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Home » What is the de Broglie wavelength of a nitrogen molecule in air at Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen )
What is the de Broglie wavelength of a nitrogen molecule in air at 300 \mathrm{~K} ? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen =14.0076 \mathrm{u} )
CBSE | Class 12 | Dual Nature of Radiation and Matter | NCERT | Physics
What is the de Broglie wavelength of a nitrogen molecule in air at 300 \mathrm{~K} ? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen =14.0076 \mathrm{u} )

Temperature of the nitrogen molecule is given as \mathbf{T}=\mathbf{3 0 0} \mathbf{K}

Atomic mass of nitrogen is 14.0076 \mathbf{u}

Hence, mass of the nitrogen molecule will be,
\mathrm{m}=2 \times 14.0076=\mathbf{2 8 . 0 1 5 2} \mathbf{u}

But, 1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{~kg}

Planck’s constant, h=6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}

Boltzmann constant, k=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}

The expression that relates mean kinetic energy \left(\frac{3}{2} k T\right) of the nitrogen molecule with the root mean is:

\frac{1}{2} m v_{r m s}^{2}=\left(\frac{3}{2} k T\right)

\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{kT}}{\mathrm{m}}}

\lambda=\frac{h}{m v_{r m s}}=\frac{h}{\sqrt{3 m k T}}

=\frac{6.63 \times 10^{-34}}{\sqrt{3 \times 28.0152 \times 1.66 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300}}

=0.028 \times 10^{-9} \mathrm{~m}=\mathbf{0 . 0 2 8} \mathrm{nm}

Therefore, 0.028 \mathrm{~nm} is the de Broglie wavelength of the nitrogen molecule.

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