91影视

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.

$\mathbf{\text{probability of finding a particle}}\mathbf{鈭�}{\mathbf{\text{(amplitude)}}}^{\mathbf{\text{2}}}$鈥︹赌︹赌︹赌︹赌︹赌�.(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

$\begin{array}{l}{\text{辫谤辞产补产颈濒颈迟测辞蹿蹿颈苍诲颈苍驳补辫补谤迟颈肠濒别,鈥塶}}_1鈭�{\text{(amplitude)}}^{\text{2}}\\ 10鈭�{\text{(amplitude)}}^{\text{2}}\end{array}$

$\text{amplitude}鈭�\sqrt{\text{10}}$鈥︹赌︹赌︹赌︹赌︹赌�.(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

$\text{A=2}\sqrt{\text{10}}$鈥︹赌︹赌︹赌︹赌︹赌�(3)

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

role="math" localid="1659078958778" $\begin{array}{c}{\text{辫谤辞产补产颈濒颈迟测辞蹿蹿颈苍诲颈苍驳补辫补谤迟颈肠濒别,鈥塶}}_{\text{2}}鈭�{\left(2\sqrt{10}\right)}^2\\ 鈭�\text{4脳(10)}\\ 鈭�\text{40}\end{array}$

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