Thermal voltage of Ge diode
The thermal voltage does not depend on the semiconductor material; it's simply
$$V_T = \frac{kT}{q},$$
where \$k\$ is the Boltzmann constant, \$k = 1.380649 × 10^{-23}\ \mathrm{J}·\mathrm{K}^{-1},\ T\$ is the temperature, and \$q\$ is the electron charge, \$q=1.602176634 × 10^{-19}\ \mathrm{C}.\$
At 25 °C, this works out to approximately 26 mV, and does not depend on material.