Question

I am looking at deriving an expression for the Gamow factor for ?-decay. I understand that the potential is the sum of the nuclear, electric and effective potentials:

$V(r) = V_N(r)+V_c(r) +\frac{l(l+1)}{2mr^2}$

The Gamow factor is given by

$G=\int_R^{r_c} \sqrt{2m(V(r)-E)} dr ? \frac{2\pi(Z-2\alpha)}{\beta}$

if we consider only the Coulomb potential $V_c(r)$

I want to also incorporate the effective potential due to the spin $\frac{l(l+1)}{2mr^2}$

In my textbook it states that the Gamow factor is a factor of $\frac{l(l+1)}{\sqrt{4mRZ\alpha}}$   

greater in the case where the effective potential is considered, but I am unable to show this.

How would I show that the Gamow factor is increased by this value when the effective potential is considered?

Thank you,

Answer 1

I am not sure, but it "smells" as if the following trick was done, see the calculus below. What I know from the ? disintegration is that more than 90% from the emitted wave is s-wave. Then, the angular term is just a correction to the rest of the potential. In this case, denoting by $V_0$ the potential without the term with angular momentum, we can write

$G=\int_R^{r_c} \sqrt{2m(V_0(r)-E) +\frac{l(l+1)}{r^2}} \ dr$

$?\int_R^{r_c} \sqrt{\sqrt {2m(V_0(r)-E)}^2 + \frac{2l(l+1) \sqrt{2m(V_0(r)-E)}}{r^2\sqrt{2m(V_0(r)-E)}} + \frac{l^2(l+1)^2}{2m(V_0(r)-E)r^4}} \$ $?\int_R^{r_c} \sqrt{\sqrt {2m(V_0(r)-E)}^2 + \frac{2l(l+1) \sqrt{2m(V_0(r)-E)}}{r^2\sqrt{2m(V_0(r)-E)}} + \frac{l^2(l+1)^2}{2m(V_0(r)-E)r^4}} \ dr$ $\approx \int_R^{r_c} \sqrt{\sqrt {2m(V_0(r)-E)}^2 + \frac{2l(l+1) \sqrt{2m(V_0(r)-E)}}{r^2\sqrt{2m(V_0(r)-E)}} + \frac{l^2(l+1)^2}{2m(V_0(r)-E)r^4}} \ dr$

drfrom which you get further

$G \approx \int_R^{r_c} \{\sqrt{2m(V_0(r)-E)} + \frac{l(l+1)}{r^2\sqrt{2m(V_0(r)-E)}}\} dr$

The integral over the first term in the integrand is known to you. The integral

$l(l+1) \int_R^{r_c} \frac{1}{r^2\sqrt{2m(V_0(r)-E)}} dr.$

Now, again, I am not sure how solves your book this integral, but I would integrate by parts:

$-\frac {l(l+1)}{r} \frac{1}{\sqrt{2m(V_0(r)-E)}}|_R^{r_c} - \int_R^{r_c} \frac {l(l+1)}{2m} \frac{dV_0(r)}{dr} \frac{m^2}{r\sqrt{2m(V_0(r)-E)}^3} dr.$

NEW EDIT: In the text below I introduced a couple of modifications (I apologize, the first draft I wrote during the night). Well, how to solve the integral I don't know, but as the angular momentum term is a small correction with respect to

the contribution of this integral should be small.

What I can estimate is the first term. It seems to me that the domain of integration is over the region where $V_0(r) + l(l+1)/r^2 \ge E$ Wherever   the square root is imaginary, moreover, we don't need this Gamow factor. Then, at $r_c$ , $\sqrt {2m(V_0(r) - E} = (1/r)\sqrt{l(l+1)} Thus,$

$-\frac {l(l+1)}{r} \frac{1}{\sqrt{2m(V_0(r)-E)}}|_R^{r_c} = -\sqrt {l(l+1)} + \frac {l(l+1)}{R} \frac{1}{\sqrt{2m(V_0(R)-E)}}$

At the inner limit R of the potential barrier, the Coulombian potential is by far greater than E, so we can for sure approximate $V_0(R) = 2Ze^2/(4\pi\epsilon _0)$ where 2 is the charge of the alpha and Z is that of the daughter nucleus. So you get

$-\frac {l(l+1)}{r} \frac{1}{\sqrt{2m(V_0(r)-E)}}|_R^{r_c} = -\sqrt {l(l+1)} + \frac {l(l+1)}{\sqrt {4mZR}} \sqrt {\frac {4\pi\epsilon _0}{e^2}}$

Now, if you calculate the 2nd term, it is by orders of magnitude greater than the first one. The factor $\sqrt {4\pi\epsilon _0/e^2}$ which is just constants, was probably swept inside the constant \alpha that you didn't tell me what it is.