Brian R Martin
First edition, March 2007 reprint
Corrections and Clarifications
(There is a separate section below for corrections to solutions to the problems)
(The notation is: line n means n lines from top, including equations; line –n means n lines from the bottom, excluding footnotes. )
p42, Eqn
(2.25): On the rhs the integral is over the vector r and for consistency the integral
should be divided by Ze, the normalization factor.
p50, lines 2–4:
The current Figure 2.8 does not illustrate the text very well. The figure below, specifically for the decay sequence (2.45) is much better.
p51-52: The argument leading to the pairing term might suggest that this can be deduced from the Fig 2.9 (i.e. the Fermi gas model) which is not correct. It should be replaced with:
The final contribution is the empirical pairing term with the form (2.52). This arises from the tendency of like nucleons in the same spatial state to couple pairwise to configurations with spin-0. When coupled like this, the wavefunctions of the two nucleons heavily overlap and so on average they are closer together than when coupled in other configurations, and hence are more tightly bound. When we have an odd number of nucleons, this term does not contribute. Thus, when both Z and N are odd, we gain binding energy by converting one of the odd protons to a neutron (or vice versa) so that it can now form a pair with its formerly odd partner. The evidence for this is that there are only four stable nuclei with odd N and Z, whereas there are 167 with even N and Z. The form used for f5 is empirical, but f(A)=a5/sqrt(A) represents the trend of the data for the pairing energies and is often used.
p82: The comments on the SNO experiment are confused (and wrong in places). The last part of the paragraph starting from the sentence 'More recent experiments .....' should be replaced with the paragraph below.
A more recent experiment (SNO) has studied the reactions:
(a) νe + d → e– + p + p , (b) νx + d → νx +p + n , (c) νx + e– → νx + e– , (3.33)
where x denotes any lepton flavour (e, μ, τ) and d is the deuteron. The cross-section for (b) is independent of the lepton flavour (this is a consequence of ‘lepton universality discussed in the next section) and hence independent of any possible oscillations. Since the observed flux is consistent with expectations, this confirms the correctness of the solar model. On the other hand, the observed flux from (a) is only about 1/3 of expectations, implying that about 2/3 of the electron neutrinos from the original decay process (3.32) have transformed to μ and/or τ neutrinos before being detected at the surface of Earth. The flux for (c) would then be due to a mixture of approximately 1/3 electron neutrinos and 2/3 μ/τ neutrinos and the observed flux, which is also below expectations for no oscillations, is consistent with this assumption.
p106, Table 3.6: the expressions for the K and K* masses are wrong. The correct expression follow directly from Eqn (3.81) a line or two above and are:
K = m + ms –(3a / 4 m ms ) and K* = m + ms +(a / 4 m ms)
Eqn (3.82) should be changed in the same way.
p169, Eqn (5.26): lhs should be –Q**2
p170, line 10: Q becomes q ; Eqn (5.30): all the Q's become q's : Eqn (5.31): the Q**2 and Q in the middle of this equation become lower case q's
Corrections
to problems and their solutions
(For completeness I have listed even trivial typographical errors. The notation is: line n means n lines from top, including equations; line –n means n lines from the bottom, excluding footnotes. Comments are in [ ] brackets.)
p30, Problem 1.1: Equation (1.35) becomes the Yukawa potential (1.35)
p110, Problem 3.11: rewording: ..... Assuming a two-component model with maximal mixing (alpha = 45 degrees) ..... mass difference of the anti-electron neutrino and its oscillating partner.
p150, Problem 4.13: The density of iron should be 1.14 x 10^4 kg m^-3. The solution is therefore changed. See below p370
p214, problem 6.6: Insert: Use sin(thetaW)^2 = 1/4.
p252, Problem 7.5: Delete the factor e from the expression for the quadrupole moment
p354, Problem C.2: denominator in expression for d should be cot and not cosec (solution is correct)
p357, Solution to Problem 1.8: the solution for R is dimensionless, not fm
p358, Solution to Problem 2.1: upper limits in integrals is a not r
p359, Solution to Problem 2.3: The numerical value should be 4.88 fm and not 6.56 fm as given.
p360, Solution to Problem 2.11: solution 269.15 becomes 269.13
p361, Solution to problem 2.13: the log term should be 0.5 x 10^7, so the final value is 2.4, not 2.6. Also, since Pb(204) is stable, the term in tau(204) should be ignored.
p365, Solution to Problem 3.8: there should be a factor hbar**2 dividing the spin terms. The final solution is correct.
p365, Solution to Problem 3.11: There is a numerical error. Using 0 < P(e to x) < 0.3 gives 0 < m^2 < 6.9x10^(-3)
p366, Solution to Problem 4.1: second minus in first equation should be a plus. The final solution is correct.
p370, Solution to Problem 4.10: the term mc^2 in the expression for n should be just m.
p370, Solution Problem 4.13: last line, because of the correction on p150 (see above), 10^33 becomes 10^30 and hence l = 2.3 x 10^3 m
p370, Solution to Problem 4.14: The target contains 5.30 x 10^24 protons and not 1.07 x 10^25, and there are two photons per interaction. Thus the number of photons produced per second is 848.
