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
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,
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, andn ,R , andT are known from the ideal gas law. The3 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" and927^@ "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)