Question

Argon (atomic radius $0.07$ nm atomic weight $40$) and Xenon (atomic radius $0.1$ nm atomic weight $140$) have the same density and at the same temperature. What is the ratio of their respective mean of having free time closest?

• A. $4.67$
• B. $2.04$
• C. $1.83$
• D. $3.67$

Answer 1

The correct option is C

$1.83$

Explanation for the correct option

C. $1.83$

Step1:

• The atomic weight of Argon$=40$
• The atomic weight of Xenon$=140$
• The atomic Radius of Argon$=0.07\mathrm{nm}$
• The atomic radius of Xenon$=0.1\mathrm{nm}$

Step2:

• $\mathrm{Mean}\mathrm{free}\mathrm{time}=\frac{1}{\sqrt{2\mathrm{n}}{\mathrm{\pi d}}^2{\mathrm{V}}_{\mathrm{rms}}}$
• where n is number density and d is the diameter of the gas atom.

Step 3:

• Mean speed of a gas atom $\mathrm{v}=\sqrt{\frac{8}{\mathrm{\pi }}\frac{\mathrm{kT}}{\mathrm{m}}}$
• where k=Boltzmann's constant, T=temperature of the sample, and m=atomic mass

Step4:

• $\tau =\frac{\lambda }{\nu }$$=\frac{\left({\displaystyle \frac{1}{\sqrt{2\mathrm{\pi }}{\mathrm{nd}}^2}}\right)}{\sqrt{{\displaystyle \frac{8}{\mathrm{\pi }}\frac{\mathrm{kT}}{\mathrm{m}}}}}$
• $\mathrm{\tau }\mathrm{\alpha }\frac{\sqrt{\mathrm{m}}}{{\mathrm{d}}^2}$where T and n are the same
• $\frac{{\mathrm{\tau }}_{\mathrm{Argon}}}{{\mathrm{\tau }}_{\mathrm{Xenon}}}=\frac{\sqrt{{\mathrm{m}}_{\mathrm{Argon}}}}{\sqrt{{\mathrm{m}}_{\mathrm{Xenon}}}}$$·\frac{{\mathrm{d}}_{\mathrm{Xenon}}^2}{{\mathrm{d}}_{\mathrm{Argon}}^2}$$=\frac{\sqrt{{\mathrm{m}}_{\mathrm{argon}}}}{\sqrt{{\mathrm{m}}_{\mathrm{Xenon}}}}$$·\frac{{\mathrm{r}}_{\mathrm{Xenon}}^2}{{\mathrm{r}}_{\mathrm{Argon}}^2}$

Step5:

• $\therefore \frac{{\mathrm{\tau }}_{\mathrm{Argon}}}{{\mathrm{\tau }}_{\mathrm{Xeon}}}=\sqrt{\frac{40}{140}}\times {\left(\frac{0.1}{0.07}\right)}^2$$=1.09$

Therefore, the ratio of their respective mean of having free time closest $=$ $1.83$ which is closest to $1.09$