p380, Solution to Problem 6.9: replace almost 20 times with about 10 times.
p380, Solution to Problem 6.8: no propagator 'means' no propagator with a W-mass factor
p382, Solution to Problem 7.3: in line-4 neutron becomes proton
In the second suggested method of explaining the spin-parity of the excited state, it would more likely that the two d5/2 protons would combine to give Jp = 0+, which would not give the desired result. So, an alternative would be to promote one of the p3/2 neutrons to the d5/2 shell, so that the final neutron and proton in d5/2 could combine to give JP = 2+, which when combined with the unpaired 3/2- neutron would give the required quantum numbers.
p384, Solution to Problem 7.7: the factor R has been omitted in the numerical solution. Thus line 1 becomes
t1/2(Th) = t1/2(Cf) R(Th) [exp(G(Th) – exp(G(Cf))] / R(Cf)
Then t1/2(Th) changes from 4.0 yrs to 3.9 yrs.
p384, Solution to Problem 7.10: The form of the indefinite integral has been incorrectly printed in the solution. (It is given correctly in the solution to Problem 7.11.) However, the value for F is 3 x 10^(-10) as given.
p391, Solution to Problem B.8: line before the end, p and E should have primes. The final solution is correct.
p391, Soln to Problem B.9: This has used the proton mass 0.983 GeV/c**2. Using the correct value 0.938 GeV/c**2 changes the final value from 0.54 m to 0.58 m.
p392, Solution to Problem B.10: on the last line the prime should be on the first E and not the second. The final solution is correct.
p50, lines 2–4:
The current Figure 2.8 does not illustrate the text very well. The figure below, specifically for the decay sequence (2.45) is much better.
p51-52: The argument leading to the pairing term might suggest that this can be deduced from the Fig 2.9 (i.e. the Fermi gas model) which is not correct. It should be replaced with:
The final contribution is the empirical pairing term with the form (2.52). This arises from the tendency of like nucleons in the same spatial state to couple pairwise to configurations with spin-0. When coupled like this, the wavefunctions of the two nucleons heavily overlap and so on average they are closer together than when coupled in other configurations, and hence are more tightly bound. When we have an odd number of nucleons, this term does not contribute. Thus, when both Z and N are odd, we gain binding energy by converting one of the odd protons to a neutron (or vice versa) so that it can now form a pair with its formerly odd partner. The evidence for this is that there are only four stable nuclei with odd N and Z, whereas there are 167 with even N and Z. The form used for f5 is empirical, but f(A)=a5/sqrt(A) represents the trend of the data for the pairing energies and is often used.
p82: The comments on the SNO experiment are confused (and wrong in places). The last part of the paragraph starting from the sentence 'More recent experiments .....' should be replaced with the paragraph below.
A more recent experiment (SNO) has studied the reactions:
(a) νe + d → e– + p + p , (b) νx + d → νx +p + n , (c) νx + e– → νx + e– , (3.33)
where x denotes any lepton flavour (e, μ, τ) and d is the deuteron. The cross-section for (b) is independent of the lepton flavour (this is a consequence of ‘lepton universality discussed in the next section) and hence independent of any possible oscillations. Since the observed flux is consistent with expectations, this confirms the correctness of the solar model. On the other hand, the observed flux from (a) is only about 1/3 of expectations, implying that about 2/3 of the electron neutrinos from the original decay process (3.32) have transformed to μ and/or τ neutrinos before being detected at the surface of Earth. The flux for (c) would then be due to a mixture of approximately 1/3 electron neutrinos and 2/3 μ/τ neutrinos and the observed flux, which is also below expectations for no oscillations, is consistent with this assumption.
p106, Table 3.6: the expressions for the K and K* masses are wrong. The correct expression follow directly from Eqn (3.81) a line or two above and are:
K = m + ms –(3a / 4 m ms ) and K* = m + ms +(a / 4 m ms)
Eqn (3.82) should be changed in the same way.
p169, Eqn (5.26): lhs should be –Q**2
p170, line 10: Q becomes q ; Eqn (5.30): all the Q's become q's : Eqn (5.31): the Q**2 and Q in the middle of this equation become lower case q's
(For completeness I have listed even trivial typographical errors. The notation is: line n means n lines from top, including equations; line –n means n lines from the bottom, excluding footnotes. Comments are in [ ] brackets.)
p30, Problem 1.1: Equation (1.35) becomes the Yukawa potential (1.35)
p110, Problem 3.11: rewording: ..... Assuming a two-component model with maximal mixing (alpha = 45 degrees) ..... mass difference of the anti-electron neutrino and its oscillating partner.
p150, Problem 4.13: The density of iron should be 1.14 x 10^4 kg m^-3. The solution is therefore changed. See below p370
p214, problem 6.6: Insert: Use sin(thetaW)^2 = 1/4.
