If you have not already done so, do .24. (a) Now consider

A blueberry pancake has diameter 15 cm and contains 6 large blueberries, each of diameter 1 cm. Find the cross section a of a blueberry and the "target" density ntar (number/area) of berries in the pancake, as seen from above. What is the probability that a skewer, jabbed at random into the pancake, will hit a berry (in terms of a and ntar, and then numerically)?

a) A certain nucleus has radius 5 fm. (1 fm = 10-15 m.) Find its cross section a in barns. (1 barn = 10-28 m2.) (b) Do the same for an atom of radius 0.1 nm. (1 nm = 10-9 m.)

A beam of particles is directed through a tank of liquid hydrogen. If the tank's length is 50 cm and the liquid density is 0.07 gram/cm3, what is the target density (number/area) of hydrogen atoms seen by the incident particles?

The cross section for scattering a certain nuclear particle by a copper nucleus is 2.0 barns. If 109 of these particles are fired through a copper foil of thickness 10 gm, how many particles are scattered? (Copper's density is 8.9 gram/cm3 and its atomic mass is 63.5. The scattering by any atomic electrons is completely negligible.)

The cross section for scattering a certain nuclear particle by a nitrogen nucleus is 0.5 barns. If 1011 of these particles are fired through a cloud chamber of length 10 cm, containing nitrogen at STP, how many particles are scattered? (Use the ideal gas law and remember that each nitrogen molecule has two atoms. The scattering by any atomic electrons is completely negligible.)

Our definition of the scattering cross section, A r sc = Nincn tar a 9 applies to an experiment using a narrow beam of projectiles all of which pass through a wide target assembly. Experimenters sometimes use a wide incident beam, which completely engulfs a small target assembly (the beam of photons from a car's headlamp, directed at a small piece of plastic, for example). Show that in this case Nsc = ninc NtarG where nin, is the density (number/area) of the incident beam, viewed head- on, and Ntar is the total number of targets in the target assembly.

Calculate the solid angles subtended by the moon and by the sun, both as seen from the earth. Comment on your answers. (The radii of the moon and sun are Rm = 1.74 x 106 m and Rs = 6.96 x 108 m. Their distances from earth are dm = 3.84 x 108 m and ds = 1.50 x 1011 m.)

In their famous experiment, Rutherford's assistants, Geiger and Marsden, detected the scattered alpha particles using a zinc sulphide screen, which produced a tiny flash of light when struck by an alpha particle. If their screen had area 1 mm2 and was 1 cm from the target, what solid angle did it subtend?

By integrating the element of solid angle (14.15), dS2 = sin 0 dO dO, over all directions, verify that the solid angle corresponding to all directions is 47r steradians.

By evaluating the necessary integral, verify the result (14.20) for the total cross section of Example 14.4 (page 571). (This is very easy if you change variables to u = cos 0.)

The differential cross section for scattering 6.5-MeV alpha particles at 120 off a silver nucleus is about 0.5 barns/sr. If a total of 1010 alphas impinge on a silver foil of thickness 1 itm and if we detect the scattered particles using a counter of area 0.1 mm2 at 120 and 1 cm from the target, about how many scattered alphas should we expect to count? (Silver has a specific gravity of 10.5, and atomic mass of 108.)

[Computer] In quantum scattering theory, the differential cross section is equal to the absolute value squared of a complex number f (6), called the scattering amplitude: da (K2 = If(0)12. The scattering amplitude can in turn be written as an infinite series h f (0) = E(ze + ei4 sin 8.e Pe (cos 0). P e=o (14.57) (14.58) Here h = 1.05 x 10-34J s is called "h bar" and is Planck's constant divided by 27r, and p is the momentum of the incident projectile. The real numbers 8e are called the phase shifts and depend on the nature of the projectile and target and on the incident energy. And Pt (cos 0) is the so-called Legendre polynomial. (P0 = 1, P1 = cos0, etc.) The partial-wave series (14.58) is especially useful at low energies where only a few of the phase shifts are different from zero. (a) Write down this series for the case of 10-MeV neutrons (mass m = 1.675 x 10-27 kg) scattering off a certain heavy nucleus for which 8 = 30, 61 = 150, and all other phase shifts are negligible. (b) Find an expression for the differential cross section (in terms of h, m, the incident energy E, and the two nonzero phase shifts) and plot it for 0 < 0 < 180. (c) Find the total scattering cross section

