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
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
#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)#