Crystal Target Scattering Angular Distribution 3D

This animation is similar to the Rutherford scattering experiment. The Rutherford scattering experiment (RSE) is famous because it established that atoms of a solid have a very compact nucleus. In the case of the RSE, the solid was a very thin gold foil and the incident particles were doubly ionized helium atoms (nucleii) which are also called alpha (`alpha`) particles. The molecular weight of gold is 197 and that of helium is 4. Therefore, if almost all the mass of the gold atom is concentrated in its small nucleus, when the alpha particle hits the gold nucleus, only a small fraction of its energy will be transferred to the nucleus. If the alpha particle hits the gold nucleus dead center, then the alpha particle will bounce back 180 degrees from its original direction. It turned out that that is what happened. It led to Rutherford making the now famous comment:

"It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you."

This result would not have been the case if the nucleus were not extremely compact. It turns out that the nucleus mean radius is of the order `r_(n)=10^(-14) ` meters. Gold has a face centered cubic (FCC) crystal structure where the cube edge is `a_(gold)=4.07cdot10^(-10)` meters. So viewed from the side, as seen in Canvas 0, the spacing between each gold atom occupies an area of 1/2 the area of a square of edge length `a_(gold)` so each gold atom occupies an area `A_(gold)=(16.56/2)cdot10^(-20) meters^2` (see Spreadsheet 0) . This means that it will be extremely unlikely that a small alpha particle would have a hard collision with a helium nucleus. In fact, the probability, P, is about

`P=(pi r_n^2)/A_(gold)=3.5 cdot 10^(-9)`

and yet a significant fraction of the alpha particles were deflected at other than small angles. Part of the reason for that is that the alpha particle did not have to actually contact the nucleus of gold since both have fairly large positive charge. The charge of gold is `Z_(gold)=79` while that of the alpha particles is `Z_(alpha)=2` where the Z values represent the number of oppositely signed electron charges, `e`, where `e=1.6cdot10^(-19)` coulombs.

With that introduction, we are now ready to work out the details of this animation. The illustration in Canvas 0 shows only a single layer of gold nucleii. An actual gold foil will have at least `n_L=200` layers so the probability of a direct hit is increased by `n_L` or to `P=0.7 cdot 10^(-6)` if `n_L=200`.

Spreadsheet 0: Gold Foil Target Spreadsheet
Canvas 1: Diagram Showing Impact Parameter
Impact Radius 50
Sphere Radius 50
Canvas 0: Diagram of Gold Crystal Geometry
Number of Unit Cell Columns or Rows

Equations for the Animation and Plots

Below I show that, when the target particle is much more massive than the beam particle as is the case for alpha particles incident on gold nucleii, the change in angle of the scattered particle velocity is just twice the incident angle. The incident angle is defined as the angle between the incident velocity and the vector between centers of the beam and target particles when they have just achieved contact. See a diagram of these angles in Canvas 3. Then, using equations similar to bouncing off of a flat wall, the final beam particle velocity vector `bb v_f` becomes:

`bb v_f=bb v_i-2 hat bb"u" ( bb v_i cdot hat bb"u") `

where `bb v_i` is the incident velocity and `hat bb"u"` is a unit vector pointing from the target particle to the beam particle. Note that this equation conserves energy

`(bb v_f cdot bb v_f) =(bb v_i cdot bb v_i)`
` -4 (hat bb"u" cdot bb v_i) ( bb v_i cdot hat bb"u")`
`+4 (bb v_i cdot hat bb"u" ) (bb v_i cdot hat bb"u") =(bb v_i cdot bb v_i)`

Also this equation is correct even when the beam particle makes multiple scatterings from the target particles even when the initial velocity is not along the x axis.
This equation gives the final velocity for each time a beam particle contacts a target particle. Then the beam particle position is iterated using this new velocity.

`x_p=x_p+v_xdt`
`y_p=y_p+v_ydt`

where `dt` is the iteration time increment. When the particle reaches the circular chamber wall, it is stopped. Then particle numbers that have reached the wall are placed in bins Vs their angle on the wall. A plot of bin occupation Vs angle is shown in Canvas 3. The same bin accumulation data is shown in Canvas 2 in a spherical color plot where the bin count color code is shown in the color bar at the right of the sphere. The learner has the option of adjusting the target opacity. This depends on both the target particle radius, `r_t`, and particle spacing `s_t` and the beam particle radius, `r_b` via the expression:

`s_t=2.5/(opacity)*(r_t+r_b)`;

where the 2.5 factor is just a parameter that seems to make the ratio of forward scatter to backscatter agree with the numerical results.

