The rms velocity of nitrogen molecule in a cylinder containing nitrogen gas at 170C is 5×1012 m/s. If the diameter of the nitrogen molecule is roughly 2.0A0 and number of molecules per unit volume is 2.56×1025 then the time taken for the collision between two nitrogen molecules

t=λVrms=1Vrms12πd2n⇒t=15×1012×11.41×3.14×2×10−102×2.56×1025t=4.4×10−20sec

The rms velocity of nitrogen molecule in a cylinder containing nitrogen gas at 170C is 5×1012 m/s. If the diameter of the nitrogen molecule is roughly 2.0A0 and number of molecules per unit volume is 2.56×1025 then the time taken for the collision between two nitrogen molecules

t=λVrms=1Vrms12πd2n⇒t=15×1012×11.41×3.14×2×10−102×2.56×1025t=4.4×10−20sec

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The rms velocity of nitrogen molecule in a cylinder containing nitrogen gas at 17⁰C is 5×10¹² m/s. If the diameter of the nitrogen molecule is roughly 2.0A⁰ and number of molecules per unit volume is 2.56× $10^{25}$ then the time taken for the collision between two nitrogen molecules

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$2 \times 10^{- 13} S$

$4.4 \times 10^{- 20} S$

$3 \times 10^{- 13} S$

$1 \times 10^{- 13} S$

answer is B.

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Detailed Solution

$t = \frac{λ}{V_{rms}} = \frac{1}{V_{rms}} \frac{1}{\sqrt{2} {πd}^{2} n}$

$\Rightarrow t = \frac{1}{5 \times 10^{12}} \times \frac{1}{1.41 \times 3.14 \times {(2 \times 10^{- 10})}^{2} \times 2.56 \times 10^{25}}$

$t = 4.4 \times 10^{- 20} \sec$

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The rms velocity of nitrogen molecule in a cylinder containing nitrogen gas at 170C is 5×1012 m/s. If the diameter of the nitrogen molecule is roughly 2.0A0 and number of molecules per unit volume is 2.56×1025 then the time taken for the collision between two nitrogen molecules

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The rms velocity of nitrogen molecule in a cylinder containing nitrogen gas at 17⁰C is 5×10¹² m/s. If the diameter of the nitrogen molecule is roughly 2.0A⁰ and number of molecules per unit volume is 2.56× $10^{25}$ then the time taken for the collision between two nitrogen molecules