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

##### 1 Answer

**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 **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

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,

#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,

#<< 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

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

And now, compare it to the wavelength at

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

If we then represent the room-temperature wavelength as **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)#