At low signal levels the RF input
resistance and audio output resistance of a detector diode are equal to
25,700,000*n/Is Ohms (current in nA).

25 million WHAT?

Ok One more, with a little help I figured it out.
It has to to with the Thermal Voltage of the diode.
Vt=KT/q
with k=1.38E-23 and q=1.6E-19. T is absolute temperature in degrees Kelvin,
k is Boltzmann's constant and q is the charge of an electron. VT is close
to 0.025 volts at 20 degrees Celsius.

With a slightly increased temperature the .025 is raised to .0257,
and .0257 / 1 nanoamp =25,700,000
Thanks, MikeK