742. Problem 39.81P (HRW) Suppose a beam of 5.0 eV protons strikes a potential energy barrier of height 6.0 eV and thickness 0.70 nm, at a rate equivalent to a current of 1000 A. (a) We have to find the time on an average that we may have to wait for one proton to tunnel through the barrier. (b) We have to find the time on an average that we may have to wait for an electron instead of a proton to tunnel through the barrier. |
Solution: Click For PDF Version The quantum transmission probability for a particle of mass m, kinetic energy E for tunnelling through a potential barrier of height U and width L is approximately given by the following function The data for the problem are as follows: Mass of a proton, , mass of an electron, , height of the potential energy barrier, , kinetic energy of protons/electrons, and width of the potential barrier, (a) The transmission probability for a proton to tunnel through the barrier will therefore be It is given that the beam of protons/electrons is equivalent to a current of 1000 A. As the charge of proton is e, the number of protons striking the barrier per second will be Let the number of protons that should strike the barrier for one proton to tunnel through the potential barrier be N. We thus have , and . Therefore, the time required for N protons to strike the barrier will be We have found that in order to see one proton to tunnel through the barrier we may have to wait for ! (b) And, the transmission probability for an electron to tunnel through the barrier will therefore be Let the number of electrons that should strike the barrier for one electron to tunnel through the potential barrier be . We thus have , and Therefore, the time required for electrons to strike the barrier will be We thus find that in order to see one electron to tunnel through the barrier we only have to wait for . We notice that the smaller mass of electron compared to that of proton makes an enormous difference in the tunnelling probability.
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