DeBroglie Wavelength: Proton at Room Temp Scattering Behavior

[edpell]

A proton at room temperature with an energy of 0.025eV has a deBroglie wavelength of about 1A (1E-10 meters). If we shoot two proton beams at each other with is low energy and large wavelength what happens? Do they scatter as if they are small hard particles of size about 1 fermi (1E-15 meters) or do they scatter as if they are big fuzzballs of size about 1A?

 

[jtbell]

It might be instructive to consider the two protons as classical particles and find the distance of closest approach: the distance at which the Coulomb potential energy equals their total initial kinetic energy.

 

[edpell]

About 2A. So if we say they are each 0.7A they never interpenetrate.

Is this correct? How do we think about the distribution of a room temperature proton? Is it 100% within 1 deBroglie wavelength? Is there a tail to the distribution so we might see an effect in the scattering?

 

[jtbell]

Check your math. I get r = about 290 Å for a potential energy of 0.05 eV (8 x 10^-21 J)

$$PE = \frac{q^2}{4 \pi \epsilon_0 r}$$

 

[edpell]

I think it is

$$PE = \frac{q}{4 \pi \epsilon_0 r}$$

 

[jtbell]

The electric potential V (in volts, in MKS units) at a distance r from a charge q is given by your formula.

When you put a second charge q₂ at that location, the electric potential energy of the system (in joules, in MKS units) is given by PE = q₂V. When the two charges are equal in magnitude (q₂ = q) this leads to my formula.

Unfortunately, in classical electromagnetism, we usually use V to refer to potential (volts), whereas in QM we often use V to refer to potential energy (joules or electron-volts), which causes confusion.