Total Transmission Across Finite Barrier Potential with E>V

In summary, when a beam of electrons with kinetic energy of 100 eV is incident on a 200 eV high and 10 nm wide barrier with a narrow momentum spread, a simple plane wave can be used as an approximation. To achieve 100% transmission and 0% reflection, the energy of the electrons must be raised to 350 eV and the barrier thickness should be equal to n*pi divided by the square root of 2m times the difference between the potential energy and the kinetic energy, all divided by the reduced Planck's constant. This solution is valid as long as the probability waves add up constructively on transmission and destructively on reflection. The wave number k1 (or 1/wavelength)
  • #1
Dahaka14
73
0

Homework Statement


A beam of electrons of KE = 100 eV is incident from the left on a barrier
which is 200 eV high and 10 nm wide. If the momentum spread is sufficiently
narrow, then a simple plane wave is a good approximation. Recall that the mass of an
electron is mc2 = 511 keV.

....._____
...1...|..2...|...3
__________|...|____________
.....x=0...x=10 nm

(periods are there to preserve picture upon post)

If the energy of the electrons is raised to 350 eV, what thickness should the barrier be
to give transmission with no reflections?

Homework Equations



[tex]T=|t|^2= \frac{1}{1+\frac{V_0^2\sin^2(k_1 a)}{4E(E-V_0)}}[/tex]

where [tex]k_1=\sqrt{2m (V_0-E)/\hbar^{2}}[/tex]

The Attempt at a Solution



I am almost positive I just use the above equation, set [tex]T=1[/tex], then solve for [tex]a[/tex] because we know every other variable. I ended up with the general equation (no substitution for values yet)

[tex]1=\frac{1}{1+\frac{V_0^2\sin^2(k_1 a)}{4E(E-V_0)}}\iff 1=1+\frac{V_0^2\sin^2(k_1 a)}{4E(E-V_0)}\iff 0=\frac{V_0^2\sin^2(k_1 a)}{4E(E-V_0)}\iff 0=\sin^2(k_1 a)\iff n\pi=\sqrt{2m (V_0-E)/\hbar^{2}}a[/tex]
[tex]\iff a=\frac{n\pi}{\sqrt{2m (V_0-E)/\hbar^{2}}}[/tex].

This solution makes me uneasy for some reason. Can someone point me out if I am wrong?
 
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  • #2
sounds reasonable, though i haven't checked your transmission equation - the transmssion condition will be when the probability waves add up constructively on transission & 100% destructively on releflection

so check this means 0 reflection..

also think about what this means for the wave number k1 (1/wavelength effectively) vs the length of the barrier...
 

What is total transmission across finite barrier potential?

Total transmission across finite barrier potential refers to the ability of a particle or wave to pass through a potential barrier with energy greater than the barrier height.

How is total transmission calculated?

Total transmission can be calculated using the transmission coefficient, which is equal to the square of the absolute value of the ratio of the transmitted wave amplitude to the incident wave amplitude.

What does it mean if total transmission is greater than 1?

If total transmission is greater than 1, it means that the particle or wave has a higher probability of passing through the barrier than being reflected by it. This is known as superluminal or superbarrier tunneling.

Can total transmission across finite barrier potential occur in classical physics?

No, total transmission across finite barrier potential is a phenomenon that can only be explained by quantum mechanics. In classical physics, a particle with energy less than the barrier height would always be reflected by the barrier.

What are some real-life applications of total transmission across finite barrier potential?

Total transmission across finite barrier potential has many applications in modern technology, such as in the development of quantum computers, tunneling diodes, and scanning tunneling microscopes. It also plays a crucial role in understanding the behavior of particles in nuclear fusion and radioactive decay.

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