## de Broglie wavelength

1. The problem statement, all variables and given/known data
What is the de Broglie wavelength of a neutron whose kinetic energy is equal to the average kinetic energy of a gas of neutrons at temperature T = 17 K?

2. Relevant equations
lambda = h/p = h/mv.

3. The attempt at a solution

Well first I tried taking the h = 6.63E-34 Jxs and dividing it by the mass times the square root of T = 17 K

Thus,

(6.63E-34)/(1.6E-27 kg)(sqrt(17)) x (1E9 m) = 100.500 nm.

I did this because we know from kinetic theory that T is proportional to the root mean square velocity v in the gas, so that v scales as sqrt(T). This means that the momentum scales in the same way, while the wavelength as 1/sqrt(T).

I'm not sure where to go from here, since the answer is not correct. =/
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

 Quote by Physics321 [b] (6.63E-34)/(1.6E-27 kg)(sqrt(17)) x (1E9 m) = 100.500 nm. I did this because we know from kinetic theory that T is proportional to the root mean square velocity v in the gas, so that v scales as sqrt(T). This means that the momentum scales in the same way, while the wavelength as 1/sqrt(T). 3. The attempt at a solution
If you look at the units alone you can see that the formula has some problems.
What are the units for sqrt(17K)? (I suppose you mean 17K)
It is true that v is proportional with sqrt(T) but it is not equal to this.
You need to take the complete formula for v (or p) from the kinetic theory of gases to get the right units (and the right answer, maybe).

You can start with KE= 3/2 kB*T where KE is the average kinetic energy.

 Tags de broglie