In deriving the cross section for scattering by a hard sphere, we used the "law of reflection," that the angles of incidence and reflection of a particle bouncing off a hard sphere are equal, as in Figure 14.10. Use conservation of energy and angular momentum to prove this law. (The definition of "hard- sphere scattering" is that a projectile bounces with its kinetic energy unchanged. That the force is spherically symmetric implies, as usual, that angular momentum about the sphere's center is conserved.)

One can set up a two-dimensional scattering theory, which could be applied to puck projectiles sliding on an ice rink and colliding with various target obstacles. The cross section a would be the effective width of a target, and the differential cross section da la would give the number of projectiles scattered in the angle d0. (a) Show that the two-dimensional analog of (14.23) is da Id() = Idbld0I. (Note that in two-dimensional scattering it is convenient to let 0 range from -11" to 7r.) (b) Now consider the scattering of a small projectile off a hard "sphere" (actually a hard disk) of radius R pinned down to the ice. Find the differential cross section. (Note that in two dimensions, hard "sphere" scattering is not isotropic.) (c) By integrating your answer to part (b), show that the total cross section is 2R as expected.

[Computer] Consider a point projectile moving in a fixed, spherical force whose potential energy is \(U(r)=\left\{\begin{matrix} -U_o & (0\leq r\leq R)\\ 0 & (R< r)\end{matrix}\right.\) where \(U_o\) is a positive constant. This so-called spherical well represents a projectile which moves freely in either of the regions r < R and R < r, but, when it crosses the boundary R = r, receives a radially inward impulse that changes its kinetic energy by \(\pm U_o(+U_o\ \text{going inward},-U_o\ \text{going outward})\). (a) Sketch the orbit of a projectile that approaches the well with momentum \(p_o\) and impact parameter \(b<R\). (b) Use conservation of energy to find the momentum p of the projectile inside the well \(r<R\). Let \(\zeta\) denote the momentum ratio \(\zeta =p_o/p\) and let d denote the projectile’s distance of closest approach to the origin. Use conservation of angular momentum to show that \(d=\zeta b\). (c) Use your sketch to prove that the scattering angle \(\theta\) is \(\theta = 2\left ( arcsin\ \frac{b}{R}-arcsin\ \frac{\zeta b}{R} \right )\) This gives \(\theta\) as a function of b, which is what you need to get the cross section. The relation depends on the momentum ratio \(\zeta\), which in turn depends on the incoming momentum \(p_o\) and the well depth \(U_o\). Plot \(\theta\) as a function of b for the case that \(\zeta=0.5\). (d) By differentiating \(\theta\) with respect to b, find an expression for the differential cross section as a function of b, and make a plot of \(d\sigma /d\Omega\) against \(\theta\) for the case that \(\zeta=0.5\). Comment. [Hint: To plot as a function of \(\theta\) you don't need to solve for b in terms of \(\theta\); instead, you can make a parametric plot of the point \((\theta,d\sigma/d\Omega)\) as a function of the parameter b running from 0 to R.] (e) By integrating \(d\sigma/d\Omega\) over all directions, find the total cross section.

One of the specific predictions of Rutherford's model of the atom was that the cross section should be inversely proportional to \(E^2\) or, equivalently, to \(v^4\) . To test this, Geiger and Marsden varied the speed v by passing the incident alphas through thin sheets of mica to slow them down. According to Rutherford's prediction, the product of \(N_{sc}\), and \(v^4\) should be the same whatever the incident speed (provided all other variables were held constant). Add a row showing \(N_{sc}v^4\) to this table of their data and see how well Rutherford's prediction was confirmed.

