$\mathbf{\text{10}}$A beam of electrons strikes a barrier with two narrow but equal-width slits. A screen is located beyond the barrier. And electrons are detected as they strike the screen. The "center" of the screen is the point equidistant from the slits. When either slit alone is open,electrons arrive per second in a very small region at the center of the screen. When both slits are open, how many electrons will arrive per second in the same region at the center of the screen?

Number of electrons when one slit is open${\text{n}}_1\text{=10/sec}$

To find the number of electrons when both slits are open${\text{n}}_{\text{2}}\text{=?}$

The following relationship can be used to describe the probability of detecting a particle in a given region.

$\mathbf{\text{probability of finding a particle}}\mathbf{\propto }{\mathbf{\text{(amplitude)}}}^{\mathbf{\text{2}}}$……………….(1)

The number of electrons that can be found in the region every second equals the probability of discovering a particle.

When there is only one slit open, useequation (1), such that

$\begin{array}{l}{\text{probabilityoffindingaparticle,\hspace{0.17em}n}}_1\propto {\text{(amplitude)}}^{\text{2}}\\ 10\propto {\text{(amplitude)}}^{\text{2}}\end{array}$

$\text{amplitude}\propto \sqrt{\text{10}}$……………….(2)

When two slits are opened at the same time, the amplitude doubles, assuming they are the same size. Therefore, using equation (2), we can write

$\text{A=2}\sqrt{\text{10}}$………………(3)

Therefore, using the equations (2) and (3), we get,

role="math" localid="1659078958778" $\begin{array}{c}{\text{probabilityoffindingaparticle,\hspace{0.17em}n}}_{\text{2}}\propto {\left(2\sqrt{10}\right)}^2\\ \propto \text{4×(10)}\\ \propto \text{40}\end{array}$

Therefore${\text{n}}_{\text{2}}\text{=40}$ when the two slits are open condition.