NRQM10sol_5

NRQM10sol_5 - Physics 216 Solutions to Problem Set 5 Spring...

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Unformatted text preview: Physics 216 Solutions to Problem Set 5 Spring 2010 1. Tritium (the isotope H 3 ), which is initially in its ground state, undergoes spon- taneous beta decay, emitting an electron of maximum energy of about 17 keV. The nucleus remaining is He 3 . NOTE : In this problem, you should neglect nuclear recoil. Note the energy of the emitted electron. What is the relevant approximation? Explain. (a) Calculate the probability that the electron of this ion is left in a quantum state of principal quantum number n = 2. Radioactive tritium H 3 decays to light helium (He 3+ ) with the emission of an electron. This electron quickly leaves the atoms and may be ignored in the following calculation. The effect of the -decay is to change the nuclear charge at t = 0 without effecting any change in the orbital electron. This is an example of a sudden time-dependent perturbation. We assume that the atom is initially in its ground state. In order to compute the probability that the He + ion is in a quantum state of principal quantum number n = 2, we note that the wave function of the He + ion just after the -decay is the same as the wave function of the tritium H 3 just before the decay. This must be true, since the wave function does not have time to re-adjust as the sudden perturbation occurs over a time-scale much shorter than any of the natural time-scales of the atomic system. Let H ( r ) be the wave function of the ground state of tritium, 1 H ( r ) = parenleftbigg Z 3 a 3 parenrightbigg 1 / 2 e Zr/ ( a , (1) where the nuclear charge (in units of e ) is Z = 1. Similarly, the wave function of the n = 2, = 0 state of the He + ion is given by: 2 He ( r ) = parenleftbigg Z 3 32 a 3 parenrightbigg 1 / 2 parenleftbigg 2 Zr a parenrightbigg e Zr/ (2 a ) , (2) where the nuclear charge (in units of e ) is Z = 2. To determine the probability that the He + ion is in the n = 2, = 0 state, one must expand the He + ion wave function with respect to the He + ion energy eigenstates, 3 | ) = summationdisplay n,,m | n m )( n m | ) . 1 The Bohr radius is is defined by a planckover2pi1 2 / ( e 2 ), where is the reduced mass of the electron nucleus system. Since m e m N (where m N is the mass of the nucleus), we shall approximate m e in what follows. Thus, we shall neglect the extremely small change to the value of in the wave function of tritium and the He + ion. 2 The case of negationslash = 0 is treated in part (b) of this problem. 3 We neglect electron spin here, although this plays no role in the computation. 1 We denote the final state of the He + ion by ( n , , m ) = (2 , , 0). Thus, the probability that the He + ion is in the n = 2, = 0 state is simply ( He | H ) = |( 2 0 0 | H )| 2 ....
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NRQM10sol_5 - Physics 216 Solutions to Problem Set 5 Spring...

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