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JEE Mains · Physics · STD 11 - 12 . kinetic theory of gases

\(5 \times 10^{-10} m\) વ્યાસ ધરાવતા અણુનો \(41^{\circ} C\) તાપમાન અને \(1.38 \times 10^5 Pa\) દબાણે, સરેરાશ મુક્તપથ _________ m થશે. ( \(k_B=1.38 \times 10^{-23} J / K\) આપેલ છે)

  1. A \(2\sqrt{2}\times10^{-10}\)
  2. B \(10\sqrt{2}\times10^{-8}\)
  3. C \(2\sqrt{2}\times10^{-8}\)
  4. D \(2\times10^{-8}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(2\sqrt{2}\times10^{-8}\)

Step-by-step Solution

Detailed explanation

\(\lambda=\frac{k_{B}T}{\sqrt{2}\pi\sigma^{2}P}\) \(=\frac{1.38\times10^{-23}\times(273+41)\times100}{\sqrt{2}\times3.14\times(5\times10^{-10})^{2}\times1.38\times10^{5}}=2\sqrt{2}\times10^{-8}\)
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