Why does quantum tunneling increase de-broglie wavelength? The picture (taken from a textbook) shows how quantum tunneling occurs with electrons.

Why does the de-Broglie wavelength of the electron change when doing this? It does not make intuitive sense to me that the speed of the electron would change after tunneling
 A: The picture in the textbook doesn't show tunneling correctly. There are at least two mistakes:


*

*Tunneling is the phenomenon of penetrating a classically-inaccessible barrier. This means that the wavefunction must be in the form of an evanescent wave inside the barrier, not an oscillating one shown in the picture.

*Assuming the potential energy around the barrier is the same, the wavelength must be restored after the particle exits the barrier (in the picture it appears to decrease: 16 px vs 20 px).


The correct depiction can be seen in the following image (taken from here):

A: Measuring the wavelengths carefully in a drawing program, I get 1.00, 0.99, and 1.10 (in arbitrary units) on the left, middle, and right. This is actually not super relevant in the end, but various people seem to be making seemingly contradictory statements about these numbers, which is probably going to confuse the discussion.
The solution of the Schrodinger equation in the classically forbidden region should be an exponential, not an exponentially damped sine wave, so the textbook illustration is wrong.
The wavelength can either increase, decrease, or stay the same after tunneling. This depends on whether the potential is lower or higher on one side or the other.
Ruslan's answer shows a supposedly correct drawing from wikipedia. This is much less wrong than the one from the unnamed textbook, but is still not right, because you can't actually depict this kind of thing using a real-valued wavefunction. Real-valued wavefunctions only work for solutions to the time-independent Schrodinger equation, not for traveling waves.
Here is one way to prove that this kind of representation can't actually be right. The incident red wave is traveling to the right. As it travels to the right, there will be times when its value at the initial boundary is zero. Since the blue segment is drawn by matching the $\Psi$ and $\Psi'$ of an exponential to those of the red wave, at this time, the blue wave would have to be zero. But then if we follow the same procedure for the second boundary, we get places where the transmitted wave should have both $\Psi$ and $\Psi'$ equal to zero. However, the transmitted wave is drawn as a sine wave, and a sine wave never does this.
The WP drawing is also not really right because there has to be a reflected wave, whereas they seem to be trying to match things on the boundaries without ever having a reflected wave. That's impossible.
The WP version is not necessarily that bad if it comes with explanatory text in the relevant WP pages explaining that it's not strictly correct. But it's not true that it's the correct picture.
