Question

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

Answer 1

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 \(28.02 * 10^(-3) \, \text{kg/mol}\).`

Now, calculate
`\(v_{\text{rms}}\)`, and then use it to find the momentum
`\(p\)` and finally the de Broglie wavelength
`\(\lambda\).`