I am looking at deriving an expression for the Gamow factor for $\alpha$-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,


1 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 $\alpha$ 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

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

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

from which you get further

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

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 $V_0 - E$, 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 $V_0(r) + l(l+1)/r^2 < E$ 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.

  • $\begingroup$ I'm afraid I am struggling to follow. I am with you up to the integration by parts, but I am confused how you go from that expression to the final one? $\endgroup$ Jan 7, 2015 at 10:43
  • $\begingroup$ @user1887919 : see my NEW EDIT. But I am afraid that your formula of $G$ has lost a $\hbar$, i.e. under the sqrt has to appear $\hbar ^2$ in the nominator. $\endgroup$
    – Sofia
    Jan 7, 2015 at 15:36

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