Another specific prediction of Rutherford's model of the atom was that the cross section should be proportional to the nuclear charge squared, that is, to Z2, where Z is the atomic number, the number of protons in the nucleus. To test this, Geiger and Marsden counted the number of scatterings off various different targets (holding all other variables fixed), with the following results: Target Gold Platinum Tin Silver Copper Aluminum Nse Z 1319 79 1217 78 467 50 420 47 152 29 26 13 Add a row to this table to show the ratio N/Z2 and see how well Rutherford's prediction was confirmed. (At the time the atomic number was not known with certainty, nor was it well understood. Rutherford had guessed, correctly, that the nuclear charge was roughly equal to half the atomic mass, and this is what they used in place of Z.) The relatively poor agreement for the case of aluminum is probably due to our neglect of the target recoil, which is more important for the lighter targets.)

One the first observations that suggested his nuclear model of the atom to Rutherford was that several alpha particles got scattered by metal foils into the backward hemisphere, 7/2 < 0 < r - an observation that was impossible to explain on the basis of rival atomic models, but emerged naturally from the nuclear model. In an early experiment, Geiger and Marsden measured the fraction of incident alphas scattered into the backward hemisphere off a platinum foil. By integrating the Rutherford cross section (14.33) over the backward hemisphere, show that the cross section for scattering with 6 > 90 should be 47 t co(E). Using the following numbers, predict the ratio Ns,(0 > 90)/Nin,: thickness of platinum foil ti 3 pm, density = 21.4 gram/cm3, atomic weight = 195, atomic number = 78, energy of incident alphas = 7.8 MeV. Compare your answer with their estimate that "of the incident a particles about 1 in 8000 was reflected" (that is, scattered into the backward hemisphere). Small as this fraction is, it was still far larger than any rival model of the atom could explain.

An important simplification in our derivation of the Rutherford cross section was that the projectile's orbit is symmetric about the direction u of closest approach. (See Figure 14.11.) Prove that this is true of almost any conservative central force, as follows: (a) Assume that the effective potential (actual plus centrifugal) behaves as in Figure 8.4; that is, it approaches zero as r coo and approaches +oo as r --> 0.13 Use this to prove that any projectile that comes in from infinity must reach a minimum value rmin and then move out to infinity again. (b) This implies that a projectile must visit any value of r between rrnin and infinity exactly twice, once on the way in and again on the way out. Prove that the values of i- at these two points are equal and opposite, and that the values of 1.fr are equal (where tfr is the polar angle defined in Figure 14.11). (c) Use these results to prove that the orbit is symmetric under reflection about the direction u.

The derivation of the Rutherford cross section was made simpler by the fortuitous cancellation of the factors of r in the integral (14.30). Here is a method of finding the cross section which works, in principle, for any central force field: The general appearance of the scattering orbit is as shown in Figure 14.11. It is symmetric about the direction u of closest approach. (See Problem 14.19.) If ir is the projectile's polar angle, measured from the direction u, then --> +0. as t --> +oo and the scattering angle is 6 = rc 21/1f. (See Figure 14.11 again.) The angle iiro is equal to f Ili dt taken from the time of closest approach to oo. Using the now-familiar trick you can rewrite this as f dr. Next rewrite 1p in terms of the angular momentum C and r, and rewrite i- in terms of the energy E and the effective potential Ueff defined in Equation (8.35). Having done all this you should be able to prove that co 0 =7r. 2 (14.61) fr. (b / r2) dr 11 (b/r)2 U(r)/E Provided this integral can be evaluated, it gives 0 in terms of b, and hence the cross section (14.23). For examples of its use, see Problems 14.21, 14.22, and 14.23.

Use the general relation (14.61) from Problem 14.20 to rederive the relation (14.24) for scattering by a hard sphere.

Use the general relation (14.61) from Problem 14.20 to rederive the relation (14.31) for Rutherford scattering.

