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 -> 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/


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)#

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