Which of the following thermometric liquids is suitable for a liquid-in-glass thermometer used in the Antarctic region?

The given ionization energy is for one mole of Titanium atoms. To find the energy required to ionize a single atom, we divide by Avogadro's number ($N_A$).

$$E_{atom} = \frac{E_{mol}}{N_A}$$ $$E_{atom} = \frac{6.36 \times 10^5 \text{ J/mol}}{6.02 \times 10^{23} \text{ mol}^{-1}}$$ $$E_{atom} = 1.056478 \times 10^{-18} \text{ J}$$

The energy of a photon ($E$) is related to its wavelength ($\lambda$), Planck's constant ($h$), and the speed of light ($c$) by the formula:

$$E = \frac{hc}{\lambda}$$

$$h = 6.63 \times 10^{-34} \text{ Js}$$ $$c = 3.0 \times 10^8 \text{ m/s}$$ $$E_{atom} = 1.056478 \times 10^{-18} \text{ J}$$ $$\lambda = \frac{(6.63 \times 10^{-34} \text{ Js}) \times (3.0 \times 10^8 \text{ m/s})}{1.056478 \times 10^{-18} \text{ J}}$$ $$\lambda = \frac{1.989 \times 10^{-25} \text{ Jm}}{1.056478 \times 10^{-18} \text{ J}}$$ $$\lambda = 1.8826 \times 10^{-7} \text{ m}$$

Since $1 \text{ nm} = 10^{-9} \text{ m}$:

$$\lambda = 1.8826 \times 10^{-7} \text{ m} \times \frac{1 \text{ nm}}{10^{-9} \text{ m}}$$ $$\lambda = 188.26 \text{ nm}$$

The calculated wavelength is approximately $188.3 \text{ nm}$.
A. $0.188 \text{ nm}$
B. $1.88 \text{ nm}$
C. $18.8 \text{ nm}$
D. $188.0 \text{ nm}$

The closest option is D.

The final answer is $\boxed{\text{D. } 188.0 \text{ nm}}$.