Answer to Question #257961 in Physics for Anand

Question #257961
What is quantum tunneling? Calculate (i) the tunneling length and (ii) the tunneling probability when an electron of energy 4.0 eV is incident on a potential barrier of height 10 eV and width of 0.2 nm.
1
Expert's answer
2021-11-01T19:25:13-0400

Quantum tunnelling is the quantum mechanical phenomenon where a particle (more accurately, its wavefunction) can propagate through a potential barrier.


Let tunneling lenght be a=0.2nm=0.2×109ma = 0.2nm = 0.2\times 10^{-9}m, the energy of the particle E=4eV6.409×1019JE = 4eV \approx 6.409 \times 10^{-19}J and the potential height U=10eV16.02×1019JU = 10eV\approx 16.02\times 10^{-19}J. The tunneling propability is then given as follows (see https://courses.physics.illinois.edu/phys485/fa2015/web/tunneling.pdf, page 13):

T=(1+sinh2(βa)4(E/V)(1E/V))1T = \left( 1 + \dfrac{\sinh^2(\beta a)}{4(E/V)(1-E/V)} \right)^{-1}


where β=2m(VE)\beta=\dfrac{\sqrt{2m(V-E)}}{\hbar}, m=9.1×1031kgm = 9.1\times 10^{-31}kg is the mass of electron and =1.055×1034Js\hbar=1.055\times 10^{-34}Js. Thus, obtain:


β=29.1×1031kg(104)eV1.055×1034Js12.54×109m1\beta=\dfrac{\sqrt{2\cdot 9.1\times 10^{-31}kg\cdot (10-4)eV}}{1.055\times 10^{-34}Js} \approx12.54\times 10^{9}m^{-1}


T=(1+sinh2(12.54×109m10.2×109m)4(4eV/10eV)(14eV/10eV))10.025T = \left( 1 + \dfrac{\sinh^2(12.54\times 10^{9}m^{-1}\cdot 0.2\times 10^{-9}m)}{4(4eV/10eV)(1-4eV/10eV)} \right)^{-1} \approx 0.025

Answer. (i) 0.2nm, (ii) 0.025.


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