Question

Question: An electron with total energy $\mathit{E}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{eV}$approaches a barrier of height ${\mathit{U}}_{\mathbf{b}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{eV}$and thickness $\mathit{L}\mathbf{=}\mathbf{750}\mathbf{}\mathbf{pm}$.What percentage change in the transmission coefficient $\mathit{T}$occurs for a 1.0% change in (a) the barrier height, (b) the barrier thickness, and (c) the kinetic energy of the incident

electron?

Short answer

(a) The transmission coefficient will decrease by $20\%$.

(b) The transmission coefficient will decrease by $10\%$.

(c) The transmission coefficient will increase by $15\%$.

Step-by-step solution

Identifying the data given in the question.

The incident energy of the electron $E=5.0\mathrm{eV}$

The height of the potential energy barrier$U_b=6.8\mathrm{eV}$

The thickness of the barrier$L=750\mathrm{pm}$

The percentage change in barrier height,$∆U_b=1\%$

The percentage change in barrier thickness, $∆L=1\%$

The percentage change in energy,$∆E=1\%$

Concept used to solve the question.

A particle for example electrons, or proton can reflect from any boundary at which its potential energy changes even when classically it would not reflect.

A potential energy barrier is a region where a traveling particle

will have increased potential energy $U_b$.

(a) Finding percentage change in the transmission coefficient with change in the barrier height.

The transmission coefficient of a particle can be given as.

$T=e^{2bL}$

Where, $L$is the length of the potential barrier.

And $b$can be given as,

$b=\sqrt{\frac{8m^2\left(U_b-E\right)}{h^2}}$

Where the mass of the particle is $m$and $E$is incident energy and $U_b$is the height of the barrier

Differentiating $b$with respect to $U_b$

$\begin{array}{rcl}\frac{db}{d{U}_b}& =& \frac{d}{d{U}_b}\left(\sqrt{\frac{8{\pi }^2\left(U_b-E\right)}{h^2}}\right)\\ & =& \frac{1}{2\sqrt{U_b-E}}\sqrt{\frac{8{\pi }^2}{h^2}}\\ & =& \frac{1}{2\left(U_b-E\right)}\sqrt{\frac{8{\pi }^2\left(U_b-E\right)}{h^2}}\\ & =& \frac{b}{2\left(U_b-E\right)}\end{array}$

The change in transmission coefficient with $U_b$can be given as

$$\begin{gathered} ∆T=\frac{dT}{d{U}_b}∆U_b \\ ∆T=\frac{d\left(e^{-2bL}\right)}{d{U}_b}∆U_b \\ ∆T=-2L{e}^{-2bL}∆U_b \\ ∆T=-2LT\frac{dT}{d{U}_b}∆U_b \end{gathered}$$

Therefore,

role="math" localid="1663224174207" $$\begin{gathered} ∆T=-2LT\left(\frac{b}{2\left(U_b-E\right)}\right)∆U_b \\ \frac{∆T}{T}=-Lb\frac{∆U_b}{\left(U_b-E\right)} \end{gathered}$$

We know

$b=\sqrt{\frac{8m^2\left(U_b-E\right)}{h^2}}$

Substituting the values,

$\begin{array}{rcl}b& =& \sqrt{\frac{8m^2\left(U_b-E\right)}{h^2}}\\ & =& \sqrt{\frac{8{\mathrm{\pi }}^2\left(9.11\times {10}^{-31}\mathrm{kg}\right)\left(6.8\mathrm{eV}-5.1\mathrm{eV}\right)\left(1.6\times {10}^{-19}\mathrm{J}/\mathrm{eV}\right)}{{\left(6.626\times {10}^{-34}\mathrm{Js}\right)}^2}}\\ & =& 6.67\times {10}^9{\mathrm{m}}^{-1}\end{array}$

So,

$\begin{array}{rcl}bL& =& \left(6.67\times {10}^9{\mathrm{m}}^{-1}\right)\times \left(750\times {10}^{-12}\mathrm{m}\right)\\ & =& 5\end{array}$

Therefore,

$\begin{array}{rcl}\frac{∆T}{T}& =& -Lb\frac{∆U_b}{\left(U_b-E\right)}\\ & =& -5\times \frac{\left(0.01\times 6.8\mathrm{eV}\right)}{\left(6.8\mathrm{eV}-5.1\mathrm{eV}\right)}\\ & =& -0.20\end{array}$

The transmission coefficient will decrease by$20\%$.

(b) Finding percentage change in the transmission coefficient with change in the barrier thickness.

The transmission coefficient of a particle can be given as.

$T=e^{-2bL}$

The change in transmission coefficient with barrier thickness can be given as

$$\begin{gathered} ∆T=\frac{dT}{dL}∆L \\ ∆T=\frac{d\left(e^{-2bL}\right)}{dL}∆L \\ ∆T=-2b{e}^{-2bL}∆L \\ ∆T=-2bT∆L \\ \frac{∆T}{T}=-2b∆L \end{gathered}$$

We already calculated

$b=6.67\times {10}^9{\mathrm{m}}^{-1}$

Substituting the values,

$\begin{array}{rcl}\frac{∆T}{T}& =& -2\times 6.67\times {10}^9{\mathrm{m}}^{-1}\times 0.01\times 750\times {10}^{-12}\mathrm{m}\\ & =& -0.10\end{array}$

The transmission coefficient will decrease by $10\%$.

(c) Finding percentage change in the transmission coefficient with change in kinetic energy of the incident electron.

The transmission coefficient of a particle can be given as.

$T=e^{2bL}$

$b=\sqrt{\frac{8m^2\left(U_b-E\right)}{h^2}}$

Differentiating $b$with respect to $E$

$\begin{array}{rcl}\frac{db}{dE}& =& \frac{d}{dE}\left(\sqrt{\frac{8{\pi }^2\left(U_b-E\right)}{h^2}}\right)\\ & =& -\frac{1}{2\sqrt{U_b-E}}\sqrt{\frac{8{\pi }^2}{h^2}}\\ & =& -\frac{1}{2\left(U_b-E\right)}\sqrt{\frac{8{\pi }^2\left(U_b-E\right)}{h^2}}\\ & =& -\frac{b}{2\left(U_b-E\right)}\end{array}$

The change in transmission coefficient with $E$can be given as

$$\begin{gathered} T=\frac{dT}{dE}∆E \\ ∆T=\frac{d\left(e^{-2bL}\right)}{dE}∆E \\ ∆T=-2L{e}^{-2bL}∆E \\ ∆T=-2LT\frac{dT}{dE}∆E \end{gathered}$$

Therefore,

$$\begin{gathered} ∆T=-2LT\left(\frac{b}{2\left(U_b-E\right)}\right)∆E \\ \frac{∆T}{T}=Lb\frac{∆E}{\left(U_b-E\right)} \end{gathered}$$

We already calculated

$bL=5$

Therefore,

$\begin{array}{rcl}\frac{∆T}{T}& =& Lb\frac{∆E}{\left(U_b-E\right)}\\ & =& 5\times \frac{\left(0.01\times 5.1\mathrm{eV}\right)}{\left(6.8\mathrm{eV}-5.1\mathrm{eV}\right)}\\ & =& 0.15\end{array}$

The transmission coefficient will increase by$15\%$.