The penetration depth model does not work for a good metal, but is ok for a moderately doped semiconductor. The pentration depth you get is smaller than an Angstrom for a good conductor. This means that the surface electrons are essentially confined to the first atomic layer, with some field entering into the next few layers, and the details of the atomic orbitals and the electron electron interactions are necessary for working out exactly how the electric field goes away.

### Penetration depth model

Here, you assume that the potential enters a self-consistent jellium, which is a uniform positive charge density plus almost free Fermi gas. The potential shifts the energy levels of the electrons, and you take this into account by filling the levels in the potential. But you take the electron electron interactions into account by modifying the potential according to the induced charge density.

If a jellium metal has a potential V imposed on it, the number density of the electrons at any point can be calculated semiclassically with no significant error, as follows. The occupied phase space volume for electrons in a box of side length $\Delta$ at an electrostatic potential $\phi$ is:

$${4\pi\over 3} (2m(E_f+e\phi))^{3\over 2} \Delta^3$$

where m is the mass of the electron, e is the magnitude of its charge, and $E_f$ is the Fermi energy. Dividing this by $h^3 = (2\pi \hbar)^3$ gives the number of occupied states. This immediately gives the electron density in a slowly varying potential

$$n(x) = {1 \over 6\pi^2} (\sqrt{ k_f^2 + {2em \phi\over \hbar^2}})^3$$

While the jellium model is no good for the low lying states, which see a nonuniform electric potential from the localized nuclei, it is still a good model material with a spherical Fermi surface, and is ok for order of magnitude estimates even away from a spherical Fermi surface. The point is that the actual number density is not what is important, it is only the variation in the number density given a certain V(x) which matters, and this can be accurate even when the number density itself is very wrong, because the states deep below the Fermi surface are not right in the model.

Anyway, expanding this density to lowest order in V and multiplying by -e gives the electronic charge density given an imposed external potential:

$$\rho(x) = - {1\over 6\pi^2} e k_f^3 + {2\over \pi} k_f {e^2\over 4\pi\hbar c} {mc\over \hbar} \phi(x)$$

The first term can be identified as the constant charge density of the electrons in the absence of a potential, and this is cancelled by the positive charge on the jellium. The remaining term gives a charge density proportional to the potential, which leads to screening of static fields.

When $\rho$ is proportional to V, $\rho= k V$ the Laplace equation becomes the Poisson equation

$$ \nabla^2 \phi = 4\pi \rho = 4\pi k \phi$$

The inverse screening length is found by solving the 1d version, and it is $\sqrt{4\pi k}$. For the particular problem above, the screening length $l_s$ is given by:

$$ l_s^{-1} = \sqrt{8 k_f \alpha \over \lambda}$$

Where $\alpha = {1\over 137}$ is the fine structure constant, $\lambda$ is the Compton wavelength of the electron times $2\pi$. The combination ${\alpha \over \lambda}$ is the Bohr radius, so this is really

$$ l_s^{-1} = \sqrt{8 k_f \over a_0}$$

The screening length is the geometric mean of the inverse Fermi momentum and the Bohr radius divided by the square root of eight.

For a good metal, the inverse Fermi momentum is about an Angstrom, although it can be much longer in a semiconductor. The result is less than an Angstrom, and is therefore unphysical. In this regime, the assumptions used to derive the self-consistent jellium break down.

So the penetration depth model is not correct for a good metal, and it requires a Fermi momentum which is about 10-100 times smaller than a lattice spacing to become good. This is the situation in a semiconductor, but not in good metals.

People spend their whole lives answering this question. The best I can do is give you a method for making a very rough estimate of the stopping power of a material. I'm going to assume a uniform slab of armor and a cubic bullet that will not deform at all. The relevant material property is the toughness. Toughness is defined as the amount of energy per unit volume a material can absorb before fracturing. It depends heavily on the rate at which the material is deformed and is typically determined empirically.

The reason I've chosen a cubic projectile is to have a uniform, time independent stress being applied to the armor. Let's also assume that the projectile applies force only to the region of the armor directly in its path. Under these many simplifications, the energy required to penetrate the armor can be estimated as $T\times t_{\text{armor}} \times A_{\text{bullet}}$, in which $T$ is the toughness of the armor, $t$ is the thickness of the armor, and $A$ is the area of the bullet. This is essentially the energy required to rip all the armor blocking the projectile out of the way.

This paper gives the high-strain rate (i.e., impact) toughness of one particular steel as roughly 2000 MPa, which translates to 2000 J/cm$^3$. So for 1 cm thick armor, a 50 gram bullet with a 1 cm cross section would need to have 2000 J of kinetic energy to make it through. Solving

$$2000 = \frac{1}{2}mv^2$$

gives $v = 282\text{m/s}$ to just penetrate.

I'm not a ballistics expert, but there are bullets that can make it through a cm of steel and bullets generally have velocities measured in the 100s of m/s so this all seems pretty reasonable.

## Best Answer

I don't think that toughness is a particularly relevant materials property for describing human skin. The property of toughness arose out of the need of engineers to describe how rigid materials such as steel components may behave differently when subjected to sudden, sharp impacts. Some steel alloys may be able to absorb considerable impact energy before they break or shatter, and those alloys are said to have high toughness. Other steel alloys of the same yield strength may be more brittle and may fail when subjected to even small impact energies. Such alloys would be said to have low toughness.

It might help if you described why you are interested in the "toughness" of skin, because there is probably a more appropriate engineering parameter to use than toughness. If what you are really interested is puncture resistance, there are probably engineering parameters to describe that.