Calculate the rms speed of smoke particles of mass `5 xx 10^(-17) kg` in their Brownian motion in air at NTP. Given `k_(B) = 1.38 xx 10^(-23) J//K`

Explanation

To calculate the root mean square (rms) speed of smoke particles in Brownian motion, we can use the formula:

\[ v_{rms} = \sqrt{\frac{3k_B T}{m}} \]

where:
- \( v_{rms} \) is the root mean square speed,
- \( k_B \) is the Boltzmann constant,
- \( T \) is the absolute temperature in Kelvin,
- \( m \) is the mass of the particle.

### Step-by-Step Solution:

1. **Identify the given values**:
- Mass of smoke particles, \( m = 5 \times 10^{-17} \) kg
- Temperature at NTP, \( T = 273 \) K
- Boltzmann constant, \( k_B = 1.38 \times 10^{-23} \) J/K

2. **Substitute the values into the formula**:
\[ v_{rms} = \sqrt{\frac{3 \times (1.38 \times 10^{-23}) \times 273}{5 \times 10^{-17}}} \]

3. **Calculate the numerator**:
- First, calculate \( 3 \times 1.38 \times 10^{-23} \):
\[ 3 \times 1.38 \times 10^{-23} = 4.14 \times 10^{-23} \text{ J/K} \]
- Now multiply this by \( 273 \):
\[ 4.14 \times 10^{-23} \times 273 = 1.13 \times 10^{-20} \text{ J} \]

4. **Calculate the denominator**:
- The denominator is simply \( 5 \times 10^{-17} \) kg.

5. **Divide the numerator by the denominator**:
\[ \frac{1.13 \times 10^{-20}}{5 \times 10^{-17}} = 2.26 \times 10^{-4} \text{ m}^2/\text{s}^2 \]

6. **Take the square root**:
\[ v_{rms} = \sqrt{2.26 \times 10^{-4}} \approx 1.5 \times 10^{-2} \text{ m/s} \]

7. **Convert to centimeters per second**:
\[ v_{rms} \approx 1.5 \text{ cm/s} \]

### Final Answer:
The root mean square speed of the smoke particles is approximately \( 1.5 \) cm/s.