- #1
Alpha&Omega
- 20
- 0
Protons and neutrons have a diameter of around [tex]10^{-15}[/tex]m. Despite this small size, physicists were able first to probe inside these particles in the 1970s, by scattering high energy electrons off them.
i). explain briefly why it is important that the electrons have high energy.
Since momentum is conserved in a collision the resulting particles will also have a high energy. This makes them easier to be detected by a detector.
ii). Assuming that the protons and neutrons were at rest, calculate the minimum momentum to which the electrons should be accelerated, in order to perform such experiments successfully
[tex]\lambda=\frac{h}{\rho} \Rightarrow \ \rho=\frac{h}{\lambda}=\frac{h}{10^{-15}}=6.63 \times 10^{-19}Jsm^{-1}[/tex]
iii). Considering such scattering collisions in the laboratory frame of reference (where the target is stationary), write formulae for the energies of the electrons and target, and evaluate then in Gev (you may ignore the rest mass of the electron)
I found this part of the question pretty difficult. Here's my attempt:
Initial state: 0---> O
I have an electron (in the "picture" above as "0") and a proton ("O"). The electron has an energy [tex]E_1[/tex] and a momentum [tex]\underline{\rho}[/tex]. I wrote this as [tex](E_1,\underline{\rho})[/tex].
Using the equation [tex]E=\rho c[/tex], [tex]E_1=\underline{\rho}c[/tex]
Final state: <-----0 O----->
Using the (energy, momentum) co-ordinates, the electron has [tex](E_2, -\underline{\rho})[/tex] and the proton has [tex](E_3, \underline{\rho})[/tex].
The initial energy I calculated using [tex]E=\sqrt{(\rho c)^2+(m_0 c^2)^2}[/tex]
[tex]\sqrt{(\frac{6.63 \times 10^{-19} c}{1.6 \times 10^{-10}})^2+(0.938)^2}[/tex]
(the mass of a proton is [tex]0.938\frac{Gev}{c^2}[/tex])
This gives E=1.557 GeV.
Hence I found the initial energy of the proton to be 0.938Gev and the energy of the electron to be 1.557-0.938=0.619Gev.
I'm unsure whether this is correct. I may have completely misunderstood the entire situation!
iv). The total energy in the centre-of-mass reference frame in such electron-nucleon collisions corresponds to the maximum possible mass of all particles in the final state and is a Lorentz-invariant quantity. Ignoring the rest mass of the electron, calculate this quality
The "maximum possible mass" suggests that all the particles are stationary. Hence I should be calcuating their rest mass.
However, since I only have a proton and an electron at the end would the answer by just the rest mass of a proton- [tex]0.938\frac{GeV}{c^2}[/tex]?
i). explain briefly why it is important that the electrons have high energy.
Since momentum is conserved in a collision the resulting particles will also have a high energy. This makes them easier to be detected by a detector.
ii). Assuming that the protons and neutrons were at rest, calculate the minimum momentum to which the electrons should be accelerated, in order to perform such experiments successfully
[tex]\lambda=\frac{h}{\rho} \Rightarrow \ \rho=\frac{h}{\lambda}=\frac{h}{10^{-15}}=6.63 \times 10^{-19}Jsm^{-1}[/tex]
iii). Considering such scattering collisions in the laboratory frame of reference (where the target is stationary), write formulae for the energies of the electrons and target, and evaluate then in Gev (you may ignore the rest mass of the electron)
I found this part of the question pretty difficult. Here's my attempt:
Initial state: 0---> O
I have an electron (in the "picture" above as "0") and a proton ("O"). The electron has an energy [tex]E_1[/tex] and a momentum [tex]\underline{\rho}[/tex]. I wrote this as [tex](E_1,\underline{\rho})[/tex].
Using the equation [tex]E=\rho c[/tex], [tex]E_1=\underline{\rho}c[/tex]
Final state: <-----0 O----->
Using the (energy, momentum) co-ordinates, the electron has [tex](E_2, -\underline{\rho})[/tex] and the proton has [tex](E_3, \underline{\rho})[/tex].
The initial energy I calculated using [tex]E=\sqrt{(\rho c)^2+(m_0 c^2)^2}[/tex]
[tex]\sqrt{(\frac{6.63 \times 10^{-19} c}{1.6 \times 10^{-10}})^2+(0.938)^2}[/tex]
(the mass of a proton is [tex]0.938\frac{Gev}{c^2}[/tex])
This gives E=1.557 GeV.
Hence I found the initial energy of the proton to be 0.938Gev and the energy of the electron to be 1.557-0.938=0.619Gev.
I'm unsure whether this is correct. I may have completely misunderstood the entire situation!
iv). The total energy in the centre-of-mass reference frame in such electron-nucleon collisions corresponds to the maximum possible mass of all particles in the final state and is a Lorentz-invariant quantity. Ignoring the rest mass of the electron, calculate this quality
The "maximum possible mass" suggests that all the particles are stationary. Hence I should be calcuating their rest mass.
However, since I only have a proton and an electron at the end would the answer by just the rest mass of a proton- [tex]0.938\frac{GeV}{c^2}[/tex]?