the foundations of modern physics

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37-1

Conceptual Questions

37.1. It shifts toward longer wavelengths. The wavelength with peak intensity is inversely proportional to the absolute temperature.

37.2. (a) Cathode rays are deflected by a magnetic field as if they are negatively charged particles. Cathodes made of any metal produce cathode rays, suggesting the rays are a fundamental entity. If they are particles, that would explain the deflection by a magnetic field, the motion away from other charged particles, and the fluorescence where they strike glass tubes. (b) They could travel in a straight line for 150 times the mean free path of known particles. They could penetrate thin foils in the same way that radio waves are known to penetrate thin foils. Waves and cathode rays both can carry energy and momentum and also cause fluorescence.

37.3. The force due to gravity is very much smaller than the electric and magnetic forces because of the very small mass of the cathode-ray particles.

37.4. It was conclusive proof that cathode rays are negatively charged.

37.5. Cathode rays emitted by various hot cathodes have the same charge-to-mass ratio independent of the material from which the cathode is made.

37.6. A proton would also pass through with no deflection. The reason is that both the force on the charged particle due to the electric field and the force on the charged particle due to the magnetic field have q in them, so if we set the magnitudes of the two forces equal to each other then q cancels.

37.7. (a) The atomic number of an element, represented by Z, is the number of protons in the nucleus. The atomic number of hydrogen is Z = 1, helium is Z = 2, and lithium is Z = 3. However, the masses of these three elements do not progress in this linear manner. The mass of helium is four times the mass of hydrogen and the mass of lithium is seven times the mass of hydrogen. If a nucleus contains only Z protons, then helium should be just twice as massive as hydrogen and lithium three times as massive as hydrogen. So there must be something else in the nucleus. (b) Thomson and his student Aston found that many elements consist of atoms of differing mass. For example, neon (Z = 10) atoms are found to have integral masses 20 u, 21 u, and 22 u. Potassium atoms (Z = 19) are found to have integral masses 39 u, 40 u, and 41 u. Because the masses of the different isotopes differ by ≈1 u, the mass of neutron must be ≈1 u, the same as the mass of a proton.

THE FOUNDATIONS OF MODERN PHYSICS

37

37-2 Chapter 37


37.8. Equation 37.6 shows that the radius r of curvature through which a charged particle is deflected by a magnetic field is

mvrqB

=

If the field B and the velocity v of the particles is the same, then the radius depends only on the mass-to-charge ratio m/q. The alpha particle is a doubly ionized He nucleus, so its charge is +2e and its mass is approximately 4 u. The cathode-ray particle has a charge of e− and a mass of about 310 u.− Therefore, we would expect that the much more massive alpha particle will be deflected much less than the cathode-ray particle.

37.9. Scientists at the time could not imagine the extremely high density of the tiny nucleus. They also had no idea what would hold protons together in a nucleus when there was a known repulsive force between protons. In addition, classical electromagnetic theory predicted that an accelerating electron would radiate away all its energy in microseconds and collapse into the nucleus!

37.10. Alpha particles scattered through large angles because they had near collisions with very massive and highly charged particles. Only small deflections were expected for an alpha particle passing through a Thomson atom because the negatively and positively charged matter in this model is mixed together in a blob so that any electric forces from this blob would mostly cancel out, resulting in only weak interaction forces.

37.11. The definition of an electron volt is the energy acquired by an electron accelerated through a potential difference of 1 V. It would gain 100 eV of energy accelerating across 100 V. But the helium nucleus has a charge of +2, so it would gain 200 eV of energy.

37.12. (a) 6Li+ (b) 13C−

Exercises and Problems

Exercises Section 37.2 The Emission and Absorption of Light 37.1. Model: Balmer’s formula predicts a series of spectral lines in the hydrogen spectrum. Solve: Substituting into the formula for the Balmer series,

2 22 2

91 18 nm 91 18 nm 410 3 nm1 11 12 62 n

λ λ. .= ⇒ = = .⎛ ⎞ −−⎜ ⎟⎝ ⎠

where n = 3, 4, 5, 6, … and where we have used n = 6. Likewise for n = 8 and n = 10, 389 0 nmλ = . and 379 9 nm,λ = . respectively.

37.2. Model: Use Equation 37.4, which is the Balmer formula. Visualize: Please refer to Figure 37.7. Solve: (a) The wavelengths in the hydrogen emission spectrum are 656.6 nm, 486.3 nm, 434.2 nm, and 410.3 nm. The formula for the Balmer series can be written

2 21 1 91.18 nm

m n λ− =

where m = 1, 2, 3, … and n = m + 1, m + 2, …. For the first wavelength, 2 2

2 2 2 21 1 91 18 nm 0 1389 0 1389

656 5 nmn m

m n n m. −− = = . ⇒ = .

.

