Question

Question:For the arrangement of Figs. 38-14 and 38-15, electrons in the incident beam in region 1 have a speed of $\mathbf{1}\mathbf{.}\mathbf{60}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$and region 2 has an electric potential of ${\mathit{V}}_{\mathbf{2}}\mathbf{-}\mathbf{500}\mathbf{}\mathit{V}$. What is the angular wave number in (a) region 1 and (b) region 2? (c) What is the reflection coefficient? (d) If the incident beam sends $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}$electrons against the potential step, approximately how many will be reflected?

Fig 38-14

Fig 38-15

Short answer

1. The angular wave number of region 1 is $1.38\times {10}^{11}{\mathrm{m}}^{-1}$
2. The angular wave number of region 2 is $7.74\times {10}^{10}{\mathrm{m}}^{-1}$
3. The reflection coefficient is $0.0794$
4. The number of reflected electrons is $2.38\times {10}^8$

Step-by-step solution

Identifying the data given in the question

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,${\mathrm{N}}_0=3.00\times {10}^9$

Concept used to solve the question

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

(a) Finding angular wave number of region 1

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

$k=\frac{2\mathrm{\pi }}{\lambda }$

Here, $\lambda$is the wavelength.

From de Broglie equation as below.

$\lambda =\frac{h}{p}$

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).

$\lambda =\frac{h}{mv}$

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

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

Substituting the known numerical values in the above equation.

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

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

(b) Finding angular wave number of region 2:

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.

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

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

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

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,

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

Hence, the angular wave number of region 2 is $7.74\times {10}^{10}{\mathrm{m}}^{-1}$.

(c) Finding the reflection coefficient

The wave function for region 1 can be given as

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

The wave function for region 2 can be given as

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

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

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

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

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

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 $0.794$.

(d) Finding the number of reflected electrons

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

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

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