**The root mean square speed of smoke particles of mass 5 * 10^-17kg in their Brownian motion in air at NTP is approximately. [Given k = 1.38 * 10^-23 JK^-1]**

Option 1 - 60 mm s-1

Option 2 - 12 mm s-1

Option 3 - 15 mm s-1

Option 4 - 36 mm s-1

2 Views|Posted 6 months ago

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Answered by

Alok Kumar Singh | Contributor-Level 10

6 months ago

Correct Option - 3

Detailed Solution:

m = 5 * 10^-17kg ; k = 1.38 * 10^-23 Jk^-1

$V_{r m s} = \sqrt[]{\frac{3 R T}{M}} = \sqrt[]{\frac{3 N k T}{N m}} = \sqrt[]{\frac{3 k T}{m}} = \sqrt[]{\frac{3 * 1.38 * 1 0^{- 23} * 300}{5 * 1 0^{- 17}}}$

$= \sqrt[]{18 * 13.8 * 1 0^{- 6}}$

$= 15.7 * 1 0^{- 3} m / s$

$= 15 m m / s$

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The root mean square speed of smoke particles of mass 5 * 10-17kg in their Brownian motion in air at NTP is approximately. [Given k = 1.38 * 10-23 JK-1]

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