This equation is satisfied when m = 2 and n = 3. For the second wavelength (486.3 nm) the equation is satisfied for m = 2 and n = 4. Likewise, for the next two wavelengths m = 2 and n = 5 and 6.

The Foundations of Modern Physics 37-3


(b) The fifth line in the spectrum will correspond to m = 2 and n = 7. Its wavelength is

( ) ( )221 12 7

91 18 nm 196 (91 18 nm) 397 1 nm45

λ . ⎛ ⎞= = . = .⎜ ⎟⎝ ⎠−

37.3. Model: Use Equation 37.4 with m = 1 and n = 2, 3, 4, and 5. This series of spectral lines is called the Lyman series. Solve: The formula for the Lyman series simplifies to

291 18 nm , where 2, 3, 4, and 5

1n

nλ −

.= =−

For n = 2, ( ) 114(91 18 nm) 1 121 6 nm.λ

−= . − = . For n = 3, ( ) 11

9(91 18 nm) 1 102 6 nm.λ−

= . − = . Likewise, for n = 4

and n = 5, 97 3 nmλ = . and 95 0 nm,λ = . respectively.

37.4. Model: Use Equation 37.4, which is the Balmer formula. Solve: (a) Equation 37.4 can be written as

2 2

2 2 2 21 1 91 18 nmn m

m n n m λ− .− = =

where m = 1, 2, 3, … and n = m + 1, m + 2, .... For λ = 102.6 nm,

2 2

2 291 18 nm 0.8887102 6 nm

n mn m

− .= =.

This equation is satisfied by m = 1 and n = 3. For λ = 1876 nm,

2 2

2 291 18 nm 0 04861876 nm

n mn m

− .= = .

This equation is satisfied by m = 3 and n = 4. (b) The ultraviolet wavelengths are < 400 nm. The infrared wavelengths are > 700 nm. So, λ = 102.6 nm is in the ultraviolet region and λ = 1876 nm is in the infrared region.

37.5. Model: Assume the metal sphere is a blackbody (so the emissivity 1).e = Visualize: First use Wein’s law (Equation 37.2) to find the temperature, then use Stefan’s law (Equation 37.1) to determine the power radiated. We are given 1 0 cmR = . and peak 2 0 m 2000 nmλ μ= . = . Solve:

6

4 8 2 4 2 4

2 90 10 nm K 1450 K2000 nm

(1)(5 67 10 Wm K )(4 )(1 0 cm) (1450 K) 315 W 0 32 kW

T

Q e ATt

σ π− − −

. × ⋅= =

= = . × . = ≈ .Δ

Assess: The sphere radiates more than 3100 W light bulbs, but it has a larger surface area than the filaments, so the answer is reasonable.

37.6. Model: Assume the blackbodies obey Wein’s law in Equation 37.2: 6peak (2 90 10 nmK)/ .Tλ = . ×

Visualize: We want to solve for T in the equation above. We are given peak 300 nm, 3000 nmλ = . Solve:

6

peak

2 90 10 nm KTλ

. × ⋅=

37-4 Chapter 37


(a) 6

32 90 10 nm K 9667 K 9 39 10 C300nm

T . × ⋅= = = . × °

(b) 62 90 10 nm K 966 7K 694 C

3000nmT . × ⋅= = . = °

37.7. Model: Assume the ceramic cube is a blackbody (so the emissivity 1).e =

Visualize: First use Stefan’s law (Equation 37.1), 4,Qt e ATσΔ = to find the temperature, then use Wein’s law

(Equation 37.2) to get the peak wavelength. We are given 26(3 0 cm 3 0 cm) 0 0054 mA = . × . = . and / 630 W.Q tΔ = Solve: Solve Stefan’s law for T:

4 4 8 2 4 2/ 630 W 1198 K

(1)(5 67 10 Wm K )(0 0054 m )Q tTe Aσ − − −

Δ= = =. × .

Now plug this temperature into Wein’s law. 6 6

peak2 90 10 nmK 2 90 10 nmK 2420 nm 2 4 m

1198 KTλ μ. × . ×= = = = .

Section 37.3 Cathode Rays and X Rays

Section 37.4 The Discovery of the Electron 37.8. Model: Current is defined as the rate at which charge flows across a cross-sectional area. Solve: Since the current is / and / ,Q t Q N eΔ Δ Δ = the number of electrons per second is

810 1 10 1

1910 nA 1 0 10 C/s 6 25 10 s 6 3 10 s

1 60 10 CN

t e

−− −

−. ×= = = . × ≈ . ×

Δ . ×

Thus, for 10 1 101 0 s, (6 3 10 s )(1 0 s) 6 3 10 .t N −Δ = . = . × . = . ×

37.9. Model: Assume the fields between the electrodes are uniform and that they are zero outside the electrodes. Visualize: Without the external magnetic field B, the electrons will be deflected up toward the positive electrode. The magnetic field must therefore be directed out of the page to exert a balancing downward force on the negative electron. Solve: In a crossed-field experiment, the magnitudes of the electric and magnetic forces on the electron are given by Equation 37.8. The magnitude of the magnetic field is

