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An ionization gauge is installed in the artificial satellite that $1\,c{m^3}$ of the atmosphere contains about a thousand million particles of gas at a height of 300 km from the earth’s surface. Find the mean free path of the gas particles at this height. Take the diameter of the particles equal to $2 \times {10^{ - 10}}\,{\text{m}}$.A. $l = 5.6\,{\text{km}}$B. $l = 2.6\,{\text{km}}$C. $l = 8.4\,{\text{km}}$D. $l = 1.1\,{\text{km}}$

Last updated date: 13th Jun 2024
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Hint: The mean free path of the molecules is the least distance travelled by the molecule after a successive collision. Recall the expression for the mean free path which is inversely proportional to the square of diameter of the molecule. Calculate the number of particles per unit volume.

Formula used:
Mean free path, $\lambda = \dfrac{1}{{\sqrt 2 \pi {d^2}n}}$
Here, d is the diameter of the particles and n is the number of particles per unit volume.

As the height of the artificial gauge is 300 km from the earth’s surface, the mean free path is under the influence of earth’s gravity. We have given that the volume is,
$V = \left( {1\,c{m^3}} \right){\left( {\dfrac{{{{10}^{ - 2}}\,m}}{{1\,cm}}} \right)^3}$
$\Rightarrow V = {10^{ - 6}}\,{{\text{m}}^3}$
Also, the diameter of the gas particles is, $d = 2 \times {10^{ - 10}}\,{\text{m}}$.
We have the expression for the mean free path of the particles,
$\lambda = \dfrac{1}{{\sqrt 2 \pi {d^2}n}}$ …… (1)
Here, d is the diameter of the particles and n is the number of particles per unit volume.

Let us find the number of particles per unit volume as follows,
$n = \dfrac{{1000 \times {{10}^6}}}{{{{10}^{ - 6}}}} = {10^{15}}$
Substituting $d = 2 \times {10^{ - 10}}\,{\text{m}}$ and $n = {10^{15}}$ in the above equation, we get,
$\lambda = \dfrac{1}{{\sqrt 2 \pi {{\left( {2 \times {{10}^{ - 10}}\,{\text{m}}} \right)}^2}\left( {{{10}^{15}}} \right)}}$
$\Rightarrow \lambda = \dfrac{1}{{1.78 \times {{10}^{ - 4}}}}$
$\Rightarrow \lambda = 5629\,{\text{m}}$
$\therefore \lambda = 5.6\,{\text{km}}$
Thus, the mean free path of the gas particles is 5.6 km.

So, the correct answer is option A.

Note: The mean free path of the molecules increases with the distance from the surface of the earth. At sea level, the mean free path is in micrometers. The mean free path also depends on the cross-sectional area of the molecules through the relation, $\lambda = \dfrac{1}{{nA}}$, where, A is the cross-sectional area. The number of molecules per unit volume can also be calculated using Avogadro’s number.