Discussion of the Animation Results

The "gold" nucleii that we use here are much larger compared to their spacing than the nucleii in a real gold foil. This is necessary in order to show at least some backscattering in a finite time. The 3D beam, which travels along the x direction, has random y and z positioning of target particles as might be expected in an experiment. Without random y and z positioning, the final scatter pattern would have false maxima.For the 3D scattering version, the target chamber is the inside of a sphere. The particle accumulations are shown as small squares of colors that correspond to the number of particles landed there divided by the maximum accumulations anywhere in the sphere. Since some of the scatter is in the z direction, all particle accumulations that do not show at the the extreme radius of the sphere have at least some velocity with a z component. We expect, with target opacity of 50%, to have about half of the beam particles go right through the target without being deflected. As you can see by looking at Canvas 2 after the Run is complete, that is reasonably valid.
Canvas 2: Beam Particle Scattering Animation
Target Opacity: 0.5
Longitude Angle: 180
Latitude Angle: 60
Single Step

Deriving the Equations for Final Velocities After Collision

In the following equations, please be aware that bold letters denote vectors which have either (x,y) or (x,y,z) components depending on whether we are considering 2D or 3D calculations. For hard sphere collisions one must consider both conservation of momentum and energy. If the collision was mediated by a potential between the particles, one would only have to consider conservation of kinetic and potential energy. Conservation of momentum requires that the final momentum be the same as the initial momentum. For a collision between particles 1 and 2:

`m_1 bb"v"_{1f}+m_2 bb"v"_{2f}=m_1 bb"v"_{1i}+m_2 bb"v"_{2i} (1)`

where the subscript f stands for final value and the subscript i stands for initial value and the arrowed bold letter ` bb"v"` stands for the velocity vector. The conservation of kinetic energy equation is as follows:

`m_1 bb"v"_{1f} cdot bb"v"_{1f} +m_2 bb"v"_{2f} cdot bb"v"_{2f}`

`=m_1 bb"v"_{1i} cdot bb"v"_{1i} +m_2 bb"v"_{2i} cdot bb"v"_{2i} (2)`

where the `cdot` between the velocity vectors stands for the inner or dot product of the two vectors. When we combine equations 1 and 2 we get equations that relate the components of the velocity vectors.

The most important concept in these calculations is that of the unit vector ` hat bb"u"` between the centers of the two particles that are colliding

` hat bb"u"=[(x_2-x_1)hat bb"x"+(y_2-y_1)hat bb"y"+(z_2-z_1)hat bb"z"]/r_{12}`

where the x, y, and z that have hats are unit vectors along their respective axes and

`r_{12}=sqrt[(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2]`

Note that the unit vector points from sphere 1 to sphere 2 so, if `x_2` is greater than `x_1`, the unit vector's x component is positive.
Applying equation 1, we know that the changes of momentum of the spheres are equal and opposite and are along the unit vector `hat bb"u"` so we can write:

`m_1 delta bb"v"_1=M delta v hat bb"u"=-m_2 delta bb"v"_2 (3)`

where `M` has units of mass and `M` and the scalar speed increment `delta v` are still to be determined. Now we may re-write equation 2 using equation 3 and obtain:

`((m_1 bb"v"_1+M delta v hat bb"u") cdot (m_1 bb"v"_1+M delta v hat bb"u"))/(m_1)+ ((m_2 bb"v"_1-M delta vhat bb"u") cdot (m_2 bb"v"_2-M delta vhat bb"u"))/(m_2) =m_1 bb"v"_{1} cdot bb"v"_{1} +m_2 bb"v"_{2} cdot bb"v"_{2}(3)`

After cancellations of the terms on its right side, equation 4 simplifies to:

`M delta v=(2m_1m_2)/(m_1+m_2)( bb"v"_1- bb"v"_2) cdot hat bb"u"`

We can now make the identifications:

`M=(2m_1m_2)/(m_1+m_2)`

and

`delta v=( bb"v"_1- bb"v"_2) cdot hat bb"u"`

where M is known as the "reduced mass" Finally using equation 3 again we can write the expressions for the final velocity vectors:

` bb"v"_{1f}= bb"v"_{1i}+M/m_1 [( bb"v"_1- bb"v"_2) cdot hat bb"u"] hat bb"u" (5)`
` bb"v"_{2f}= bb"v"_{2i}-M/m_2 [( bb"v"_1- bb"v"_2) cdot hat bb"u"] hat bb"u"`

When `m_2/m_1 gtgt 1` then `M/m_1` becomes 2. We use equation 5 with `M/m_1=2` to compute the scatter direction of the beam particles.
Canvas 3: Diagram Showing Incident and Scattering Angles
Impact Radius b: 50

Solid Angle for Hard Sphere Collision

Let `r_1` and `r_2` be the sphere radii and let the collision be in the (x,y) plane. For the following calculations let

`r=r_1+r_2`

The next task for us is to relate these final speeds and energy changes to the collision parameter, `b`. First we can write the following equation for the unit vector `u` between the two spheres:

`bb u=1/r[-(sqrt(r^2-b^2))bb (hat x)+b bb hat(y)]`

Thus

`u_x=-sqrt(1-(b^2/(r^2))), u_y=b/r`

Without loss of generality, let's just have `bb v_1=v_1 bb (hat x)` and `bb v_2=0` and then use equation 5 to compute the final velocity.

