# Determine how many times the root-mean-square velocity of a dust grain with a mass of m = 1.75.10-12 kg

**Determine how many times the root-mean-square velocity of a dust grain with a mass of m = 1.75.10-12 kg suspended in air is less than the mean square-law velocity of motion of nitrogen molecules?**

Data: m1 (dust grain mass) = 1.75 * 10 ^ -12 kg.

Constants: M (molar mass of nitrogen) = 28 g / mol (0.028 kg / mol); Na (Avogadro’s number) = 6.02 * 10 ^ 23 mol-1.

The ratio of the root mean square velocities of a grain of dust and nitrogen molecules will have the form: Ek1 / Ek2 = 0.5 * m1 * V12 / (0.5 * m2 * V2 ^ 2) = m1 * V12 / (m2 * V2 ^ 2), whence V2 ^ 2 / V1 ^ 2 = m1 / m2 and V2 / V1 = √ (m1 / m2) = √ (m1 * Na / M).

Let’s calculate: V2 / V1 = √ (1.75 * 10 ^ -12 * 6.02 * 10 ^ 23 / 0.028) = 6.13 * 10 ^ 6 times.

Answer: Less than 6.13 * 10 ^ 6 times.