So the Wood-Saxon density distribution for protons in a nucleus is given as follows: $$\rho_p(r) = \frac{\rho_{p0}}{1+e^{[(r-R_0)/a]}}$$

Here $r$ is the distance to the centre of the nucleus, and $a$ is the so called skin depth. $\rho_{p0}$ is a constant such that the integral of the distribution equals the number of protons in the nucleus.

Now in a recent exam, the following question was asked:

"In the Woods-Saxon formula, $a$ is the measure of the width of the edge region of the nucleus (also denoted as the “skin depth”). The density at the edge is decreased to 20% of the density in the centre of the nucleus at a radius $R_{20}$ and to 80% of the density at a radius $R_{80}$. What is the distance between $R_{20}$ and $R_{80}$ expressed in units of $a$?"

Now I plugged in the following in Mathematica, $\rho_p(0)=x\rho_p(r)$, where $x$ denotes the percentage. After solving this for $r$ and subtracting $r$ for $x=0.20$ and $x=0.80$, I came up with the following final answer: $$a \ln\left(0.0625 + \frac{1}{5.333... + 4.266.... \exp{R_0/a}}\right).$$

However the final answer given in the exam was $a \ln(0.0625)$. What am I missing here?

  • Is $R_0>>a$ in general?
  • Was the professor wrong?
  • Something else?
  • 2
    $\begingroup$ Is general $R_0 \gt \gt a$? - yep $R_0$ describes the nuclear radius and $a$ describes the width of the fall of region. In most cases nuclei are pretty much spherical hence to a good approximation $R_0 \gt \gt a$ in most cases. $\endgroup$ Apr 20, 2017 at 11:57

1 Answer 1

  • Is $a \ll R_0$ in general? Most of the time, yes (it is not true near drip lines). By the way, if you consider symmetric nuclei only so you have $$\rho(r) = \rho_0 \left[1+\exp\left(\frac{r-R_0}{a}\right)\right]^{-1}$$ you can do a Sommerfeld expansion to recover $A$ (number of nucleons).

  • Was the professor wrong? No, he was right! Indeed, you are looking for the expression of $\Delta R = R_{80}-R_{20}$, let us express $R_{x}$ where $x$ denotes the percentage.

$$x \rho_{p_0} = \rho_{p_0} \left[1+\exp\left(\frac{R_x-R_0}{a}\right)\right]^{-1}$$

$$x = \frac{1}{1+\exp\left(\frac{R_x-R}{a}\right)}$$

$$1+\exp\left(\frac{R_x-R}{a}\right) = \frac{1}{x}$$

$$\exp(R_x - R_0)/a = 1/x - 1$$

$$R_x = a \ln(1/x - 1) + R_0$$

Finally we have

$$\Delta R = a\left[\ln(1/0.8 -1) +R_0 - \ln(1/0.2 - 1) - R_0\right]$$

Knowing that $\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)$, we recover the expression of your professor:

$$\Delta R = a\ln\left(\frac{1/0.8-1}{1/0.2-1}\right) = a\ln(0.0625)$$

  • Something else? $a$ is often called the diffuseness parameter.
  • $\begingroup$ Thanks, this explains a lot! In your calculation of $R_x$, you made use of the assumption $a<<R_0$ as well right? Otherwise there would still be a dependence on $R_0$., if I'm not mistaken. $\endgroup$ Apr 20, 2017 at 15:22
  • $\begingroup$ No, i don't use this assumption. The $R_0$ dependence disappears when you calculate the difference. I edited my answer to give you more details. You can valid my answer if it satisfies you. $\endgroup$
    – T. Auerrac
    Apr 20, 2017 at 15:41
  • $\begingroup$ I totally agree on the algebra, the only point I'm stuck on is definitions. $R_x$ is defined as the distance where $\rho_p$ has decreased to $x$ percent of it's value at the centre. You take the value at the centre to be $\rho_{p0}$, and I take it to be $\rho_p(o) = \rho_{p0} / [1 + \exp[...]]$. Hence my confusion. The only way I can reconcile this is by assuming $a<<R_0$. $\endgroup$ Apr 20, 2017 at 15:41
  • $\begingroup$ Consider this picture : Wood-Saxon density profiles. When you take $80\%$ of the density profile at the center, you get $0.8\rho_{p_0}$, simplifying with the other side of the equation. $\endgroup$
    – T. Auerrac
    Apr 20, 2017 at 16:05
  • $\begingroup$ Thanks! But this doesn't hold for low A nucleons right? Like in this plot (I took these of the Nuclear Charge Density Archives) $\endgroup$ Apr 20, 2017 at 16:16

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