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 Kis a constant, rather than 1/r2. 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 mvnrn= nh_and mvn2/rn= Krn. Solving for rnone obtains rn= [nh_/Km]1/2. The potential energy is given by ∫∞=-=n2/)(2nrnKrFdrrU. The angular frequency is ?= v/r= K/m. The total energy is En= =Kr=Kr+mv2n2n2n2121nh_K/m= nh_?. 79 ··· The frequency of revolution of an electron in a circular orbit of radius ris frev= v/2pr, where vis the speed. (a) Show that in the nth stationary state (b) Show that when n1= n, n2= n – 1, and nis much greater than 1, n1hf33p2meZk=422rev
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