The de Broglie wavelength of an oxygen molecule at $27^{\circ} \mathrm{C}$ is $x imes 10^{-12} \ma

Question

The de Broglie wavelength of an oxygen molecule at $27^{\circ} \mathrm{C}$ is $x 	imes 10^{-12} \mathrm{~m}$. The value of $x$ is (take Planck's constant $=6.63 	imes 10^{-34} \mathrm{~J} . \mathrm{s}$, Boltzmann constant $=1.38 	imes 10^{-23} \mathrm{~J} / \mathrm{K}$, mass of oxygen molecule $=5.31 	imes 10^{-26} \mathrm{~kg}$ )

Professor PC Thomas Classes · Answer

The thermal kinetic energy of the molecule is $K = \frac{3}{2}kT$. Its momentum is $p = \sqrt{2mK} = \sqrt{3mkT}$. The de Broglie wavelength is $\lambda = \frac{h}{p} = \frac{h}{\sqrt{3mkT}}$. Substituting the given values: $\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{3 	imes 5.31 	imes 10^{-26} 	imes 1.38 	imes 10^{-23} 	imes 300}}$. The denominator evaluates to $\sqrt{900 	imes 7.3278 	imes 10^{-49}} = \sqrt{6.595 	imes 10^{-46}} \approx 2.568 	imes 10^{-23}$. Thus, $\lambda = \frac{6.63 	imes 10^{-34}}{2.568 	imes 10^{-23}} \approx 2.58 	imes 10^{-11} 	ext{ m} = 25.8 	imes 10^{-12} 	ext{ m} \approx 26 	imes 10^{-12} 	ext{ m}$. Thus, $x = 26$.