91影视

Electron Energy in region $1,\mathrm{E}=800\mathrm{eV}$

Height of potential step, ${\mathrm{U}}_1={\mathrm{U}}_{\mathrm{b}}=600\mathrm{eV}$

Number of electrons sent by electron beam,${\mathrm{N}}_0=5.00x{10}^5$

Planck鈥檚 constant, $\mathrm{h}=6.626脳{10}^{-34}\mathrm{Js}$

Mass of electron, $\mathrm{m}=9.1脳{10}^{-31}\mathrm{kg}$

The energy,$\mathrm{E}=800脳1.6脳{10}^{-19}\mathrm{J}$

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 region 1 can be given using the formula,

$\mathrm{k}=\frac{2\mathrm{蟺}}{\mathrm{h}}\sqrt{2\mathrm{mE}}$

Here, E is the electron energy, m is the mass of the electron, and h is plank constant.

Substituting the values into the formula.

$\begin{array}{rcl}\mathrm{k}& =& \frac{2脳3.14}{6.626脳{10}^{-34}\mathrm{Js}}\sqrt{2脳9.1脳{10}^{-31}\mathrm{kg}脳800脳1.6脳{10}^{-16}\mathrm{J}}\\ & =& 1.45脳{10}^{-11}{\mathrm{m}}^{-1}\end{array}$

Hence, the angular wave number of region 1 is $1.45脳{10}^{11}{\mathrm{m}}^{-1}$.

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

$\mathrm{k}=\frac{2\mathrm{蟺}}{\mathrm{h}}\sqrt{2\mathrm{m}\left(\mathrm{E}-{\mathrm{U}}_{\mathrm{b}}\right)}$

Here, E is the electron energy, m is the mass of the electron, h is plank constant, and ${\mathrm{U}}_{\mathrm{b}}$ is the potential energy or potential step.

Substituting the values into the formula,

$\begin{array}{rcl}{\mathrm{k}}_{\mathrm{b}}& =& \frac{2脳3.14}{6.626脳{10}^{-34}\mathrm{Js}}\sqrt{2脳9.1脳{10}^{-31}\mathrm{kg}脳\left(800-600\right)脳1.6脳{10}^{-16}\mathrm{J}}\\ & =& 7.25脳{10}^{10}{\mathrm{m}}^{-1}\end{array}$

Hence the angular wave number of region 2 is$7.25脳{10}^{10}{\mathrm{m}}^{-1}$ .

The wave function for region 1 can be given as

${\mathrm{蠄}}_1\left(\mathrm{x}\right)={\mathrm{Ae}}^{\mathrm{ikx}}+{\mathrm{Be}}^{-\mathrm{ikx}}$

The wave function for region 2 can be given as

${\mathrm{蠄}}_2\left(\mathrm{x}\right)={\mathrm{Ce}}^{{\mathrm{ik}}_{\mathrm{b}}\mathrm{x}}$

By using the boundary condition, we have two equations

$A+\mathrm{B}=C........\left(1\right)$ localid="1663142091494" $\mathrm{Ak}-\mathrm{Bk}={\mathrm{Ck}}_{\mathrm{b}}.......\left(2\right)$

As already calculated

$$\begin{gathered} \mathrm{k}=1.45脳{10}^{11}{\mathrm{m}}^{-1} \\ {\mathrm{k}}_{\mathrm{b}}=7.25脳{10}^{10}{\mathrm{m}}^{-1} \end{gathered}$$

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

$$\begin{gathered} \mathrm{A}\left(1.45脳{10}^{11}{\mathrm{m}}^{-1}\right)+\mathrm{B}\left(1.45脳{10}^{11}{\mathrm{m}}^{-1}\right)=\mathrm{C}\left(7.25脳{10}^{10}{\mathrm{m}}^{-1}\right) \\ 2\mathrm{A}-2\mathrm{B}=\mathrm{C}.........\left(3\right) \end{gathered}$$

From equations (1) and (3)

$\frac{\mathrm{B}}{\mathrm{A}}=\frac{1}{3}$

The reflection coefficient is

$\begin{array}{rcl}\mathrm{R}& =& \frac{{\left|\mathrm{B}\right|}^2}{{\left|\mathrm{A}\right|}^2}\\ & =& {\left|\frac{1}{3}\right|}^2\\ & =& \frac{1}{9}\\ & =& 0.111\end{array}$

Thus, the reflection coefficient is $0.111$.

The number of reflected electrons can be given by the formula

${\mathrm{N}}_{\mathrm{R}}={\mathrm{RN}}_0$

Here, ${\mathrm{N}}_{\mathrm{R}}$is the number of reflected electrons, R is the reflection coefficient, and ${\mathrm{N}}_0$is the incident electrons.

Substituting the values into the formula

$\begin{array}{rcl}{\mathrm{N}}_{\mathrm{R}}& =& \left(\frac{1}{9}\right)\left(5.00脳{10}^5\right)\\ & =& 5.56脳{10}^4\end{array}$

Hence, the number of reflected electrons is $\begin{array}{rcl}& & 5.56脳{10}^4\end{array}$.