91Ó°ÊÓ

Electron beam speed in region 1,${\mathrm{v}}_1=1.60\mathrm{x}{10}^7\mathrm{m}/\mathrm{s}$

The electric potential of region 2, ${\mathrm{V}}_2=-500\mathrm{V}$

Number of electrons sent by electron beam,${\mathrm{N}}_0=3.00×{10}^9$

When a particle's potential energy changes at a boundary, it can reflect even though it would not normally reflect under classical theory.

The angular wave number of the region1 can be given using the formula,

$k=\frac{2\mathrm{π}}{\mathrm{λ}}$

Here, $\mathrm{λ}$is the wavelength.

From de Broglie equation as below.

$\mathrm{λ}=\frac{h}{p}$

Here, h is plank constant and p is momentum.

The momentum is given by

$p=mv........\left(3\right)$

Where, m is mass and v is speed.

Substitute for into equation (2).

$\mathrm{λ}=\frac{h}{mv}$

Substituting $\frac{h}{mv}$for $\mathrm{λ}$into equation (1).

$$\begin{gathered} k=\frac{2\mathrm{Ï€}}{h/mv} \\ k=\frac{2\mathrm{Ï€}mv}{h} \end{gathered}$$

Substituting the known numerical values in the above equation.

$\begin{array}{rcl}\mathrm{k}& =& \frac{2×3.14×9.1×{10}^{-31}\mathrm{kg}×1.60×{10}^7\mathrm{m}/\mathrm{s}}{6.626×{10}^{-34}\mathrm{Js}}\\ & =& 1.38×{10}^{11}{\mathrm{m}}^{-1}\end{array}$

Hence the angular wave number of region 1 is $1.38×{10}^{11}{\mathrm{m}}^{-1}$.

The electron energy in region 1 can be given as

$E=\frac{1}{2}m{v}^2$

Here, m is mass and v is speed.

$\begin{array}{cc}\mathrm{E}=\frac{1}{2}×9.31×{10}^{31}\mathrm{kg}×(1.60×{10}^7& \mathrm{m}/\mathrm{s}{)}^2\\ =1.17×{10}^{16}\mathrm{J}& \\ =728.8\mathrm{eV}& \end{array}$

Since the electric potential of region 2 is ${\mathrm{V}}_2=-500\mathrm{V}$.

Therefore, the potential step for region 2 is

${\mathrm{U}}_b=-500eV$

The angular wave number of region 2 can be given using the formula

$k=\frac{2\mathrm{Ï€}}{h}\sqrt{2m\left(E-U_b\right)}$

Here, $E$is the electron energy, $m$is the mass of the electron, $h$is plank constant, and $U_b$is the potential energy or potential step.

Substituting the values into the formula,

$\begin{array}{rcl}{k}_b& =& \frac{2×3.14}{6.626×{10}^{-34}\mathrm{Js}}\sqrt{2×9.31×{10}^{-31}\mathrm{kg}×\left(728.8-500\right)×1.6×{10}^{-16}\mathrm{J}}\\ & =& 7.74×{10}^{10}{\mathrm{m}}^{-1}\end{array}$

The wave function for region 1 can be given as

${\mathrm{ψ}}_1\left(x\right)=A{e}^{ikx}+B{e}^{ikx}$

The wave function for region 2 can be given as

${\mathrm{ψ}}_2\left(x\right)=C{e}^{i{k}_{b}x}$

By using the boundary condition, there are two equations as below.

$$\begin{gathered} A+\mathrm{B}=C.........\left(1\right) \\ Ak-Bk=C{k}_b........\left(2\right) \end{gathered}$$

Solving equations (1) and (2) gives

$\frac{B}{A}=\frac{k-k_b}{k-k_b}$

As already calculated

$$\begin{gathered} \mathrm{k}=1.38×{10}^{11}{\mathrm{m}}^{-1} \\ {\mathrm{k}}_{\mathrm{b}}=7.74×{10}^{10}{\mathrm{m}}^{-1} \end{gathered}$$

Substituting the values of wave number in equation 2 (ii)

$\begin{array}{rcl}\frac{\mathrm{B}}{\mathrm{A}}& =& \frac{1.38×{10}^{11}{\mathrm{m}}^{-1}-7.74×{10}^{10}{\mathrm{m}}^{-1}}{1.38×{10}^{11}{\mathrm{m}}^{-1}-7.74×{10}^{10}{\mathrm{m}}^{-1}}\\ & =& 0.282\end{array}$

The reflection coefficient is

$\begin{array}{c}R=\frac{{\left|B\right|}^2}{{\left|A\right|}^2}\\ ={\left|0.282\right|}^2\\ =0.0794\end{array}$

Hence, the reflection coefficient is $0.794$.

The number of reflected electrons can be given by the formula

$N_R=R{N}_0$

Here, $N_R$is the number of reflected electrons, $R$reflection coefficient, and $N_0$incident electrons.

Substituting the values into the formula

$\begin{array}{rcl}{N}_R& =& \left(0.0794\right)\left(3.00×{10}^9\right)\\ & =& 2.38×{10}^8\end{array}$

Hence, the number of reflected electrons is $2.38×{10}^8$.