A-level Physics (Advancing Physics)/Boltzmann Factor/Worked Solutions

1u = 1.66 x 10−27 kg

g = 9.81 ms−2

'''1. A nitrogen molecule has a molecular mass of 28u. If the Earth's atmosphere is 100% nitrous, with a temperature of 18 °C, what proportion of nitrogen molecules reach a height of 2 km?'''

$$\epsilon = mgh = 28 \times 1.66 \times 10^{-27} \times 9.81 \times 2000 = 9.12 \times 10^{-22}\mbox{ J}$$

$$\frac{n}{n_0} = e^{\frac{-9.12 \times 10^{-22}}{1.38 \times 10^{-23} \times 291}} = 0.797$$

So, 79.7% reach a height of at least 2 km.

2. What proportion of the molecules in a box of hydrogen (molecular mass 2u) at 0 °C have a velocity greater than 5ms−1?

$$\epsilon = \frac{1}{2}mv^2 = 0.5 \times 2 \times 1.66 \times 10^{-27} \times 5^2 = 4.15 \times 10^{-26}\mbox{ J}$$

$$\frac{n}{n_0} = e^{\frac{-4.15 \times 10^{-26}}{1.38 \times 10^{-23} \times 273}} = 0.99999$$

(Practically all of them.)

3. What is the temperature of the hydrogen if half of the hydrogen is moving at at least 10ms−1?

$$\epsilon = \frac{1}{2}mv^2 = 0.5 \times 2 \times 1.66 \times 10^{-27} \times 10^2 = 1.66 \times 10^{-25}\mbox{ J}$$

$$0.5 = e^{\frac{-\epsilon}{kT}}$$

$$\ln{0.5} = -\ln{2} = \frac{-\epsilon}{kT}$$

$$T = \frac{\epsilon}{k\ln{2}} = \frac{1.66 \times 10^{-25}}{1.38 \times 10^{-23} \times \ln{2}} = 0.0174\mbox{ K}$$

(almost absolute zero)

'''4. Some ionised hydrogen (charge -1.6 x 10−19 C)is placed in a uniform electric field. The potential difference between the two plates is 20V, and they are 1m apart. What proportion of the molecules are at least 0.5m from the positive plate (ignoring gravity) at 350°K?'''

$$\epsilon = 0.5 \times 20 \times 1.6 \times 10^{-19} = 1.6 \times 10^{-18}\mbox{ J}$$

$$\frac{n}{n_0} = e^{\frac{-1.6 \times 10^{-18}}{1.38 \times 10^{-23} \times 350}} \approx 0$$

They would all fly onto the positive plate.