# Simple Tunneling

#### Electron wave

An electron travels toward negative values of $x$ having a kinetic energy $E=2 eV$. Please write the equation of this electron wave and describe its characteristic values.

The general equation of the wave travelling in the negative x direction (towards left) is

$\psi(x,t)=A\exp[i(kx+\omega t)]$ ($A$ is the wave amplitude)

For a fixed point of the wave that has a constant value of $\psi$, the exponential has to maintain its value, so that for time t increasing, $x$ has to decrease, for the wave to travel towards left).

For wave number and frequency we write:

$k=2\pi/\lambda$, $\omega=2\pi/T$, $\nu=\lambda/T=\omega/k$

From the de Broglie relation

$\lambda =h/p=h/\sqrt{2mE}=\frac{6.62*10^{-34}}{\sqrt{2*9.1*10^{-31}*2*1.6*10^{-19}}}=8.67*10^{-10} m$ $k=7.24*10^9 (1/m)$

and

$v=\sqrt{2E/m}=\sqrt{\frac{2.2*1.6*10^{-19}}{9.1*10^{-31}}}$

$\omega=kv=6.07*10^{15} (1/s)$

#### Tunneling Proton

A proton that is tunneling through a potential $V=10 MeV$. It has a kinetic energy $E_k= 2MeV (<V)$. Please find the tunneling probability if the width of the potential $V$ is $10^{-14} m$. What would be the new width of the potential barrier if the tunneling probability is to increase 10 times?

The real transmission coefficient for tunneling through a square barrier can be found elsewhere. A simpler reasoning (with a simpler result, but with sufficient precision) is the following:

Consider that you have a proton situatied to the left of the barrier. The proton is traveling towards right. Its associated wave form is complex $\psi_1(x)=A\exp(ik_1*x)$

Now the proton enters the barrier. Its wave function has to decrease exponentially inside the barrier.

$\psi_2(x)=A*\exp(-k_2x)$

Then the proton exits the barrier and it is again a wave

$\psi_3(x)=B\exp(ik_1x)$

(To the left and to the right of the barrier since the proton is situated in the free space the wave number is the same $k_1$)

The probability of tunneling is thus simply the ratio of its wavefunction amplitudes

$P=|\psi_3/\psi_1|^2=|B/A|^2=|\psi_2(L)/\psi_2(0)|^2=\exp(-2L*k_2)$

Inside the barrier the wavenumber is:

$k_2=2\pi/\lambda_2=\frac{{2\pi}\sqrt{2m(U-Ek)}}{\hbar}=…=6.2*10^{14}$

Thus the probability for $L=10^{-14} m$ is

$P=\exp(-2*10^{-14}*6.2*10^{14})=4.12*10^{-6}$

If the tunneling probability is increased 10 times we obtain:

$P_1/P_2=(1/10)=\frac{\exp(-2L_2*k_2)}{\exp(-2a*k_2)}=\exp[-2k_2*(L-a)]$

$l-a=\frac{ln(10)}{2k_2}=1.85*10^{-15} m$

$a=(1.4-0.185)*10^{-15}=1.215*10^{-15} m$