p252, Problem 7.5: Delete the factor e from the expression for the quadrupole moment
p354, Problem C.2: denominator in expression for d should be cot and not cosec (solution is correct)
p357, Solution to Problem 1.8: the solution for R is dimensionless, not fm
p358, Solution to Problem 2.1: upper limits in integrals is a not r
p359, Solution to Problem 2.3: The numerical value should be 4.88 fm and not 6.56 fm as given.
p360, Solution to Problem 2.11: solution 269.15 becomes 269.13
p361, Solution to problem 2.13: the log term should be 0.5 x 10^7, so the final value is 2.4, not 2.6. Also, since Pb(204) is stable, the term in tau(204) should be ignored.
p365, Solution to Problem 3.8: there should be a factor hbar**2 dividing the spin terms. The final solution is correct.
p365, Solution to Problem 3.11: There is a numerical error. Using 0 < P(e to x) < 0.3 gives 0 < m^2 < 6.9x10^(-3)
p366, Solution to Problem 4.1: second minus in first equation should be a plus. The final solution is correct.
p370, Solution to Problem 4.10: the term mc^2 in the expression for n should be just m.
p370, Solution Problem 4.13: last line, because of the correction on p150 (see above), 10^33 becomes 10^30 and hence l = 2.3 x 10^3 m
p370, Solution to Problem 4.14: The target contains 5.30 x 10^24 protons and not 1.07 x 10^25, and there are two photons per interaction. Thus the number of photons produced per second is 848.
p380, Solution to Problem 6.9: replace almost 20 times with about 10 times.
p380, Solution to Problem 6.8: no propagator 'means' no propagator with a W-mass factor
p382, Solution to Problem 7.3: in line-4 neutron becomes proton
In the second suggested method of explaining the spin-parity of the excited state, it would more likely that the two d5/2 protons would combine to give Jp = 0+, which would not give the desired result. So, an alternative would be to promote one of the p3/2 neutrons to the d5/2 shell, so that the final neutron and proton in d5/2 could combine to give JP = 2+, which when combined with the unpaired 3/2- neutron would give the required quantum numbers.
p384, Solution to Problem 7.7: the factor R has been omitted in the numerical solution. Thus line 1 becomes
t1/2(Th) = t1/2(Cf) R(Th) [exp(G(Th) – exp(G(Cf))] / R(Cf)
Then t1/2(Th) changes from 4.0 yrs to 3.9 yrs.
p384, Solution to Problem 7.10: The form of the indefinite integral has been incorrectly printed in the solution. (It is given correctly in the solution to Problem 7.11.) However, the value for F is 3 x 10^(-10) as given.
p391, Solution to Problem B.8: line before the end, p and E should have primes. The final solution is correct.
p391, Soln to Problem B.9: This has used the proton mass 0.983 GeV/c**2. Using the correct value 0.938 GeV/c**2 changes the final value from 0.54 m to 0.58 m.
p392, Solution to Problem B.10: on the last line the prime should be on the first E and not the second. The final solution is correct.
(For
completeness I have listed even trivial typographical errors. The
notation is: line n means n lines from top, including equations; line
–n means n lines from the bottom, excluding footnotes. Comments are in
[ ] brackets.)
p19, Eqn (1.33): c**4 becomes c**2
p22, line 10: GsubF becomes 2sqrt(2) GsubF
p38, Eqn (2.17): q in the exponent of the exponential should be q (ie bold)
p42, Eqn (2.31) This is for the nuclear not the charge radius
p99, Figure 3.11(a): one of the -1/2 should 1/2
p107, Eqn (3.88): The expression on the lhs should be the square of the sum of the spins, not the sum of the squared spins
p107, Eqn (3.90) For consistency, the term b should be divided by hbar**2
p108, Eqn (3.91) For consistency, the first term b should be divided by hbar**2
p108, Table 3.7: in the expression for sigma*, the tem 1/m should be 1/(m**2)
p108, Table 3.7: in the expression for omega, the tem 3b/(ms**2) should be 3b/(4ms**2)
p156, line -10: many becomes several
p159, caption Fig. 5.5: the labels 2-- become 2++, as correctly given in Table 5.2
p161, line 12: c^4 becomes c^2
p194, Eqn (6.24): rhs should be 4.2 x 10(-3) approx equal to 0.6 alpha
p199, Eqn (6.35): GsubF becomes 2sqrt(2) GsubF
p199, Eqn (6.36): lhs should be divided by sqrt(2)
p226, line 1: dipole becomes quadrupole
p227, Figure 7.4: the 1F state is split into 1F5/2 and a 1F7/2 states. The former is incorrectly written as 1F1/2
p233, Eqns (7.34) and (7.35): delete the factor e from the rhs of these equations
p235, Eqn (7.40): in the denominator, (2j+1) becomes (2j+3)
p236, line 6: Neils becomes Niels
p243, Eqn (7.60): lhs should be divided by sqrt(2)
p273, Eqn (8.45a): H on rhs becomes He
p333, Eqn (A.14): following from the above, the denominator in the last tem becomes 2mL^2
p334, line 7: (L / pi) becomes (pi / L). The rest of the derivation is correct.
p393, line 6: Fernlow becomes Fernow
1
Last updated February 2007