Question

The de - Broglie wavelength of a neutron at `27^@C` is `lambda`. What will be its wavelength at `927^@C` ??

Answer 1

It would be approximately half.

This physically says that at higher temperatures, the particle-wave moves faster. A faster particle-wave has a larger linear momentum, which means it has a larger average kinetic energy.

Therefore, since temperature is directly proportional to average kinetic energy of linear motion, the higher the temperature, the smaller the wavelength of the particle-wave.

This makes sense because as `lambda -> 0`, we reach the classical limit, where the particle-wave behaves less like a wave and more like a particle. This is also known as the high-temperature limit.

This agrees with the correspondence principle, that as `lambda->0`, discrete energies converge to become a continuum of energy:

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We first consider that the de Broglie wavelength is given by:

`lambda = h/(mv) = h/p`

where:

• `m` is the mass of the particle in `"kg"`.
• `v` is its speed in `"m/s"`.
• `h = 6.626 xx 10^(-34) "J"cdot"s"` is Planck's constant.
• `p` is the momentum of the particle in `"kg"cdot"m/s"`.
• `lambda` is the wavelength in `"m"`.

Now, it is important to recognize that the kinetic energy of a particle with mass (so not photons) is given by:

`K = 1/2 mv^2 = p^2/(2m)`

And so, `p = sqrt(2mK)` is the forward momentum. Therefore, we can rewrite `lambda` in terms of `K`:

`lambda = h/sqrt(2mK)`

Next, consider that the translational average kinetic energy is given by the equipartition theorem in the classical limit:

`<< K >>_(tr) = K/N = 3/2 nRT`

where `N` is the number of particles in an ensemble, and `n`, `R`, and `T` are known from the ideal gas law. The `3` indicates the three dimensions in Cartesian coordinates, `x,y,z`.

Assuming that

• A neutron has only translational motion,
• We are at a suitably high temperature that the "classical limit" applies at both `27^@ "C"` and `927^@ "C"`,

then for a single neutron, `N = 1`, so that

`<< K >>_(tr) = K/1 = 3/2 nRT`.

We then have

`n = "1 particle" xx "1 mol"/(6.022 xx 10^23 "particles")`

`= 1.6605 xx 10^(-24) "mols"` of particles.

This can then be related back to the wavelength. Consider the wavelength at `27^@ "C" = "300.15 K"`:

`lambda_"300.15 K" = h/sqrt(3mnR cdot "300.15 K")`

And now, compare it to the wavelength at `"1200.15 K"`:

`lambda_"1200.15 K" = h/sqrt(3mnR cdot "1200.15 K")`

If we then represent the room-temperature wavelength as `lambda` and the high-temperature wavelength as `lambda'`, then the high-temperature wavelength is about half of the room-temperature wavelength:

`color(blue)(lambda') ~~ h/sqrt(3mnR cdot "300.15 K" cdot 4)`

`~~ 1/2 h/sqrt(3mnR cdot "300.15 K")`

`~~ color(blue)(lambda/2)`