Question

32. An electron has a kinetic energy of 12.0 eV. The electron is incident upon a rectangular barrier of height 20.0 eV and width 1.00 nm. If the electron absorbed all the energy of a photon of green light (with wavelength 546 nm) at the instant it reached the barrier, by what factor would the electron’s probability of tunneling through the barrier increase?

Answer 1

@douglaswinslowcooper
@ybarrap

Answer 2

The sum of all the probabilities is going to equal 1 so part of the 1 goes to Transmission, the rest goes to Reflection.
\[\large R+T=1\]
The equation for transmission is defined as \(\LARGE T=e^{-2CL}\) where \(\LARGE C=
\dfrac {\sqrt{2m(U-E)}}{h-bar}\).

Based on the given information, if I multiply the top and bottom by the speed of light, then the equation now becomes \(\LARGE C= \dfrac {\sqrt{2mc^2(U-E)}}{(h-bar) *c}\).

K=12.0 eV
L=1.0 nm
U=20.0 eV
\(\lambda\)=546 nm

Using this information, the energy of the photon is \(\large E=hf={hc\over \lambda}\).

The energy of the electron is 511 MeV so doing the substitution, C=14.51/nm.

If I do the proportionality
\(\huge \dfrac{e^{-2CL}}{e^{-2C'L}}=e^{-2L(C-C')}\).
I got C'=12.28/nm but that is incorrect.
The final (correct) answer is actually 1.2255 X10^10 m ^-1.

Answer 3

h-bar is planck's constant divided by 2pi.

Answer 4

@LastDayWork
@Mashy

Answer 5

Sorry. I do not have a clue, except that 1228 is nearly 1226.