# Solved Problems on the Particle Nature of Matter

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Order NowCharles Asman, Adam Monahan and Malcolm McMillan Department of Physics and Astronomy University of British Columbia, Vancouver, British Columbia, Canada Fall 1999; revised 2011 by Malcolm McMillan Given here are solutions to 5 problems on the particle nature of matter. The solutions were used as a learning-tool for students in the introductory undergraduate course Physics 200 Relativity and Quanta given by Malcolm McMillan at UBC during the 1998 and 1999 Winter Sessions. The solutions were prepared in collaboration with Charles Asman and Adam Monaham who were graduate students in the Department of Physics at the time. The problems are from Chapter 3 The Particle Nature of Matter of the course text Modern Physics by Raymond A. Serway, Clement J. Moses and Curt A. Moyer, Saunders College Publishing, 2nd ed., (1997).

Coulombâs Constant and the Elementary Charge When solving numerical problems on the particle nature of matter it is useful to note that the product of Coulombâs constant k = 8.9876 Ă 109 m2 / C2 (1) and the square of the elementary charge e = 1.6022 Ă 10â19 C is ke2 = 1.4400 eV nm = 1.4400 keV pm = 1.4400 MeV fm where eV = 1.6022 Ă 10â19 J (4) (3) (2)

Breakdown of the Rutherford Scattering Formula: Radius of a Nucleus Problem 3.9, page 39 It is observed that Î± particles with kinetic energies of 13.9 MeV or higher, incident on copper foils, do not â4 obey Rutherfordâs (sin Ï/2) scattering formula. âą Use this observation to estimate the radius of the nucleus of a copper atom. Assume that the nucleus remains ïŹxed in a head-on collision with an Î± particle. Solution Rutherfordâs (sin Ï/2) scattering formula governs the scattering of an Î± particle by an atomic nucleus when the interaction between the Î± particle and the nucleus is due entirely to the Coulomb force. â4

At the distance d of closest approach in a head-on collision, the kinetic energy of the Î± particle is zero and therefore, by conservation of energy, the incident kinetic energy K of the Î± particle is equal to the potential energy of the system: k(Ze)(2e) K= (5) d where Z is the atomic number of the nucleus. Rutherfordâs formula breaks down when the Î± particle is close enough to the nucleus to be inïŹuenced by the strong nuclear force. The distance d calculated with the lowest kinetic energy K at which the Rutherfordâs formula breaks down is an estimate of the radius of the nucleus. For copper, Z = 29 and K = 13.9 MeV, so d = 6.00 fm. The value of the radius calculated using 1.3A1/3 fm with A = 64 is 5.2 fm.

Bohrâs Model of the Atom: Radii and Speeds Problem 3.14, page 39 âą Use text Eq. (3.35): rn = a0 n2 52.9n2 n = = pm me ke2 Z Z Z 2 2 to calculate the radius of the ïŹrst, second and third Bohr orbits of hydrogen; a0 = 2 /(me ke2 ) = 52.9 pm is the Bohr radius of hydrogen. That is, a0 = r1 when Z = 1. âą Determine the electronâs speed in the same three orbits. Is a relativistic correction necessary? Explain. Solution Eq. (6) gives the radius rn of the nth Bohr orbit of a single electron orbiting a ïŹxed nucleus of charge +Ze. For hydrogen (Z = 1), Eq. (6) yields r1 r2 r3 It follows using text Eq. (3.24): me vr = n that the speed vn of an electron in the nth Bohr orbit is vn = so v1 v2 v3 = = = 7.30 Ă 10â3 c 3.65 Ă 10â3 c 2.43 Ă 10â3 c (12) (13) (14) nme a0 = 7.30 Ă 10â3 c n (11) (10) = = = 52.9 pm 212 pm 476 pm (7) (8) (9)

These speeds are all very much less than c so relativistic corrections are small. Energy corrections are of the order of Â”eV.