Consider the scattering of a particle with energy E by a fixed, repulsive 1/r3 force field, with potential energy U = y/r2. Use the relation (14.61) from Problem 14.20 to find 0 in terms of b and hence show that the differential cross section is y 7r2(7r 0) dS-2 E 02(27r 0)2 sin6 (14.62) To refresh your memory as to how to find ri.nin you might look at Figure 8.5 (the case E > 0). You should be able to solve your equation for 0 in terms of b to get b in terms of 0 and thence the cross section.

Consider the scattering of two particles of equal mass (for example, scattering of protons off protons). In this case, elab = 40,m. (See Problem 14.27.) (a) Use this result in (14.45) to prove that when the projectile and target have equal masses (dado. 4 cos viab ( ) 6/2 dS2 labcm (14.63) (b) Write down the lab cross section for the scattering of two equal-mass hard spheres. (We know that in the CM frame, the differential cross section is R2/4 where R = R1 + R2.) By integrating over all directions, verify that the total cross section in the lab frame is 7 R2, as it has to be.

Using Figure 14.15, prove Equation (14.53), that tan Blab = sin 0,. (14.64)

Using suitable trig identities, rewrite (14.64) to give cos Blab as a function of cos 9,., and verify the derivative (14.55) that was essential for finding the relation between the CM and lab cross sections

(a) By setting the mass ratio X = 1 in Equation (14.64) of Problem 14.25, prove that in equal-mass scattering, Olab = 10cm. (b) Redraw Figure 14.15 for the case that A = 1 (m1 = m2) and explain why the maximum value of Ow, is 7/2.

It is often interesting to know about the momentum of the recoiling target particle in the lab frame. Let us denote by *lab the recoil angle, defined as the angle between the recoil momentum Plab2 and the incident direction (the angle below the dashed line in Figure 14.14). _(a) Show that in Figure ,1ab , 14.15 the recoil momentum is represented by the vector DC. Deduce that (7r Ocm)/2. (b) Show that, in the special case of equal masses (m = m2), Blab = 7r/2; that is, the angle between the two outgoing particles in an equal-mass, elastic collision is 90. (c) Prove this last result directly using just conservation of momentum and energy (in an elastic collision).

An elastic collision is defined as one in which the total kinetic energy of the two particles is the same before and after the collision. (a) Show that in the CM frame, the individual kinetic energies of the two particles are separately conserved in an elastic collision. (b) Explain clearly why the same result is obviously not true in the lab frame. (Think about the energy of the target particle.) (c) Let A E denote the energy gained by the target particle in the collision (and hence the energy lost by the projectile). Using Figure 14.15, show that the fractional energy lost by the projectile (in the lab frame) is DE = 4X sin2 (Oc./2) E (1 + A)2 where, as usual, A is the mass ratio m 1/ ni2. (Note that in Figure 14.15 the line DC represents the recoil momentum of the target.) (d) For a given mass ratio X, what sort of collision gives the largest fractional energy loss? What value of X maximizes this energy loss? (Your answer is important in situations where one wants a particle to lose energy as quickly as possible as in a nuclear reactor, for example.)

If you have not already done so, do Problem 14.24. (a) Now consider the scattering of two equal- mass hard spheres, A and B, with B initially stationary. Write down the standard expression (14.42) for the number of projectiles A scattered into a solid angle dO at a chosen angle O. Call this number N (A into d Q at 0). Now suppose that we monitor for the number of target particles B recoiling into the same solid angle dS2 at the same angle 0. Find N(B into dSl at 0), the number of B's that will be observed. How does this compare with the number of A's?

[Computer] Consider the elastic scattering of a projectile that is heavier than the target, that is, m1 > m2 or X > 1. (a) Draw the analog of Figure 14.15 for this case. Show clearly that there are two different values of the CM angle Ocm corresponding to each value of Blab. (b) What are the two CM angles that correspond to Old, = 0? In terms of this example, explain why there is this two-fold ambiguity in Ocn, when m1 > m2. (c) Plot Blab as a function of Ocm for the case that X = 2. (d) Use your picture from part (a) to find an expression for the maximum possible value of ()lab for a given value of Check that your answer is correct for the case that X = 1.