END; include ("common_short.php"); physics_header("",""); print <<Physics H133: Problem Set #\$ps_num Here are some hints, suggestions, and comments on the problem set.

## Two-Minute Problems

Remember to give a good explanation, no longer than two sentences.

1. Q12T.3: Is this a bound system? Sign of binding energy?
2. Q13T.4: What decay channels are possible for A=26? How do the various terms in the Bethe-Weizsaecker formula change during the decay? Is the solution unique?

## Chapter Q12 and Q13 Problems

• Q12S.2: If you neglect tunneling, what keeps the alpha particle from entering the nucleus? At which point does the attractive strong interaction overwhelm the Coulomb repulsion? How much energy does the alpha need to make it there? After you solve the problem this way, consider qualitatively the possibility of tunneling and how it affects the minimum kinetic energy.
• Q12S.9: (a) What is the energy and mass of the Earth-Moon system at infinite separation at rest? How are they different in the actual real-life situation? (b) Straightforward, but interesting. Aside: If by splitting uranium you can convert 0.7% of its mass into energy, how many tons of uranium would you need to explode to separate Earth and Moon to infinity?
• Q13S.1: (a) Write down and check the balance equations for charge, lepton number and baryon number for each of the 4 processes. (b) Use conservation laws to first figure out the nature of y, then x.
• Q13S.3: Consider the various contribution to the binding energy per nucleon in Eq. (Q13.17).
• Q13R.1: Draw first an energy level diagram for Mg(26) without worrying about Coulomb energy. Would it be stable? How should you modify the energy level diagram to achieve stability of Mg(26) without making Mg(25) or Mg(24) unstable?
• Q12R.1: (a) Does Eq. (Q12.5) apply if there are neutrons only and also gravity if we assume incompressible balls? How can we figure out how many neutrons there are, given that the mass is 1.4 times the sun's mass (which is given in the inside front cover). (b) This should have the same form as the gravitational potential energy we've used before, except now there is only one mass involved. (c) Just plug in at this point.