Question: An electron beam strikes a barrier with a single narrow slit, and the electron flux number of electrons per unit time per unit area detected at the very center of the resulting intensity pattern is . Next, two more identical slits are opened, equidistant on either side of the first and equally “illuminated” by the beam. What will be the flux at the very center now? Does your answer imply that more than three times as many electrons pass through three slits than through one? Why or why not?

The wavelength of a particle can be obtained using the formula $\mathit{\lambda }\mathbf{=}\frac{\mathbf{h}}{\mathbf{p}}\mathbf{=}\frac{\mathbf{h}}{\mathbf{m}\mathbf{v}}$.

There is more destructive and beneficial interference between matter waves as the number of slits rises. As a result, the combined amplitude of additional matter waves increases (remember how a diffraction grating is compared to a double slit in terms of the interference pattern). Now, much like the intensity and the electric field , the probability density is proportional to the wave amplitude squared $\left({\psi }^2\right)$.

Consequently, when the wave amplitude triples, the probability density will increase by 9 times, which will be seen on the screen as 9 more electrons striking the same region per second, or flux.

It is important to note that the initial beam intensity (the number of electrons assaulted per second of the beam) has not changed; the only change we have made is to increase the number of slits. However, this has changed how the electrons are distributed across the screen; as a result, more of them are now restricted to smaller regions.

Therefore, to balance off the destructive interference and preserve the overall amount of energy, the constructive interference occurred, and as we can see, it does so in a non-linear (squared) way.