Question

shows the probability density for an electron that has passed through an experimental apparatus. What is the probability that the electron will land in a $$\begin{gathered} 0.010-mm-widestripat\left(a\right)x=0.000mm,\left(b\right)x=0.500mm, \\ \left(c\right)x=1.000mm,and \\ \left(d\right)x=2.000mm?x\left(mm\right)-3-2-101230.50 \\ p\left(x\right)=0c\left(x\right)02\left(mm-1\right) \end{gathered}$$

Short answer

The probability of electron will land in a$0.010mmmm$while strong at a position of$x-0.000m{m}^2$

Step-by-step solution

Ex39.13 the probability density of the electron at the position x-0.000 mm is as knowspx=0.50mm4      =0.50 mm41m210-3mm2      =0.50x103m-2Substitute 0.50x103m-1 for px1x10-5m forδx in the equation of solveConvert the units for the small width of the strip from mm to mThe probability of electron will land in a 0.010mm mm while strong at a position of  x-0.000mm2

$$\begin{gathered} \delta x=0.010mm \\ =0.010mm\left(\frac{1m}{1x{10}^{2}mm}\right) \\ =1x{10}^{-2}m \end{gathered}$$

From the figure