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 $$k = 1.38 \times 10^{-23}$$ J K$$^{-1}$$]

We have smoke particles of mass $$m = 5 \times 10^{-17}$$ kg undergoing Brownian motion in air at NTP. At NTP the temperature is $$T = 273$$ K. The Boltzmann constant is $$k = 1.38 \times 10^{-23}$$ J K$$^{-1}$$.

The root mean square speed from kinetic theory is given by $$v_{\text{rms}} = \sqrt{\frac{3kT}{m}}$$.

Substituting the values: $$v_{\text{rms}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 273}{5 \times 10^{-17}}}$$

We first compute the numerator: $$3 \times 1.38 \times 273 = 3 \times 376.74 = 1130.22$$. So the numerator is $$1130.22 \times 10^{-23}$$.

Now $$\frac{1130.22 \times 10^{-23}}{5 \times 10^{-17}} = \frac{1130.22}{5} \times 10^{-23+17} = 226.04 \times 10^{-6} = 2.26 \times 10^{-4}$$.

Therefore $$v_{\text{rms}} = \sqrt{2.26 \times 10^{-4}} = \sqrt{2.26} \times 10^{-2} \approx 1.503 \times 10^{-2}$$ m/s $$\approx 15$$ mm s$$^{-1}$$.

Hence, the correct answer is Option 3.