Question about Woods-Saxon density distribution 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?

 A: *

*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.

