# De Broglie wavelength of helium atoms

1. Oct 1, 2014

### leehufford

1. The problem statement, all variables and given/known data
a) In the double slit interference pattern for helium atoms, the kinetic energy of a beam of atoms is 0.020 eV. What is the de Broglie wavelength of a helium atom with this kinetic energy? b) Also, estimate the de Broglie wavelength of the atoms from the fringe spacing (about 10 micrometers) and compare to the wavelength from (a). The distance from the double slit to the scanning slit is 64 cm. Slit spacing is 8 micrometers.

2. Relevant equations
This is my first post since the new physics forums formatting, and I can't find the symbols, so I suppose I'll do this with words:

wavelength = plank's constant/momentum = plank's constant/sqrt(2mK).

3. The attempt at a solution

So I used 4.136e-21 MeV*s for plank's constant, then for mass I used [2(938.3 MeV/c^2) + 2(0.511) MeV/c^2) + 2(939.6) MeV/C^2)]. (Two protons, two neutrons and two electrons). I get 3.374e-19 m. This wavelength seems way too small, but my real problem is that the wavelength I get from the diffraction grating formula is very different than the wavelength here.

For b, I use 8 micrometers for the slit spacing (given). The formula is wavelength = d*y/D, where d is the slit spacing, y is the fringe maximum distance and D is the space between slit and detector, so wavelength = (8 micrometers)(10 micrometers)/(6.4e5 micrometers) = 1.25e-4 micrometers = 1.25e-10 meters, which is very, very far off from the wavelength from part a. Obviously there is a fundamental misunderstanding going on here, it's sort of driving me crazy. Any help would be much appreciated. Thanks!

-Lee

2. Oct 1, 2014

### leehufford

I've been looking at this problem for hours... I cannot tell which wavelength is wrong. I feel like the smaller wavelength is probably right, but I'm not sure where I made my mistake with the larger wavelength. If my original post is unclear, feel free to ask for clarification.

Lee

3. Oct 2, 2014

### unscientific

Your units are wrong for part (a):

$$\lambda = \frac{h}{\sqrt {2 m E} }$$

Using units you gave:

$$\frac{eV*s}{\sqrt{\frac{eV}{c^2} * eV } } = s*c^2$$

I propose using SI units for everything (especially mass of helium). It's fine to ignore the mass of the electrons (why?).