# Quantum Tunneling Minimum Energy

1. May 18, 2010

### fiontie

Good day, everyone!
Lately I faced the necessity of solving a problem from a field I know literally nothing about. So I just made an online research but without success. Any help (hints, good sources, relevant equations) would be greatly appreciated!

1. The problem statement, all variables and given/known data

Gold nucleus (Au Z=79 A=197) of radius R is bombarded by an alpha particle (He z=2 A=4) of radius r. Find the minimum energy of the alpha particle required to penetrate inside the nucleus.

2. Relevant equations

Would be great to know.

3. The attempt at a solution

(the following most likely has nothing to do with a solution and is not to be read :) )

A nucleus and a particle are repulsed by the Coulomb force when colliding, so the problem refers to overcoming the Coulomb barrier. Its approximate height for an arbitrary nucleus is

$B = \frac{Zz}{A^{\frac{1}{3}}}$

What I am missing here is a "minimum energy required to penetrate". In quantum mechanics, the particle with energy far lower than a barrier can still penetrate, isn't it just a matter of possibility? At what point does the penetration become impossible?

Thus, we can find the possibility of overcoming our barrier (which is its transparency coefficient $D$). Given the formula for a rectangular barrier

$D = \exp{\left(-\frac{2}{\hbar} d \sqrt{2m(U-E)}\right)}$

where $d$ is a barrier width, $U$ is a barrier height and $E$ is a particle energy, we can get one for an arbitrary barrier by breaking it into thin rectangular stripes and integrating over them:

$D = \exp{\left(-\frac{2}{\hbar} \int_{x_1}^{x_2} dx \, \sqrt{2m(U-E)} \right)}$

Now let's apply this formula to our case. The potential energy of a particle on a distance $r \geq R$ is defined by the energy of a Coulomb interaction (I just stumbled across this formula and am not sure where it comes from):

$U(r) = E_{C} \frac{R}{r}$
$E_0 = m_\alpha c^2$ is a rest energy of an alpha particle

The barrier boundaries are $R$ and $r_\alpha$, where $r_\alpha = R\frac{E_C}{E_\alpha}$ is a distance where the energy of a particle becomes equal to the energy of a Coulomb repulsion.

So finally,

$D = \exp{\left(-\frac{2}{\hbar c} \sqrt{2 E_0} \int_{R}^{r_\alpha} dr \, \sqrt{E_{C} \frac{R}{r} - E_\alpha} \right)}$

I'm actually clueless how this can help.
And I'm also not sure how to make use of an alpha particle size.