What Are the Turning Points in Alpha Decay Potential Barrier?

[arierreF]

Homework Statement

A nuclei of a atomic number Z decays into a alpha particle (a He nucleus with Z =2) and a daughter nucleus with $(Z_{d}$).
The decay may be described as the tunneling of an alpha-particle through a barrier caused by the Coulomb potential between the daughter and the alpha-particle.

The potential is: $V(r) = \frac{1}{4\piε_{0}}\frac{2Z_{d}e^{2}}{r}$
Knowing that the probability of transmission , T, is proportional to

$T \propto e^{{-2* \int_a^b \sqrt{\frac{2m(V(r)-E)}{\hbar }} \,dr }}$

Calculate the turning points, $a$ and $b$.
Notes: The diagram for the potential barrier is shown inf the link:
https://www.google.pt/search?q=alph...taneous_Decay_Processes_-_Alpha_Decay;624;354Im stuck in this problem. I know that i can calculate the turning points at V(r) = E(r), but i do not have E(r). Can u give me a tip for solve this problem?

 

[arierreF]

I found the second turning point doing the following:

E(r_{2})=V(r_{2})

so


$V(r_{2}) = \frac{1}{4\piε_{0}}\frac{2Z_{d}e^{2}}{r_{2}}$

then we $r_{2} = \frac{1}{4\piε_{0}}\frac{2Z_{d}e^{2}}{E(r_{2})}$