Question

What is the average de Broglie wavelength of oxygen molecules in air at a temperature of 27°C? Use the results of the kinetic theory of gases The mass of an oxygen molecule is 5.31 x 1026 kg

Answer 1

Answer:

$\lambda = 2.57 \times 10^{-11} m$

Explanation:

Average velocity of oxygen molecule at given temperature is

$v_{rms} = \sqrt{\frac{3RT}{M}}$

now we have

M = 32 g/mol = 0.032 kg/mol

T = 27 degree C = 300 K

now we have

$v_{rms} = \sqrt{\frac{3(8.31)(300)}{0.032}$

$v_{rms} = 483.4 m/s$

now for de Broglie wavelength we know that

$\lambda = \frac{h}{mv}$

$\lambda = \frac{6.6 \times 10^{-34}}{(5.31\times 10^{-26})(483.4)}$

$\lambda = 2.57 \times 10^{-11} m$