Partial width for elastic scattering

[Yaste]

Homework Statement

A beam of neutrons hit a target of heavy nuclei with spin ##J_N = 0## with resonance when the energy of the incident beam is 250eV, in the cross section distribution with a maximum of 1300 barns. The width of the maximum is 20 eV. Find the partial width of resonance for the elastic scattering.

2. Homework Equations
$$\sigma_{tot} = \dfrac{\pi}{k^2}\dfrac{g\Gamma_i \Gamma}{(E-E_0)^2+\Gamma^2/4}$$

$$g = \dfrac{2J+1}{(2s_1+1)(2s_2+1)}$$

The Attempt at a Solution

The only thing I don't know is the ##g## in the previous equation. That's the only thing I need in order to solve the problem. Any tips would be appreciated.

 

[mark726]

First, let's define the variables in the equations:

- ##\sigma_{tot}## = total cross section
- ##k## = wave number
- ##g## = spin degeneracy factor
- ##\Gamma_i## = partial width of the initial state
- ##\Gamma## = total width of the resonance
- ##E## = energy of the incident beam
- ##E_0## = resonance energy

Now, let's look at the given information:

- ##J_N = 0##, which means that the target has a spin of zero.
- The incident beam has an energy of 250eV.
- The maximum cross section is 1300 barns, and it has a width of 20 eV.

From these, we can deduce that the resonance energy ##E_0## is 250eV, and the total width ##\Gamma## is 20eV. We also know that the spin degeneracy factor ##g## is related to the spins of the target and the incident particle, but since the target has a spin of zero, we can simplify the equation to:

$$g = \dfrac{2J_N+1}{(2s_2+1)}$$

Since we don't know the spin of the incident particle, we can't determine the exact value of ##g##. However, we can make an educated guess based on the fact that the spin of the target is zero and the total width is relatively small compared to the resonance energy. In this case, we can assume that the incident particle also has a spin of zero, which means that ##s_2 = 0## and ##g = 1##.

Now, we can use the given information to solve for the partial width of resonance for the elastic scattering. Plugging in the values in the first equation, we get:

$$1300 = \dfrac{\pi}{k^2}\dfrac{\Gamma_i \Gamma}{20^2+\Gamma^2/4}$$

Since we don't know the value of ##k##, we can't solve for ##\Gamma_i## and ##\Gamma## directly. However, we can use the given information to find the ratio between them. We know that at the maximum cross section, the total width is equal to the width of the maximum, which means that:

$$20 = \dfrac{\Gamma}{2