Oxygen Molecule de Broglie Wavelength Calculation

This solution determines the de Broglie wavelength ($\lambda$) for an oxygen molecule at $27^\circ\text{C}$, relating it to thermal energy.

Temperature Conversion

Convert the temperature from Celsius to Kelvin:

$ T(\text{K}) = T(^\circ\text{C}) + 273.15 $

Given $ T = 27^\circ\text{C} $:

$ T = 27 + 273.15 = 300.15 \text{ K} $

We use $ T = 300 \text{ K} $ for calculation, which is standard for this temperature.

De Broglie Wavelength Formula

The de Broglie wavelength associated with the thermal motion of a molecule is given by:

$ \lambda = \frac{h}{\sqrt{3mk_BT}} $

Where:

• $ h $ = Planck's constant = $ 6.63 \times 10^{-34}\text{ J.s} $
• $ m $ = mass of oxygen molecule = $ 5.31 \times 10^{-26}\text{ kg} $
• $ k_B $ = Boltzmann constant = $ 1.38 \times 10^{-23}\text{ J/K} $
• $ T $ = Absolute temperature = $ 300 \text{ K} $

Calculation Steps

1. Calculate the term $ 3mk_BT $:

   $ 3mk_BT = 3 \times (5.31 \times 10^{-26}\text{ kg}) \times (1.38 \times 10^{-23}\text{ J/K}) \times (300 \text{ K}) $

   $ 3mk_BT \approx 6.586 \times 10^{-46} \text{ kg}^2\text{m}^2/\text{s}^2 $

   (Note: $ \text{J} = \text{kg} \cdot \text{m}^2/\text{s}^2 $)

2. Calculate the square root: $ \sqrt{3mk_BT} $.

   $ \sqrt{3mk_BT} \approx \sqrt{6.586 \times 10^{-46}} \text{ kg.m/s} $

   $ \sqrt{3mk_BT} \approx 2.566 \times 10^{-23} \text{ kg.m/s} $

3. Calculate the de Broglie wavelength $ \lambda $.

   $ \lambda = \frac{h}{\sqrt{3mk_BT}} = \frac{6.63 \times 10^{-34}\text{ J.s}}{2.566 \times 10^{-23}\text{ kg.m/s}} $

   $ \lambda \approx 2.584 \times 10^{-11} \text{ m} $

4. Determine the value of $ x $. The wavelength is given in the format $ x \times 10^{-12}\text{ m} $.

   $ \lambda = 2.584 \times 10^{-11} \text{ m} = 25.84 \times 10^{-12} \text{ m} $

   Thus, $ x \approx 25.84 $.

Final Answer Determination

The calculated value $ x \approx 25.84 $ is closest to option 3 ($ 26 $).