WJEC Physics for AS: Student Bk

85 Terms & definitions Exercise 1.7 1. A proton, an electron and a helium nucleus are each accelerated through a potential difference of 500 V . State the increase in kinetic energy of each (a) in eV and (b) in J . 2. The masses of atomic particles are often expressed in terms of the atomic mass unit , u which has a value of 1.660 × 10 –27 kg . The neutron has a mass of 1.008 665 u . Calculate the mass of a neutron (a) in kg , (b) in MeV / c 2 [ c = 2.998 × 10 8 m s –1 ]. 3. The radioactive isotope 13 7 N decays by the emission of a positron, when one of the protons in its nucleus transforms into a neutron. The equation is: p  n + e + + X where X is an unidentified particle. The half life of the decay is 10.1 minutes. (a) Identify X and justify your choice in terms of relevant conservation laws. (b) State which of the interactions controls this decay. Justify your choice. (c) An isolated proton cannot decay into a neutron in this way. Explain which conservation law would be violated. (b) Conservation of baryon and quark numbers Similarly to lepton number we define a baryon number, B . Each baryon, e.g. proton, has B = 1 ; antibaryons, have B = – 1 ; leptons and mesons have B  =  0 . Again, baryon number is always conserved. The following reaction could not happen: p + π –  n + n even if the p and π – had enough kinetic energy and even though it conserves charge and lepton number. Why not? Because the total baryon number on the left is 1 and on the right it is 2! The conservation of baryon number is really a special case of the conservation of quark number, Q . Looking again at our ‘impossible reaction’ and assigning a quark number of -1 to antiquarks, we can tally the quarks as follows: Left-hand side: Q = 3 + ( 1 – 1 ) = 3 Right-hand side: Q = 3 + 3 = 6 On the other hand the Q totals for the p + p  p + n + π + are the same on both sides. Looking more closely we see that the individual quark numbers, U (up) and D (down) are also conserved, with the usual convention that an antiquark has a value – 1 : on each side U = 4 and D  =  2 . Individual quark numbers are conserved in strong and e-m inter- actions, but can be changed by ±1 in weak interactions . Looking again at neutron decay: n ( udd )  p + e – +  ν e Writing this in terms of quarks: udd  uud + e – +  ν e , U = 1 2 0 0 D = 2 1 0 0 We see that U changes from 1 to 2 and D from 2 to 1. The total quark number Q is 3 on both sides but one of the quarks has changed its flavour from down to up. Study point The number of mesons is not conserved. The most common π + decay mode ( >99% ) is: p +  m + + ν μ . The reason for non-conservation of mesons is that they each consist of a q  q pair so Q = 0 . 1.7.7 Self-test For the decay: p +  m + + ν μ , (a) Explain what interaction is responsible. (b) Show what conservation laws are demonstrated. The different types of quark, up, down, etc., are said to possess different flavours . This strange (!) use of the word is possibly related to the German sour milk product which is often flavoured with fruit. Examtip Indications of a weak force are: 1. Leptons are involved: they don’t feel the strong force. 2. Neutrinos are involved: they don’t feel the e-m force. 3. If quarks are involved, a change in quark flavour occurs. 4. If it is a decay, the lifetime is greater than ~ 10 –10 s . Particles and nuclear structure