Bohrâs Model of the Atom: Photon Emission Problem 3.24, page 140

A hydrogen atom originally at rest in the n = 3 state decays to the ground state with the emission of a photon. âą Calculate the wavelength of the emitted photon. âą Estimate the recoil momentum of the atom. Where does this energy come from? Solution Text Eq. (3.33): 1 =R Î» where R= 1 1 2 â n2 nf i (15)

ke2 = 1.0973 Ă 107 mâ1 = 1/91.13 nm 2a0 hc gives the wavelength Î» of the photon emitted when a hydrogen atom initially at rest in the ni state decays to the nf state. It follows that in the decay from n = 3 state to the ground state (n = 1), the wavelength Î» of the emitted photon is 9 = 103 nm. (17) Î»= 8R The energy E and magnitude p of the momentum of the emitted photon are E= p= hc = 12.0 eV Î» (18)

E = 12.0 eV/c. (19) c Momentum is conserved in the process and therefore p is also the magnitude of the momentum of the recoiling hydrogen atom. The kinetic energy K of the recoiling hydrogen atom is K= p2 = 77.2 neV. 2mH (20)

Eq. (17) assumes that all the transition energy is carried oïŹ by the photon and that K = 0. Comparison of Eqs. (18) and (20) shows that this is a reasonable assumption.

Bohrâs Model of the Atom: Muonic Atom Problem 3.33, page 141 A muon is a particle with a charge equal to that of an electron and a mass equal to 207 times the mass of the electron. Muonic lead is formed when 208 Pb captures a muon to replace an electron. Assume that the muon moves in such a small orbit that it sees a nuclear charge Z=82. âą According to Bohrâs theory, what are the radius and energy of the ground state of muonic lead? Use the concept of reduced mass in solving the problem. Solution Eq. (6) gives the radius of the nth Bohr orbit and text Eq. (3.36): En = â ke2 2a0 Z2 n2 =â 13.58Z 2 n2 eV (21)

gives the energy of the nth Bohr level for a single electron orbiting a ïŹxed nucleus of charge +Ze.

The model may be extended further when the electron is replaced by a particle of mass m and charge e. In this case, Eqs. (6) and (21) are replaced by rn = me Â” Â” me Â”= 52.9n2 Z 13.58Z 2 n2 mmN m + mN pm (22)

En = â where

eV

(23)

(24)

where mN is the mass of the nucleus; Â” is the reduced mass of the orbiting particle and the nucleus. A negative muon has charge e and mass m = 207me . Muonic lead is formed when 208 Pb captures a negative muon to replace an electron. Assuming that the muon moves in such a small orbit that it only feels a positive charge corresponding to Z = 82 (i.e., that screening eïŹects caused by the other electrons surrounding the nucleus are negligible), then Eqs. (22) and (23) with n = 1 give the radius r1 and ground state energy E1 of the ground state of muonic lead according to Bohr theory: r1 = 3.1 fm E1 = â19 MeV. Nuclear physics studies indicate that the nuclear radius rnucleus is given approximately by rnucleus = 1.3 A 3 fm = 7.7 fm when A = 207. 1

(25) (26)

(27)

The Bohr model prediction (Eq. (25)) is of the right order of magnitude but too small; the assumption that the muon feels a positive charge corresponding to Z = 82 is incorrect. The radius r of the ground state of muonic lead is equal to the nuclear radius R when the eïŹective charge in the Bohr model corresponds to Z = 33. The calculations indicate that muonic atoms can be used to probe the atomic nucleus.

Bohrâs Model of the Atom: Positronium Problem 3.35, page 141 Positronium is a hydrogen-like atom consisting of a positron (a positively charged electron) and an electron revolving around each other. âą Using the Bohr model, ïŹnd the allowed radii (relative to the center of mass of the two particles) and the allowed energies of the system. Use the concept of reduced mass in solving the problem. Solution Eqs. (22) and (23) with m = mN = me and Z = 1 give the Bohr model values for the allowed radii rn and energies En of positronium: rn = 105.8n2 pm (28) En = â 6.79 eV n2 (29)