Answer:
[tex]\lambda = 2.57 \times 10^{-11} m[/tex]
Explanation:
Average velocity of oxygen molecule at given temperature is
[tex]v_{rms} = \sqrt{\frac{3RT}{M}}[/tex]
now we have
M = 32 g/mol = 0.032 kg/mol
T = 27 degree C = 300 K
now we have
[tex]v_{rms} = \sqrt{\frac{3(8.31)(300)}{0.032}[/tex]
[tex]v_{rms} = 483.4 m/s[/tex]
now for de Broglie wavelength we know that
[tex]\lambda = \frac{h}{mv}[/tex]
[tex]\lambda = \frac{6.6 \times 10^{-34}}{(5.31\times 10^{-26})(483.4)}[/tex]
[tex]\lambda = 2.57 \times 10^{-11} m[/tex]