` bb"v"_{1f}= bb"v"_{1i}+M/m_1 [( bb"v"_1) cdot hat bb"u"] hat bb"u" `

Taking the dot product of `bb v_1` with `hat bb u` we obtain:

` bb"v"_{1f}= bb v_{1i}+M/m_1 [( v_1)/r[-(sqrt(r^2-b^2))]] hat bb"u" `
` bb"v"_{2f}= -M/m_2 [( v_1)/r[-(sqrt(r^2-b^2))]] hat bb"u" `

Inserting the results for ux and uy into this equation we have:

`v_(1fx)=v_(1i)(1-2[1-(b/r)^2])), v_(1fy)=-v_(1i)(2sqrt[1-(b/r)^2])b/r)`

Then the expression for the tangent of the angle of `bb v_1` is:

`tan(phi_(sc))=-{2(b/r)sqrt[1-(b/r)^2]}/{1-2[1-(b/r)^2]}`

The scatter direction of particle 2 is along the `bb u` vector and the scatter direction of particle 1 is along the unit vector:

` bb u_(1f)= {bb v_{1i}+M/m_1 [( v_1)/(r)[-(sqrt(r^2-b^2))]] hat bb u}/ (sqrt((bb (v_(1f)) cdot bb (v_(1f))))) (6)`

Proof that scatter angle is twice the incident angle:

Think of `b/r` as the base, x, of a right triangle of hypotneuse length 1. Then the height, y, will be `sqrt[1-(b//r)^2]`. Then
`phi_1(x,y)=ArcTan(y/x)` which is easy to recognize as the arc tangent of the right triangle.
Then using the same notation
`tan(phi_(sc))=-(2yx)/(1-2x^2)` which is our equation for the scatter angle
The equation for the tangent of twice an angle tan(2a) is
`tan(2a) =(2tan(a))/(1-tan(a)^2)`
Letting `tan(a)=y/x` we have:
`tan(2a)=(2(y/x))/(1-(y/x)^2)=(2xy)/(x^2-y^2)=(2xy)/(x^2-(1-x^2))=(2xy)/(2x^2-1)=-(2xy)/(1-2x^2)`
which is same as the expression for `tan(phi_(sc))`

Computing the Solid Angle Vs Impact Parameter

From the angle result the number of particles that will scatter into a given solid forward or backward solid angle increment `delta Omega` can be computed similarly to the famous Rutherford scattering experiment of charged alpha particles from heavy nucleii in gold foil. First note that from the definition of a solid angle:

`(d Omega)/(d theta)=2 pi theta`

and then we can use an expression for `theta` to compute `(d theta)/(db)` and obtain:

`(d Omega)/(d b)=2 pi theta (d theta)/(db)`

Now we need the expression for `(d theta)/(db)`

`(d theta)/(db)=(d)/(db)[tan^-1 (v_(1y)/v_(1x))]`

The derivative of `tan^-1(x)` is `1/(1+x^2) so we need the derivative:

`(d theta)/(db)=(d)/(db)[1/(1+(v_(1y)/v_(1x))^2)]`

Following through with the derivatives of v with respect to b we have:

`(d v_(1y))/(db)=-v_(1ix) [sqrt(r^2-b^2)/(r^2)-b^2 /(r^2 sqrt(r^2-b^2))]`

We may now use equation 5 to compute the direction of `bb v_(1f)` on the assumption that `m_2 gt gt m_1` which means that `M/(m_1)=2`:

`v_(1fx)=v_(1ix)-2v_(1ix)(r^2-b^2)/(r^2)`
`v_(1fy)=-2v_(1ix)b sqrt(r^2-b^2)/(r^2)`
`theta_(xi)=tan^-1(v_(1fy)/v_(1fx)))`

w. where the scattered angle, `theta_(xi)=0`, for the case where `b=r` (grazing incidence) and decreases to `-pi` for rear end collisions. `theta_(xi)` is thus the angle with respect to the initial direction of `bb v_1`. Looking at the diagram in Canvas 1, it is easy to show that the value of `theta_(xi)` is twice the value of the incident angle `theta_(iota)` where

`theta_(iota)=tan^-1(b/sqrt(r^2-b^2))`

so that, when `m_2` is very large, the angle of exit with respect the `bb u` is just twice the angle of incidence with respect to `bb u`. From the angle result the number of particles that will scatter into a given solid forward or backward solid angle increment `delta Omega` can be computed similarly to the very important Rutherford scattering experiment where charged alpha particles scattered from heavy nucleii in gold foil. First note that from the definition of a solid angle:

`(d Omega)/(d theta)=2 pi theta`

and then we can use our expression for `theta_xi` to compute `(d theta)/(db)` and obtain:

`(d Omega)/(d b)=2 pi theta (d theta_xi)/(db)`

Now we need the expression for `(d theta_xi)/(db)`

`(d theta_xi)/(db)= 2(d theta_iota)/(db)=2/(sqrt(r^2-b^2)`

so that

`(d Omega)/(d b)=(4 pi tan^-1(b/sqrt(r^2-b^2)))/(sqrt(r^2-b^2)`

`(d Omega)/(d b)=(4 pi theta_iota)/(sqrt(r^2-b^2)`

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