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

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Oct 15, 2017

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 $\lambda \to 0$, 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 $\lambda \to 0$, discrete energies converge to become a continuum of energy:

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

$\lambda = \frac{h}{m v} = \frac{h}{p}$

where:

• $m$ is the mass of the particle in $\text{kg}$.
• $v$ is its speed in $\text{m/s}$.
• $h = 6.626 \times {10}^{- 34} \text{J"cdot"s}$ is Planck's constant.
• $p$ is the momentum of the particle in $\text{kg"cdot"m/s}$.
• $\lambda$ is the wavelength in $\text{m}$.

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

$K = \frac{1}{2} m {v}^{2} = {p}^{2} / \left(2 m\right)$

And so, $p = \sqrt{2 m K}$ is the forward momentum. Therefore, we can rewrite $\lambda$ in terms of $K$:

$\lambda = \frac{h}{\sqrt{2 m K}}$

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

${\left\langleK\right\rangle}_{t r} = \frac{K}{N} = \frac{3}{2} n R T$

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 ${27}^{\circ} \text{C}$ and ${927}^{\circ} \text{C}$,

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

${\left\langleK\right\rangle}_{t r} = \frac{K}{1} = \frac{3}{2} n R T$.

We then have

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

$= 1.6605 \times {10}^{- 24} \text{mols}$ of particles.

This can then be related back to the wavelength. Consider the wavelength at ${27}^{\circ} \text{C" = "300.15 K}$:

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

And now, compare it to the wavelength at $\text{1200.15 K}$:

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:

$\textcolor{b l u e}{\lambda '} \approx \frac{h}{\sqrt{3 m n R \cdot \text{300.15 K} \cdot 4}}$

$\approx \frac{1}{2} \frac{h}{\sqrt{3 m n R \cdot \text{300.15 K}}}$

$\approx \textcolor{b l u e}{\frac{\lambda}{2}}$

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