Proton Structure: De Broglie Wavelength & Resolution

[Federica]

Homework Statement

An accelerator produces an electron beam with energy E=20 GeV. The electrons diffused at θ=6° are detected. Neglecting their recoil motion, what is the minimum structure in the proton that can be resolved?

Relevant Equations

De Broglie wavelength, λ=h/p and particle energy in natural units $$E =\sqrt{m^2 + p^2} $$

Hi all. I'd personally consider the De Broglie wavelength λ=h/p, with p being the momentum of the electron beam. I get $$\lambda \simeq 0.6 \times 10^{-18} m $$ and since the radius of a quark is $\leq 10^{-19} m$, the proton structure can't be resolved. I'm quite sure there's something wrong with my reasoning, since I didn't use the information on θ.

 

[mfb]

The relevant momentum is not the total momentum of the electron, but the momentum transfer between electron and proton.

Quarks are probably elementary without a radius, but that's not what you want to resolve to look inside protons: If you see quarks you see a substructure already.

 

[Federica]

The relevant momentum is not the total momentum of the electron, but the momentum transfer between electron and proton.

Quarks are probably elementary without a radius, but that's not what you want to resolve to look inside protons: If you see quarks you see a substructure already.

thank you for your reply. But how can I find the momentum transfer between proton and electron if I have to neglect the recoil motion? I assume this means the energy of the proton is always equal to its mass (if c is equal to 1)

 

[mfb]

Think of the electron momentum before and after the collision.

 

[Federica]

I know how to do this as long as I can consider the recoil motion. Here's my reasoning: I call $E$ and $E'$ the energies of the incoming and outgoing electron respectively, while $p$ and $p'$ are their momenta. $E_{r}$ is the recoil motion, $m_e$ and $m_p$ are the masses of the electron and proton. I have:

$$ (E+m_p)^2 - p^2 = (E' + E_r)^2 - (\overrightarrow{p}' + \overrightarrow{p_r})^2 $$

which gives

$$ EM = E'E_r -p'p_rcos\theta = E'(E + M - E') - pp'cos\theta - p^2 $$

$$ E' = \frac{E}{1+\frac{E}{M}(1-cos\theta)} $$

and by $E -E'$ I can get the momentum transfer between electron and proton. But since I have to neglect the recoil motion, I get:

$$ (E+M)^2 - p^2 = (M+E')^2 - p'^2 $$

$$E = E'$$

which implies the momentum transfer is equal to zero. What's wrong with my reasoning?

 

[mfb]

If you assume that the proton doesn't get energy (but can get momentum) then the electron has the same energy before and after. From the change in flight direction you can calculate its change in momentum.

As the proton is heavy it won't get much energy from a change in momentum.