Question

The root mean square speed of smoke particles of mass \( 5 \times 10^{-17} \) kg in their Brownian motion in air at NTP is approximately. (Given \( k = 1.38 \times 10^{-23} \, {JK}^{-1} \))

• A. \( 60 \, {mm/s} \)
• B. \( 12 \, {mm/s} \)
• C. \( 15 \, {mm/s} \)
• D. \( 36 \, {mm/s} \)

Hint:

In Brownian motion, the root mean square speed depends on temperature and particle mass. The formula \( v_{{rms}} = \sqrt{\frac{3kT}{m}} \) is essential for solving such problems.

Answer 1

The Correct Option is C

Step 1: The root mean square speed (\( v_{{rms}} \)) of particles in Brownian motion is given by the equation: \[ v_{{rms}} = \sqrt{\frac{3kT}{m}} \] where: - \( k = 1.38 \times 10^{-23} \, {JK}^{-1} \) is the Boltzmann constant, - \( T \) is the temperature in Kelvin, - \( m = 5 \times 10^{-17} \, {kg} \) is the mass of the smoke particle. Step 2: Since the temperature at NTP (Normal Temperature and Pressure) is \( T = 273 \, {K} \), we can substitute these values into the equation. \[ v_{{rms}} = \sqrt{\frac{3 \times (1.38 \times 10^{-23}) \times 273}{5 \times 10^{-17}}} \] Step 3: Simplify the expression: \[ v_{{rms}} = \sqrt{\frac{1.242 \times 10^{-20}}{5 \times 10^{-17}}} \] \[ v_{{rms}} = \sqrt{2.484 \times 10^{-4}} = 0.0157 \, {m/s} \] Converting to millimeters per second: \[ v_{{rms}} = 15.7 \, {mm/s} \approx 15 \, {mm/s}. \] Thus, the root mean square speed is approximately \( 15 \, {mm/s} \).