are again limited by the angular momentum condition L = nh _ , what are then the radii of these orbits? Show that for this case the kinetic energies of the stationary states are given by E = nh _ ? , where ? is the angular frequency of the electron about the proton. We have mv n r n = nh _ and mv n 2 / r n = Kr n . Solving for r n one obtains r n = [ nh _ / Km ] 1/2 . The potential energy is given by = - = n 2 / ) ( 2 n r n Kr Fdr r U . The angular frequency is ? = v / r = K/m . The total energy is E n = = Kr = Kr + mv 2 n 2 n 2 n 2 1 2 1 nh _ K/m = nh _ ? . 79 ··· The frequency of revolution of an electron in a circular orbit of radius r is f rev = v /2 p r , where v is the speed. ( a ) Show that in the n th stationary state ( b ) Show that when n 1 = n , n 2 = n – 1, and n is much greater than 1, n 1 h f 3 3 p 2 m e Z k = 4 2 2 rev

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