Assuming the nitrogen milecule is moving with rms velcoty at 400k the debroglie wavelft of nitrogen is close to

Question

asked Jul 6, 2024 96.1k views

1 Answer

Leo Fabrikant · Answer 1 · 2024-07-12T03:14:48+0000

Final answer:

The de Broglie wavelength of a nitrogen molecule can be determined using its rms velocity and the de Broglie relation, which involves Planck's constant and the particle's momentum.

Step-by-step explanation:

The de Broglie wavelength
$($\lambda$)$ of a particle is given by the de Broglie wavelength formula:

$\lambda = (h)/(p)$

where:

$\lambda$ is the de Broglie wavelength,

$h$ is Planck's constant
$($6.626 * 10^(-34) \, \text{m}^2 \cdot \text{kg} / \text{s}$),$

$p$ is the momentum of the particle.

The momentum
$($p$)$ of a particle is given by:

$p = m \cdot v$

where:

$m$ is the mass of the particle,

$v$ is the velocity of the particle.

For a nitrogen molecule
$($N_2$)$ , the mass
$($m$)$ is the total mass of the molecule. The root mean square velocity
$($v_{\text{rms}}$)$ is related to the velocity of the particles in a gas, and it is given by:

$v_{\text{rms}} = \sqrt{(3 \cdot k \cdot T)/(m)}$

where:

$k$ is the Boltzmann constant
$($1.381 * 10^(-23) \, \text{J/K}$),$

$T$ is the temperature in kelvin.

Let's calculate it for nitrogen
$($N_2$)$ at a temperature of 400 K.

First, we find
$v_{\text{rms}}$ :

$v_{\text{rms}} = \sqrt{(3 \cdot k \cdot T)/(m)}$

Substitute the values:

$v_{\text{rms}} = \sqrt{\frac{3 \cdot (1.381 * 10^(-23) \, \text{J/K}) \cdot 400 \, \text{K}}{m_(N_2)}}$

The molar mass of
$N_2$ is approximately
$28.02 \, \text{g/mol}$ , so the mass of one nitrogen molecule
$($m_(N_2)$) is $28.02 * 10^(-3) \, \text{kg/mol}$.$