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Chapter 6: Q4CQ (page 223)
The equations for and T in the barrier essentially the same as light through a transparent film. It is possible to fabricate a thin film that reflects no light. Is it possible to fabricate one that transmits no light? Why? Why not?
The answer is no.
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What fraction of a beam of electrons would get through a wide electrostatic barrier?
The potential energy barrier in field emission is not rectangular, but resembles a ramp, as shown in Figure 6.16. Here we compare tunnelling probability calculated by the crudest approximation to that calculated by a better one. In method 1, calculate T by treating the barrier as an actual ramp in which U - E is initially, but falls off with a slop of M. Use the formula given in Exercise 37. In method 2, the cruder one, assume a barrier whose height exceeds E by a constant (the same as the average excess for the ramp) and whose width is the same as the distance the particle tunnels through the ramp. (a) Show that the ratio T1/T2 is . (b) Do the methods differ more when tunnelling probability is relatively high or relatively low?
A beam of particles of energy incident upon a potential step of,is described by wave function:
The amplitude of the wave (related to the number of the incident per unit distance) is arbitrarily chosen as 1.
For the E>U0 potential barrier, the reflection, and transmission probabilities are the ratios:
Where A, B, and F are multiplicative coefficients of the incident, reflected, and transmitted waves. From the four smoothness conditions, solve for B and F in terms of A, insert them in R and T ratios, and thus derive equations (6-12).
Fusion in the Sun: Without tunnelling. our Sun would fail us. The source of its energy is nuclear fusion. and a crucial step is the fusion of a light-hydrogen nucleus, which is just a proton, and a heavy-hydrogen nucleus. which is of the same charge but twice the mass. When these nuclei get close enough. their short-range attraction via the strong force overcomes their Coulomb repulsion. This allows them to stick together, resulting in a reduced total mass/internal energy and a consequent release of kinetic energy. However, the Sun's temperature is simply too low to ensure that nuclei move fast enough to overcome their repulsion.
a) By equating the average thermal kinetic energy that the nuclei would have when distant,. and the Coulomb potential energy they would have when fm apart, roughly the separation at which they stick, show that a temperature of about K would be needed.
b) The Sun's core is only about . If nuclei can鈥檛 make it "over the top." they must tunnel. Consider the following model, illustrated in the figure: One nucleus is fixed at the origin, while the other approaches from far away with energy. As decreases, the Coulomb potential energy increases, until the separation is roughly the nuclear radius . Whereupon the potential energy is and then quickly drops down into a very deep "hole" as the strong-force attraction takes over. Given then , the point , where tunnelling must begin. will be very large compared with , so we approximate the barrier's width as simply b. Its height, , we approximate by the Coulomb potential evaluated at . Finally. for the energy which fixes , let us use . which is a reasonable limit, given the natural range of speeds in a thermodynamic system.Combining these approximations, show that the exponential factor in the wide-barrier tunnelling probability is
c)Using the proton mass for , evaluate this factor for a temperature of. Then evaluate it at. about that of an incandescent filament or hot flame. and rather high by Earth standards. Discuss the consequences.
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