Question

Consider a potential barrier of height $\mathbf{30}\mathbf{}\mathit{e}\mathit{V}$. (a) Find a width around$\mathbf{}\mathbf{1}\mathbf{.}\mathbf{000}\mathbf{}\mathit{n}\mathit{m}$for which there will be no reflection of $\mathbf{35}\mathbf{}\mathit{e}\mathit{V}$electrons incident upon the barrier. (b) What would be the reflection probability for $\mathbf{36}\mathbf{}\mathit{e}\mathit{V}$electrons incident upon the same barrier? (Note: This corresponds to a difference in speed of less than$\mathbf{1}\mathbf{(}\mathbf{1}\mathbf{/}\mathbf{2}\mathbf{)}\mathbf{}\mathbf{\%}\mathbf{.}$

Short answer

The width is $1.098nm$and the reflection probability is$0.475.$

Step-by-step solution

Definition:

Tunneling defines the penetration of a barrier of high energy by a low-energy wave or particle. For no reflection,$\mathbf{}\mathit{E}\mathbf{>}\mathit{U}\mathbf{.}$ This is called resonant transmission.

$E=U_{0+}\frac{n^2{\pi }^2{\varnothing }^2}{2m{L}^2}$

Reflection probability or reflection coefficient is defined as the ratio of the amplitude of the reflected wave to that of the incident wave.

$R=\frac{{\sin }^2\left[{\displaystyle \frac{\sqrt{2m(E+U_0)L}}{\varnothing }}\right]}{{\sin }^2\left[{\displaystyle \frac{\sqrt{2m(E+U_0)L}}{\varnothing }}\right]+4\left({\displaystyle \frac{E}{U_0}}\right)\left[\left({\displaystyle \frac{E}{U_0}}\right)+1\right]}$

Given/known parameters

$E=35eV$and$U=30eV$

Solution

(a)$\frac{n}{L}=\frac{\sqrt{2m(E-U)}}{\mathrm{\pi }}$

$\frac{n}{L}=\frac{\sqrt{2\times 9.1\times {10}^{-31}\times \left(35-30\right)\times 1.6\times {10}^{-19}J/eV}}{3.14\times 1.05\times {10}^{-34}}$

$\frac{n}{L}=3.64\times {10}^9$

If $L=1nm,n=3.64$. Rounding off to the nearest integral value,$n=4.$This gives $L=1.098nm.$

(b)Putting values in reflection formula:

$R=\frac{{\sin }^2\left[{\displaystyle \frac{\sqrt{2\times 9.1\times {10}^{-31}(36-30)\times 1.6\times {10}^{-19}}\times 1.098}{1.05\times {10}^{-34}}}\right]}{i{n}^2\left[{\displaystyle \frac{\sqrt{2\times 9.1\times {10}^{-31}(36-30)\times 1.6\times {10}^{-19}\times }1.098}{1.05\times {10}^{-34}}}+4\left({\displaystyle \frac{36}{30}}\right)\left[\left({\displaystyle \frac{36}{30}}\right)-1\right]\right]}$

$R=0.475$

Explanation and Conclusion

For a barrier width of$1.098nm,$no reflection will take place.

Reflection coefficient for$36eV$beam is $0.475$. It is quite small, which means the window for resonant transmission can be small.