91Ó°ÊÓ

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.

$E=U_{0+}\frac{n^2{\mathrm{Ï€}}^2{âˆ\dots }^2}{2m{L}^2}$

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

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

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

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

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

$\frac{n}{L}=3.64×{10}^9$

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:

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

$R=0.475$

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.