Angular distribution of Neutrons from the Photo-Disintegration of the Deuteron

this paper was written on 1949. at that time, deuteron just discovered 20 years. this paper presents a method on detecting the diffraction cross section of the neutron from a disintegrated deuteron by gamma ray of energy 2.76MeV. and by this, they found the photo-magnetic to photo-electric cross section ration. the ratio is 0.295 ± 0.036.

the photo-electric dipole transition and photo-magnetic dipole transition can both be induced by the gamma ray. Photo carry 1 angular momentum, the absorption of photon will excited the spherical ground state $^1S$ into $^3P$. the 2 mechanisms of the disintegrations results 2 angular distributions of the neutrons. by examine the angular distribution, they find out the ratio.

the photo-magnetic cross section is isotropic and the photo-electric cross section is follow a of a $sin^2$ distribution. the average intensity of neutron detected on a angle is:

$I(\gamma ) = \int_{\gamma_1}^{\gamma_2} {(a + b sin^2(\gamma)) sin(\gamma) d\gamma } / \int_{\gamma_1}^{\gamma_2} {sin(\gamma) d\gamma }$

where a is the contribution from the photo-magnetic interaction and b is from photo-electric interaction. and $\gamma_1$ and $\gamma_2$ are the angle span by the finite size of the target and detector. the integration is straight forward and result is:

$I(\gamma) = a+b( 1 - 1/3 ( cos^2(\gamma_1) + cos(\gamma_1) cos(\gamma_2) + cos^2 ( \gamma_2) )$

and the author guided us to use the ration of 2 angle to find the ration of a and b. and the ration of a and b is related to the probability of the magnetic to the electric effect by

$a/b = 2/3 \tau$

. and the photo-magnetic to photo-electric cross section ratio is:

$\tau/(\tau+1)$

the detector was described in detail on 4 paragraphs. basically, it is a cylindrical linear detector base on the reaction $B^{10} ( n,\alpha)Li^7$. it was surrounded by paraffin to slow down fast nuetrons.

on the target, which is heavy water, $D_2 O$, they use an extraordinary copper toriod or donut shape container. it is based on 3 principles:

• The internal scattering of neutron
• Departure from point source
• The angular opening of the γ – ray source

they place the γ – ray source along the axis of the toriod, move it along to create different scattering angle.

they tested the internal scattering of the inside the toriod and found that it is nothing, the toriod shape does not have significant internal scattering.

they test the reflection of neutron form surrounding, base on the deviation from the inverse-square law. and finally, they hang up there equipment about 27meters from the ground and 30 meters from buildings walls. (their apparatus’s size is around 2 meters. They measured 45, 60, 75 and 90 degree intensity with 5 degree angular opening for each.

Differential Cross Section III

a single diagram can illustrate everything — the relation between the yield ( the number of particle detected per second in the detector ) and the differential cross section.

the upper drawing is a big view, and the lower one is a zoom-in. the red-circles on the upper drawing are same as the lower one.

From the upper one, the number of particles that scatted by the target and go to direction (θ,φ) is :

$\frac {N_0}{S} \times ( \Delta \sigma \times N) = n$

the left hand side can be interpolated at follow:

$\frac{N_0}{S}$ is the number of particle per second per unit area, or the flux.

$\Delta \sigma \times N$ is the total area of  the cross section that deflect or scatter particle to direction (θ,φ).

The left hand side can also be viewed as :

$N_0 \times \frac { \Delta \sigma \times N } {S}$

where the fraction after multiplication is the chance of getting scattered to the direction (θ,φ). ( the requirement of the flux $N_0$ is in HERE. )

and the right hand side is the number of particle detected at direction (θ,φ).

since both $n$ and $\Delta \sigma$ depend on (θ,φ). thus we can differential it and get the angle dependent of these 2.

$\frac { N_0 N}{S} \frac {d \sigma}{d \Omega} = \frac {d n}{d \Omega}$

if we set the detector moving as radius R. thus the detector area is:

$D_A = R^2 d \Omega$

Therefore, the number of particles will be detected per second on the detector with some area (= Yield ) is:

$Y = \frac {d n}{d \Omega} R^2 d \Omega = \frac { N_o N}{S} R^2 \frac{ d \sigma}{d \Omega} d \Omega$

finish. Oh, the unit of the differential cross section is barn = $10^{-28} m$, recall that a nucleus radius is about $10^{-14} m$

the relation between the differential cross section to the nuclear potential was discussed on HERE.

Differential Cross Section

In nuclear physics, cross section is a raw data from experiment. Or more precisely differential cross section, which is some angle of the cross section, coz we cannot measure every scatter angle and the differential cross section gives us more detail on how the scattering going on.

The differential cross section (d.s.c.) is the square of the scattering amplitude of the scatter spherical wave, which is the Fourier transform of the density.

$d.s.c = |f(\theta)|^2 = Fourier ( \rho (r), \Delta p , r )$

Where the angle θ come from the momentum change. So, sometime we will see the graph is plotted against momentum change instead of angle.

By measuring the yield of different angle. Yield is the intensity of scattered particle. We can plot a graph of the Form factor, and then find out the density of the nuclear or particle.

However, the density is not in usual meaning, it depends on what kind of particle we are using as detector. For example, if we use electron, which is carry elected charge, than it can feel the coulomb potential by the proton and it reflected on the “density”, so we can think it is kind of charge density.

Another cross section is the total cross section, which is sum over the d.s.c. in all angle. Thus, the plot always is against energy. This plot give us the spectrum of the particle, like excitation energy, different energy levels.