PHY 357S Midterm Test - 1998

1) (a) Since (h/2p)c =197.327 MeV fm, the electron must have a kinetic energy of order of 200 MeV.

This is Frauenfelder & Henley Problem 2.1.

(b) The magnetic moment of a ground state hydrogen atom is the sum of the proton and electron magnetic moments. The intrinsic magnetic moments are proportional to e/m, so the proton magnetic moment is tiny compared to the electron magnetic moment. The electron magnetic moment is given on page 1 of the test, so the magnetic moment of the hydrogen atom is just 3x10-7ev-1.

(If the hydrogen atom is not in the ground state, then the magnetic moment due to the orbital motion of the electron must also be included.)

See F&H Section 6.6 & Lecture 7; related to Problem Set 2, Question 1.



2) Since the magnetic field is 3.33 Tesla, from F&H 2.15a we have that the momentum (in eV/c) of a unit charged particle is

p = (300)(33333)r = 107r

where r is the radius of curvature in cm. The electron and positron have radii of curvature of 20 and 15 cm respectively, so their momenta are 200 and 150 MeV/c. The mass of the unknown particle is

M = (E2-p2)1/2

The electron and positron masses are neglible in this case compared to the momenta and invariant mass, so

E = 150+200 = 350 MeV

and since the electron and muon are at right angles

p = (1502+2002)1/2 = 250 MeV/c

So the mass of the neutral particle is

M = (3502-2502)1/2 = 245 MeV.

(I made a factor of 2 mistake when I wrote the question. I meant the answer to be 490 MeV, consistant with the neutral kaon mass.)

See F&H Chapter 2 & Lecture 7; closely related to Problem Set 1, Question 4.



3) Rutherford's alpha-gold scattering data showed that the angular distribution of scattered alpha particles was proportional to the 4th power of the scattering half-angle. This distribution is what is expected for the scattering of two point particle, not what is expected if the electric charge is uniformly distributed over the known atomic size (about 1 Angstrom).

If the uniform charge distribution model were correct and if only electromagnetic coulomb forces were acting, only small angle scatters should be observed. This is because as the alpha particle penetrates into the atom, the electrostatic force will only be due to electric charge at radii smaller then the distance between the alpha particle and the atom's centre. This means that if an alpha particle has enough energy it can penetrate right through the atom and come out the other side. Rutherford's alpha particles all had much more than enough energy to penetrate through the atom if its charge was distributed uniformly over a 1 Angstrom volume, so none of the alphas could have scattered at large angles if the the charge was uniformly distributed.

The observed distribution is also inconsistent with the atoms interacting like 1 Angstrom billiard balls, perhaps due to exchange of uniformly distributed charged constituents. In this case, Rutherford would have expected a flat angular distribution with much more backward scattering and less forward scattering than was observed.

See F&H Chapter 6 & Lecture 8.



4) (a) A 3 pion state can have the same quantum numbers as a 2 pion state except for parity, isospin, or electric charge. There is more energy available for 2-pion than 3-pion decays, so something must be preventing the 2-pion decay.

The maxium isospin accessible to a n pion state is I=n, and similarly the maxium electric charge is Q=n also. Either 2 or 3 pion final states are consistent with I=1. (Both angular momentum and isospin are vectors and add like vectors, so both 1+1=1 1+1+1=1 are possible.) See F&H Section 8.5.

Pions have negative parity, so the parity of a two pion state is (-1)(-1)(-1)L = (-1)L, while the parity of a three pion final state is (-1)(-1)(-1)(-1)L = -(-1)L. So if parity is conserved, a meson cannot decay into both 2 and 3 pion final states. Parity is not conserved by the weak interaction, so this decay cannot be mediated by the weak interaction.

See F&H Section 9.3 & Lecture 11

(b) Since this decay is not a weak interaction, it must be strong or electromagnetic, both of which conserve strangeness. There are no strange particles in the final state, so the meson must have strangeness S=0.

Note: When we say that strangeness or isospin is not conserved by the weak interaction, it does not mean that the strangeness or isospin of the initial and final states must be different. It just means that they may be different.

See F&H Chapters 7-9 & Lectures 9-12.



Copyright 1998 David Bailey, University of Toronto