Question

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)

Answer 1

Answer:

use the formula ∆x × ∆m∆V=nh/4π

Answer 2

Answer:

The de Broglie wavelength of a nitrogen molecule in air = 0.028 nm.

Explanation:

As per the question,

Temperature of the the nitrogen molecule = 300 K

Atomic mass of nitrogen = 14.0076 u

Therefore,

Molecular mass of nitrogen molecule, N₂ = 2 × 14.0076  = 28.0152 u

According to relation of root mean square velocity, we have

$V_{rms}=\sqrt{\frac{3kT}{m}}$

V(rms) = root mean square velocity

k = Boltzmann constant = 1.38 × 10⁻²³ J K⁻¹

And

De Broglie wavelength is given by:

$\lambda=\frac{h}{mv_{rms}}$

Where,

h = Planck’s constant = 6.63 × 10⁻³⁴ Js

After putting the value of v(rms) in this equation, we get

$\lambda=\frac{h}{\sqrt{3mkT}}$

Put the value of given data in this equation, we get

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

λ = 0.028 nm

Hence, the de Broglie wavelength of a nitrogen molecule in air = 0.028 nm.