Approximating current in photoelectric effect

I am currently doing an experiment looking at the energy distribution of photoelectrons in the photoelectric effect.

I have done some data collection for different wavelengths of light, and have the following results:

I understand there is no direct link between photocurrent and backing voltage, however I found this experiment done at Princeton which first uses the equation for the density of states of a free electron gas to approximate the current:

In this case, we will make only a first order correction to the linear
approximation. We assume that the conduction band of potassium can be
approximated as a free electron gas (this is a reasonable
approximation for most good conductors because the ‘sea of electrons’
in the conduction band is not significantly perturbed by the nuclei).
Fig. 4 shows an energy diagram for a one-dimensional conduction band;
we set the energy to be zero at the bottom of the conduction band. EF
stands for the Fermi energy of the free electron gas and is equivalent
to the bandwidth. In 3 dimensions, the density of states of a free
electron gas is:

$$D(E) = frac{V}{2pi^2} (frac{2m}{hbar})^{3/2} E^{½}$$

In order to determine how many electron states can be reached at a
given photon energy for a given reverse bias, we simply integrate the
density of states from the lowest possible energy level that can be
photoexcited to the Fermi energy E_F . The photocurrent should be
directly proportional to the number of states that can be
photoexcited. You should be able to work this out to derive an
equation for the current of the form:

$$ I = A(B^{3/2} – (V-C)^{3/2})$$

where B should correspond to E_F . Note that in this equation V is the
reverse bias, ie. it is the negative of the applied voltage.

I have tried just about everything to understand how this equation is derived but seem unable to.

If I integrate from E = 0 to E = E_f, I just don’t understand where the voltage comes into it. To me it seems as though there should only be one part:

$$ N = int_{0}^{E_F} frac{(2m)^{3/2}Volume}{2pihbar}E^{1/2} dE$$
$$ N = frac{2}{3}frac{(2m)^{3/2}Volume}{2pihbar} E_F^{3/2}$$

In short, I want to understand how voltage relates to the integral of the Fermi energy, and how this can be made proportional to current.