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)
Given: The temperature is 300 K and atomic mass of nitrogen is 14.0076 u .
The root mean square speed is given as,
υ r m s = 3 k T m (1)
The de Broglie wavelength of the nitrogen molecule is given as,
λ = h m υ r m s (2)
By substituting the value of root mean square speed in the equation (2), we get
λ = h 3 m k T
Where, the de Broglie wavelength is λ , the atomic mass of nitrogen is m , Boltzmann constant is k , Planck’s constant is h and the temperature is T .
By substituting the given values in above formula, we get
λ = 6.63 × 10 − 34 3 × 28.0152 × 1.66 × 10 − 27 × 1.38 × 10 − 23 × 300 = 0.028 × 10 − 9 m = 0.028 nm
Thus, the de Broglie wavelength of nitrogen molecule is 0.028 nm .
Given: The temperature is
The root mean square speed is given as,
The de Broglie wavelength of the nitrogen molecule is given as,
By substituting the value of root mean square speed in the equation (2), we get
Where, the de Broglie wavelength is
By substituting the given values in above formula, we get
Thus, the de Broglie wavelength of nitrogen molecule is
