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

`\lambda = 2.57 * 10^(-11) m`

Step-by-step explanation:

Average velocity of oxygen molecule at given temperature is

`v_(rms) = \sqrt{(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{(3(8.31)(300))/(0.032)`

`v_(rms) = 483.4 m/s`

now for de Broglie wavelength we know that

`\lambda = (h)/(mv)`

`\lambda = (6.6 * 10^(-34))/((5.31* 10^(-26))(483.4))`

`\lambda = 2.57 * 10^(-11) m`