Subatomic Physics Solutions Manual

The following sections are included:

• Preface

• Contents

1.2 The range of the gravitational and electromagnetic interactions is infinite. The range of the hadronic (strong) interaction is of the order of ħ/𝑚𝑐, or about 1 fm = 10^-13 cm. The range of the weak interaction is given by ħ/𝑚_𝑊𝑐 or 2.5 X 10^-3 fm = 2.5 X 10^-16 cm …

2.1 We need ƛ = ħ/p = (ħc)/(pc) ≈ 1 fm; pc ≈ ħc/1 fm ≈ 200 MeV fm/1 fm = 200 MeV. Since the electrons are highly relativistic at this momentum, we find E_𝐾 ≈ 200 MeV …

3.1 From Fig. 3.5 we notice that the logarithm of the energy loss, d𝐸/dx, is approximately a straight line for the range of kinetic energy required. That is, d𝐸/dx satisfies a power law …

4.4 From Fig. 3.7 we get that for a 14 keV gamma ray, the absorption length is 2 X 10^-2 g/cm². The density of NaI is 3.7 g/cm³. Therefore, the absorption length is 2 X 10^-2 g cm^-2/3.7 g cm^-3 = 5.4 X 10^-3 cm. This thickness is approximately the optimum one …

5.1 No. The gravitational force on a particle of zero mass and energy E is proportional to its effective mass, E/c². The force on such a particle exerted by a mass M, is GME/(cr)² …

6.1 The maximum energy loss and momentum transfer occur when, in the c.m., the alpha travels after the collision at 180° relative to its initial direction. If the momentum of the alpha particle before the collision is 𝑝 in the laboratory system, then the velocity of the c.m. is 𝑝/(𝑀+𝑚) ≈ (𝑝/𝑀)(1 ‒ 𝑚/𝑀), where 𝑀 is the mass of the alpha particle and 𝑚 that of the electron …

7.1

8.1 (a) ψ′ *𝐹ψ→ψ^{𝑅′ *} 𝐹ψ^𝑅= ψ′^* 𝑈^†𝐹𝑈ψ = ψ′ *𝐹^𝑅ψ with 𝐹^𝑅 = 𝑈^†𝐹𝑈 …

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11.1 The neutron-proton mass difference is 1.29 MeV. The maximum kinetic energy available to the electron and anutrino is therefore 1.29 MeV ‒ 𝑚_𝑒c² = 1.29 ‒ 0.51 = 0.78 MeV. The maximum recoil energy occurs when the electron and antineutrino are emitted in the same direction, opposite to that of the proton …

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15.1 Like protons, the u and d quarks will lose energy through scattering from bound electrons; see Chapter 3. The electrons can be excited to higher states, including the continuum. If they are sufficiently energetic, scattering from nuclei can also occur. A combination of such quarks (with integral charge) may come to rest inside the solid and form an atom with electrons in the solid…

16.1 The binding energy of an atom can be approximated by the ionization energy. The last electron is shielded from the nucleus by all the other electrons, so that it sees an effective charge of e rather than Ze. We can use the hydrogen atom to estimate the ionization energy of an atom. For hydrogen, the binding energy of the electron is 13.5 eV. This is also approximately the ionization energy of atoms and is much less than the average binding energy per nucleon in a nucleus, about 8 MeV. The correction is therefore of the order of 1 x 10^-5

17.3 . On averaging the second teerm over all nucleons in a closed shell…

18.1Assume that , the charge density, is located close to the origin, and that the potential ϕ is slowly varying with the density. Then we can expand ϕ…

19.1 The Coulomb barrier is given by