91Ó°ÊÓ

The electron beam speed in region 1,$\mathrm{v}=900\mathrm{m}/\mathrm{s}$

The height of the potential step ${\mathrm{V}}_{\mathrm{b}}=-1.25\mathrm{μ³Õ}$

The electric current ${\mathrm{l}}_0=500\mathrm{A}$

Potential step

A region where a particle's potential energy rises at the expense of its kinetic energy is described by this term.

Classical physics states that if a particle's initial kinetic energy

exceeds the region's capacity for reflection, it should never do so.

However, there is a reflection by quantum physics.

Where $\mathbf{R}$is the reflection coefficient and transmission coefficient can be given as,$\mathbf{T}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{R}$

The electron energy in region 1 can be given as

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

Where $m$is mass and $v$is speed

localid="1663150143863" $\begin{array}{rcl}E& =& \frac{1}{2}×9.1×{10}^{-31}\mathrm{kg}×{\left(900\mathrm{m}/\mathrm{s}\right)}^2\\ & =& 3.69×{10}^{-25}\mathrm{J}\\ & =& 2.306\mathrm{μ\pm ð³Õ}\end{array}$

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

$k=\frac{2\mathrm{π}}{\mathrm{λ}}$

Where $\mathrm{λ}$is wavelength

From de Broglie equation we know

$\mathrm{λ}=\frac{h}{p}$

Where $h$is plank constant and $p$is momentum

We know

$p=mv$

Where $m$is mass and $v$is speed

Substituting $\mathrm{λ}=\frac{h}{mv}$in the wave number formula

$\begin{array}{rcl}k& =& \frac{2\mathrm{Ï€}}{{\displaystyle \raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$mv$}\right.}}\\ & =& \frac{2\mathrm{Ï€³¾\pm ¹}}{\mathrm{h}}\end{array}$

Substituting the values,

$\begin{array}{rcl}k& =& \frac{2×3.14×9.31×{10}^{-31}kg×900m/s}{6.626×{10}^{-34}}\\ & =& 7.77×{10}^{6}m/s\end{array}$

We know, In region 2 $V_b=-1.25\mathrm{μ³Õ}$

$\begin{array}{rcl}{U}_b& =& e{V}_b\\ & =& 1.25\mathrm{μ}eV\end{array}$

Therefore, the angular wave number in region 2 is

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

Where $E$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.

$\begin{array}{rcl}{k}_b& =& \frac{2×3.14}{6.626×{10}^{-34}\mathrm{Js}}×\sqrt{2×9.1×{10}^{-31}\mathrm{kg}×\left(2.306-1.25\right)\mathrm{μ\pm ð³Õ}×1.6×{10}^{-16}\mathrm{J}/\mathrm{μ\pm ð³Õ}}\\ & =& 5.528×{10}^6{\mathrm{m}}^{-1}\end{array}$The reflection coefficient can be given as

$R=\frac{{\left|B\right|}^2}{{\left|A\right|}^2}$

Where,

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

As we already calculated

$$\begin{gathered} \mathrm{k}=7.77×{10}^6{\mathrm{m}}^{-1} \\ {\mathrm{k}}_{\mathrm{b}}=5.258×{10}^6{\mathrm{m}}^{-1} \end{gathered}$$

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

localid="1663151160800" $$\begin{gathered} \frac{B}{A}=\frac{7.77×{10}^6{\mathrm{m}}^{-1}-5.258×{10}^6{\mathrm{m}}^{-1}}{7.77×{10}^6{\mathrm{m}}^{-1}+5.258×{10}^6{\mathrm{m}}^{-1}} \\ \frac{B}{A}=0.1928 \end{gathered}$$

The reflection coefficient is

$\begin{array}{c}R=\frac{{\left|B\right|}^2}{{\left|A\right|}^2}\\ ={\left|0.1928\right|}^2\\ =0.0372\end{array}$

The transmission coefficient is

$\begin{array}{c}T=1-R\\ =1-0.0372\\ =0.9628\end{array}$

Therefore, the current on the other side of the step boundary can be given by the formula

$$\begin{gathered} I_t=T{I}_0 \\ {I}_t=0.9628×5.00\mathrm{mA} \\ {I}_t=4.81\mathrm{mA} \end{gathered}$$

Hence the current on the other side of the step boundary is $I_t=4.81\mathrm{mA}$