The probability density for an electron that has passed through an experimental apparatus. If $1.0\times {10}^6$electrons are used, what is the expected number that will land in alocalid="1649312933417" $0.010mm$wide strip atlocalid="1649312941736" $\left(a\right)x=0.000mm$and localid="1649312949893" $\left(b\right)2.000mm$?

The expression which relate the expected number of particles, probability density, and the total number of particles available is,$N(in\delta xatx)=N\left(prob\right(in\delta xatx\left)\right)$

Here,$Prob(in\delta xatx)$is the probability of any particle ends up in the at position $x$and $N$is total number of particles.

The figure which represent the probability density for an electron that has passed through an experimental apparatus is given below.

Here, $P\left(x\right)$is the probability density of an electron and $x$is the position of the electron.

Determine the expected number of electron landing in the narrow strip when the position of $x=0.000mm$using the formula.

localid="1648896853985" $N(in\delta xatx)=N\left(Prob\right(in\delta xatx\left)\right)$

Here, localid="1648896868788" $Prob(in\delta xatx)$is the probability of any electron ends up in the strip position $x$and $N$is total number of electrons

The probability of any electron ends up in the strip position $x$is,

localid="1649312975174" $Prob(in\delta xatx)=P\left(x\right)\delta x$

Here localid="1648896939090" $P\left(x\right)$is the probability density andlocalid="1648896948996" $\delta x$is the narrow width.

substitute localid="1648896972257" $P\left(x\right)\delta x$for localid="1648896987333" $Prob(in\delta xatx)$inlocalid="1648897035959" $N(in\delta xatx)=N\left(Prob\right(in\delta xatx\left)\right)$to solve forlocalid="1648897049152" $N(in\delta xatx)$.

thus the expression of the expected number is, localid="1648897064957" $N(in\delta xatx)=NP\left(x\right)\delta x$

Convert the units of the narrow width of the strip for mm to m.

$\delta x=0.010mm$

$=0.010mm\left(\frac{1m}{1\times {10}^{3}mm}\right)$

$=1\times {10}^{-5}m$

Convert the units for the probability density from $m{m}^{-1}$to $m^{-1}$

$p\left(x\right)=0.333m{m}^{-1}$

$$\begin{gathered} =0.333m{m}^{-1}\left(\frac{1m^{-1}}{{10}^{-3}m{m}^{-1}}\right) \\ =0.333\times {10}^3{m}^{-1} \end{gathered}$$

Substitute $1.0\times {10}^6$electrons for N ,$0.333\times {10}^3{m}^{-1}$for$p\left(x\right),1\times {10}^{-5}$m for localid="1648896421358" $\delta x,$and$0.000mm$for x in localid="1648897115366" $(in\delta xatx)=NP\left(x\right)\delta x$

localid="1648897145273" $N(in0.010mmat0.000mm)=(1.0\times {10}^{6}electrons)(0.333\times {10}^3{m}^{-1})(1\times {10}^{-5}m)$

localid="1649312988526" $=3330electrons$

Rounding off to two significant figures, the expected number of electrons landing in the narrow strip when the position of$x=0.000mm$is$3300electrons$

Determine the expected number of electrons landing in the narrow strip when the position of $x=2.000mm$using the expression $N(in\delta xatx)=NP\left(x\right)\delta x$.

From the figure, the probability density of the electrons is $0.11m{m}^{-1}$. Convert the units for the probability density from $m{m}^{-1}$to $m^{-1}$.

$$\begin{gathered} P\left(x\right)=0.11m{m}^{-1}\left(\frac{1m^{-1}}{{10}^{-3}m{m}^{-1}}\right) \\ =0.110\times {10}^3{m}^{-1} \end{gathered}$$

Substitute $1.0\times {10}^6$electrons for $N,0.110\times {10}^3{m}^{-1}$for $P\left(x\right)$, $1\times {10}^{-5}m$for $\delta x$and $0.000mm$for $x$in $N(in\delta xatx)=NP\left(x\right)\delta x$.

$N(in0.010mmat2.000mm)=(1.0\times {10}^{6}electrons)(0.110\times {10}^3{m}^{-1})(1\times {10}^{-5}m)$

$=1100electrons$.

Therefore, the expected number of electrons landing in the narrow strip when the position of$x=2.000mm$is$1100electrons$.