For the silver–lead pair, the constants a and b are given as 2.50 ^...

The thermo-emf $E$ of a thermocouple with the cold junction at 0^{}^{{}^{\circ}}\text{C} and the hot junction at t^{}^{{}^{\circ}}\text{C} is given by the relation:

$E = at + \dfrac{1}{2}bt^2$

The neutral temperature $t_n$ is the temperature of the hot junction at which the thermo-emf is maximum, which means $\dfrac{dE}{dt} = 0$.

$\dfrac{dE}{dt} = a + bt_n = 0 \implies t_n = -\dfrac{a}{b}$

Given a = 2.50 \mu\text{V }^{}^{{}^{\circ}}\textC^{-1} and b = 0.012 \mu\text{V }^{}^{{}^{\circ}}\textC^{-2}:

t_n = -\dfrac{2.50}{0.012} = -208.3^{}^{{}^{\circ}}\text{C}

Since the colder junction is kept at 0^{}^{{}^{\circ}}\text{C}, the hot junction must be at a temperature t > 0^{}^{{}^{\circ}}\text{C}. Because t_n < 0^{}^{{}^{\circ}}\text{C}, the hot junction can never reach the neutral temperature. Thus, a neutral temperature will not exist for this setup.

The inversion temperature $t_i$ is the temperature of the hot junction at which the thermo-emf becomes zero (other than 0^{}^{{}^{\circ}}\text{C}).

$E = 0 \implies at_i + \dfrac{1}{2}bt_i^2 = 0 \implies t_i = -\dfrac{2a}{b} = 2t_n$

t_i = 2 \times (-208.3^{}^{{}^{\circ}}\text{C}) = -416.6^{}^{{}^{\circ}}\text{C}

Since this temperature is below absolute zero (-273.15^{}^{{}^{\circ}}\text{C}), it is physically impossible to achieve. Thus, an inversion temperature will not exist.

Therefore, both the neutral temperature and the inversion temperature will not exist.

Answer: a neutral temperature will not exist; an inversion temperature will not exist