JEE MAIN - Physics (2020 - 9th January Evening Slot - No. 16)
Two gases-argon (atomic radius 0.07 nm,
atomic weight 40) and xenon (atomic radius
0.1 nm, atomic weight 140) have the same
number density and are at the same
temperature. The raito of their respective mean
free times is closest to :
2.3
1.83
4.67
3.67
Explanation
$$\lambda = {1 \over {\sqrt 2 \pi {d^2}n}}$$
Mean free time, t = $${\lambda \over v}$$
Also v $$ \propto $$ $$\sqrt {{T \over M}} $$
$$ \therefore $$ t $$ \propto $$ $${{\sqrt M } \over d}$$
$${{{t_{Ar}}} \over {{t_{xe}}}}$$ = $${{d_{Xe}^2} \over {d_{Ar}^2}} \times \sqrt {{{{M_{Ar}}} \over {{M_{Xe}}}}} $$
= $${\left( {{{0.1} \over {0.07}}} \right)^2} \times \sqrt {{{40} \over {140}}} $$
= 1.09
$$ \therefore $$ Nearest possible answer is 1.83.
Mean free time, t = $${\lambda \over v}$$
Also v $$ \propto $$ $$\sqrt {{T \over M}} $$
$$ \therefore $$ t $$ \propto $$ $${{\sqrt M } \over d}$$
$${{{t_{Ar}}} \over {{t_{xe}}}}$$ = $${{d_{Xe}^2} \over {d_{Ar}^2}} \times \sqrt {{{{M_{Ar}}} \over {{M_{Xe}}}}} $$
= $${\left( {{{0.1} \over {0.07}}} \right)^2} \times \sqrt {{{40} \over {140}}} $$
= 1.09
$$ \therefore $$ Nearest possible answer is 1.83.
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