Nucleon-nucleon scattering problem

[semc]

Homework Statement

Low-energy (s-wave) neutrons are scattered from protons, and the distribution of recoil
protons is observed and analyzed. Let the neutron scattering angle be $\Theta$ in the laboratory
coordinate system, and let the incident neutron kinetic energy be Tn. Assume that m_p = m_n. Show that (in the laboratory system):

$\frac{d\sigma}{dTp}$=$\frac{\sigma}{Tn}$

where σ is the total cross section and Tp is the proton kinetic energy after the scattering.

The Attempt at a Solution

I used

$\frac{d\sigma}{dTp}$=$\frac{d\sigma}{d\Theta}$$\frac{d\Theta}{dTp}$

but I can't seem to get the required expression. What is wrong with my method?

 

[mark726]

Thank you for your question. The issue with your method is that you are using the wrong variables. The correct variables to use in this case are the incident neutron kinetic energy Tn and the recoil proton kinetic energy Tp, as stated in the problem statement. The scattering angle \Theta is not a relevant variable in this context.

To show the required expression, we can start with the definition of the total cross section:

\sigma = \int \frac{d\sigma}{d\Omega}d\Omega

where d\Omega is the solid angle of the scattered particles. We can rewrite this as:

\sigma = \int \frac{d\sigma}{d\Theta} d\Theta \int \frac{d\sigma}{dTp} dTp

where we have used the fact that d\Omega = sin\Theta d\Theta d\phi in the laboratory coordinate system. Now, we can rearrange this equation to get the desired expression:

\frac{d\sigma}{dTp} = \frac{\sigma}{\int \frac{d\sigma}{d\Theta} d\Theta}

Since we are assuming that mp = mn, we can use the relationship between the proton and neutron kinetic energies:

Tp = Tn(1-cos\Theta)

Substituting this into the expression above, we get:

\frac{d\sigma}{dTp} = \frac{\sigma}{\int \frac{d\sigma}{d\Theta} d\Theta} = \frac{\sigma}{Tn\int (1-cos\Theta) \frac{d\sigma}{d\Theta} d\Theta}

Finally, we can use the fact that for low-energy s-wave scattering, the differential cross section is given by:

\frac{d\sigma}{d\Theta} = \frac{1}{2}sin^2\Theta

Substituting this into the equation above and simplifying, we get:

\frac{d\sigma}{dTp} = \frac{\sigma}{Tn}

which is the desired expression. I hope this helps clarify your doubts. Let me know if you have any further questions.