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

1 Answer
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 -> 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:

http://hyperphysics.phy-astr.gsu.edu/http://hyperphysics.phy-astr.gsu.edu/


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)