Question

3. The process of alpha-radioactivity is an example of a barrier tunneling phenomenon, which can be envisioned as an alpha particle trapped inside a nucleus by a "Coulomb Barrier" produced by the combination of the attractive nuclear force that exists inside the nucleus, and the repulsive Coulomb force. A very crude picture of this situation is shown in the picture below, where we make the approximation that the alpha particle must tunnel through a square potential barrier to escape the nucleus. For the purposes of this example, assume that the height of the barrier is 20 MeV, and that the length Lis 15 fm. (a) (20 points) Calculate the probability that an alpha particle can penetrate the barrier and leave the nucleus if its energy is 6 MeV, and if it is 4 MeV. (Possibly useful hint: The quantity /2mg 0.411 MeVifm) (b) (5 points extra credit) Inside a nucleus, we can imagine that the alpha particle bounces back and forth many times, and encounters the barrier about 2x 1019 times per second. The mean lifetime for such a nucleus then depends on the number of times the alpha particle encounters the barrier, and the probability per encounter that the alpha particle escapes. Estimate the mean lifetime of the radioactive nucleus if the alpha particle has an energy of 6 MeV, and 4 MeV. U(x)-20 MeV Eg L 15 fm Figure 5. Crude picture of the Coulomb Barrier encountered by an alpha particle involved in alpha-radioactivity.

Answer 1

- Potential kidthd= 15 tm No=20Mev, 15x105 E,= 6 Mev ama (Vore) and 5,34Mev ka kid = 2ma (1-5) d? Tama 22 kg d = Jama Jure, a kad = 0.411 Mev" fmJia menu istom Kid = 0.411.*514 x 15 = 23.06 Probability = exp (-2k, d) Pi = exp(-2% 23.06) 1P, = 9.204 X1021 / => for E,=6 Mevl

kod = 1 ama (No-E2) Kad = Jama suoraad = 0.411 Meutz fm' sio mere 15 tm kad = 0.411 4 x Kos is = 24.66 Pe= exp (-2 ked) = exp(-2X 24.66) Pe = 3.80411 X1022 For E=4Mev

px time = Lite time = - - pif Ti = 9.204x152' 2 x 10! I Ti = 5.432 sec L and - = Т. – P. 5 3.807X1022x2x18 Tz = 131.33 sec

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