The potential energy barrier in field emission is not rectangular, but resembles a ramp, as shown in Figure 6.16. Here we compare tunnelling probability calculated by the crudest approximation to that calculated by a better one. In method 1, calculate T by treating the barrier as an actual ramp in which U - E is initially$\varphi$, but falls off with a slop of M. Use the formula given in Exercise 37. In method 2, the cruder one, assume a barrier whose height exceeds E by a constant $\mathit{\varphi }\mathbf{/}\mathbf{2}$(the same as the average excess for the ramp) and whose width is the same as the distance the particle tunnels through the ramp. (a) Show that the ratio T₁/T₂ is ${\mathit{e}}^{\frac{\sqrt{\mathbf{8}\mathbf{m}{\mathbf{\varphi }}^{\mathbf{3}}}}{\mathbf{3}\sout{\mathbf{h}}\mathbf{M}}}$ . (b) Do the methods differ more when tunnelling probability is relatively high or relatively low?

Tunnelling is a phenomenon when a particle propagates through a potential barrier when the potential energy of the barrier is higher than the kinetic energy of the particle.

Tunneling Probability is the ratio of squared amplitudes of the waves after crossing the barrier to the incident waves.

The potential energy is modelled as,

$U\left(x\right)-E=\varphi -Mx$

If the particle enters where x = 0 and exits where E = U, or $x=\varphi /M$.

Hence, by method 1, tunnelling probability is given by

role="math" localid="1660047829428" $\begin{array}{rcl}{T}_1& \cong & e^{\frac{-2}{\sout{h}}\left\{{\int }_0^{\varphi /M}\sqrt{2m\left(\varphi /Mx\right)}dx\right\}}\\ & =& e^{\frac{-2}{\sout{h}}\left\{{\overline{)\frac{{\left(2m\left(\varphi /Mx\right)\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{3Mm}}}_0^{\varphi /M}\right\}}\\ & =& e^{\frac{-2}{\sout{h}}\left\{\frac{\sqrt{8m{\varphi }^3}}{3M}\right\}}\end{array}$

Also, by method 2, tunnelling probability is given by

$\begin{array}{rcl}{T}_2& =& e^{\frac{-2}{\sout{h}}\left\{\frac{\sqrt{2m{\varphi }^3}}{M}\right\}}\\ & =& e^{\left\{\frac{\sqrt{8m{\varphi }^3}}{\sout{h}M}\right\}}\end{array}$

Now the required ratio, T₁/T₂ can be calculated as,

$\begin{array}{rcl}{T}_1/T_2& =& e^{\left\{\frac{-2}{\sout{h}}\frac{\sqrt{8m{\varphi }^3}}{3M}+\frac{\sqrt{8m{\varphi }^3}}{\sout{h}M}\right\}}\\ & =& e^{\frac{\sqrt{8m{\varphi }^3}}{3M}}\end{array}$

Hence, the ratio T₁/T₂ is obtained as $\begin{array}{rcl}& & e^{\frac{\sqrt{8m{\varphi }^3}}{3M}}\end{array}$.

As it can be clearly seen from the ratio obtained in the previous step, the ratio will have the largest value, i.e., the probabilities differ the most when the argument of the exponential is large or when the tunnelling probability is small.