What is the de-Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u).

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Solution

Temperature, T = 300 K
Atomic mass of nitrogen = 14.0076 u.

∴

Mass of nitrogen molecule, m = 2 \times 14.0076 u

= 2 \times 14.0076 \times 1.6606 \times 10^{- 27} kg = 46.52 \times 10^{- 27} kg

Now, de-Broglie wavelength is given by,

λ = \frac{h}{mv} = \frac{h}{p}

\Rightarrow

λ = \frac{h}{\sqrt{2 {mE}_{k}}} = \frac{h}{\sqrt{2 m (\frac{3}{2} k_{B} T)}} = \frac{h}{\sqrt{3 {mk}_{B} T}}

(∵ E_{k} = \frac{3}{2} k_{g} J)

(where k_B is Boltzmann constant)

\Rightarrow

λ = \frac{6.63 \times 10^{- 34}}{\sqrt{3 \times 46.52 \times 10^{- 27} \times 1.38 \times 10^{- 23} \times 300}}

= \frac{6.63}{240.37} \times 10^{- 9} m = 0.0276 nm

= 0.276 Å

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