The de Broglie wavelength of an oxygen molecule at $27^{\circ}C$ is $x imes 10^{- 12} ext{ }m$. T

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

Accepted Answer

$\lambda = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2\text{ }m\left( \frac{3}{2}kT \right)}}$

$${\lambda = \frac{h}{\sqrt{3mkT}} }{= \frac{6.63 \times 10^{- 34}}{\sqrt{3 \times 5.31 \times 10^{- 26} \times 1.38 \times 10^{- 23} \times 300}} }{= 2.58 \times 10^{- 11} = 25.8 \times 10^{- 12} }$$So, $x = 26$

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