The equations equal to each other sin k 2 l φ k2 k1

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the equations equal to each other:sin(k2L+φ) =-k2k1cos(k2L+φ)Now, use the sum-to-product formulae fromhigh school trigonometry (and we make thedefinitionk2L= Φ for brevity):sin Φ cosφ+ sinφcos Φ =-k2k1(cos Φ cosφ-sin Φ sinφ)Now, we multiply both sides of the equationbyradicalBigk21+k22, and remember that we solved
doyle (dwd596) – HW08webmanual – markert – (58195)2for tanφabove, thus getting rid of the termsinvolvingφsin Φk1+k2cos Φ =-k2k1(k1cos Φ-sin Φk2)(k22-k21) sin Φ = 2k1k2cos Φtan Φ =2k1k2k22-k21Now, we recall from our initial definitionsthatk2=2mE¯handk1=radicalbig2m(U0-E)¯h,so we havek22-k21=2m¯h2(E-(U0-E))=2m¯h2(2E-U0)andk1k2=2m¯h2radicalbigE(U0-ESo, finally putting it all together, we have:tanparenleftBiggL2mE¯hparenrightBigg=2radicalbigE(U0-E)2E-U0Which is an equation that is only valid forspecial values of E. Putting in the parametersfrom the problem, and using a numerical util-ity to find the values of E satisfying the aboveequation, we find the allowed bound energies:E1= 0.2165 eVE2= 0.86437 eVE3= 1.93739 eVE4= 3.4233 eVE5= 5.29768 eVE6= 7.5038 eVE7= 9.79196 eVWe know that the electron starts at the0.86437 eV energy state and absorbs a photonwith energyhcλ= 1.24082 eV, which is verynearly equal toE3-E2= 1.93739 eV =0.86437 eV = 1.07302 eV, so we conclude thatthe answer most consistent with the setup isthat the energy absorbed from the photoncauses the electron to gain the energy requiredto transition fromn= 2 ton= 3.00210.0pointsUse equationTe-2α ato calculate the or-der of magnitude of the probability that aproton will tunnel out of a nucleus in one col-lision with the nuclear barrier if it has energy6 MeV below the top of the potential barrierand the barrier thickness is 4×10-15m.Correct answer: 0.0135731.Explanation:Let :a= 4×10-15m,U0-E= 6 MeV,mc2= 938 MeV,and¯hc= 1.974×10-13MeV·m.α=radicalBigg2m(U0-E)¯h2=radicalbig2mc2(U0-E)¯hcThen,T=e-2α a= expbracketleftBigg-2radicalbig2mc2(U0-E)¯hcabracketrightBigg= expbracketleftbigg-2 (4×10-15m)×

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