33

6/ (200 V)/(8 0 10 m) 5 0 10 T

5 0 10 mE V dBv v

−−Δ . ×= = = = . ×

. ×

Thus 3(5 0 10 T, out of page).B −= . ×

37.10. Model: Assume the fields between the electrodes are uniform and that they are zero outside the electrodes. Visualize: Please refer to Figures 37.12 and 37.13. Solve: (a) The speed with which a particle can pass without deflection is

37

3/ (600 V)/(5 0 10 m) 6 0 10 m/s

2 0 10 TE V dvB B

−

−Δ . ×= = = = . ×

. ×

(b) The radius of cyclotron motion in a magnetic field is 31 7

19 39 11 10 kg 6 0 10 m 0 17 m 17 cm1 60 10 C 2 0 10 T

m vre B

−

− −

⎛ ⎞⎛ ⎞. × . ×⎛ ⎞= = = . =⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟. × . ×⎝ ⎠ ⎝ ⎠⎝ ⎠



Section 37.5 The Fundamental Unit of Charge 37.11. Model: Assume the electric field (E = ΔV/d) between the plates is uniform. Visualize: Please refer to Figure 37.14. To balance the weight, the electric force must be directed toward the upper electrode, which is more positive than the lower electrode. Solve: Since ( ) 34

drop 3 ,m V Rπρ ρ= = the equation drop dropm g q E= gives

193 19 3

4 3 243 3

( / )(15 ) [(25 V)/(0 012 m)](15)(1 60 10 C) 1 4163 10 m 0 52 m(860 kg/m )(9 8 m/s )

V d eR Rgπ π

μρ

−−Δ . . ×= = = . × ⇒ = .

.

37.12. Model: Assume the electric field (E = ΔV/d) between the plates is uniform. Visualize: Please refer to Figure 37.14. Solve: (a) The mass of the droplet is

3 3 6 3 16 16drop

4 4(885 kg/m ) (0 4 10 m) 2 37 10 kg 2 4 10 kg3 3

m V Rπ πρ ρ − − −⎛ ⎞= = = . × = . × ≈ . ×⎜ ⎟⎝ ⎠

(b) In order for the upward electric force to balance the gravitational force, the charge on the droplet must be

drop drop

16 2drop 18 18

drop 3(2 37 10 kg)(9 8 m/s ) 1 28 10 C 1 3 10 C

20 V/11 10 m

GEq F m g

m gq

E

−− −

−

= =

. × .= = = . × ≈ . ××

(c) Because the electric force is directed toward the electrode at the higher potential (or more positive plate), the charge on the droplet is negative. The number of surplus electrons is

18droplet

191 28 10 C 81 60 10 C

qN

e

−

−. ×= = =. ×

37.13. Model: The charge on an object is an integral multiple of a certain minimum charge value. Solve: From smallest charge to the largest charge, the measured charges on the drops are 192.66 10 C, 3.99−× ×

19 1910 C, 9.31 10 C,− −× and 1910.64 10 C.−× The differences between these values are

19 191.33 10 C, 2.66 10 C,− −× × 192.66 10 C,−× and

191.33 10 C.−× These differences are a multiple of 191.33 10 C.−× Thus, the largest value of the

fundamental unit of charge that is consistent with the charge measurements is 191.33 10 C.−×

Section 37.6 The Discovery of the Nucleus

Section 37.7 Into the Nucleus 37.14. Model: The electron volt is a unit of energy and is defined as the kinetic energy gained by an electron or proton if it accelerates through a potential difference of 1 volt. Assume the speeds are low enough to neglect special relativity. Solve: (a) Converting electron volts to joules,

196 121 6 10 J7 0 MeV 7 0 10 eV 1 12 10 J

1 0 eV

−−. ×. = . × × = . ×

.

Using the definition of kinetic energy 212 ,K mv= the speed of the neutron is

127

272 2(1 12 10 J) 3 7 10 m/s

1 67 10 kgKvm

−

−. ×= = = . ×

. ×

(b) Likewise, the speed of the helium atom is 12

727 27

2(15 MeV) 2(2 40 10 J) 2 7 10 m/s4(1 67 10 kg) 4(1 67 10 kg)

v−

− −. ×= = = . ×

. × . ×

37-6 Chapter 37


(c) The mass of the particle is 16

312 7 2 7 2

2 2(1 14 keV) 2(1 82 10 J) 9 1 10 kg(2 0 10 m/s) (2 0 10 m/s)

Kmv

−−. . ×= = = = . ×

. × . ×

The particle is an electron.

37.15. Model: The electron volt is a unit of energy and is defined as the kinetic energy gained by an electron or proton if it accelerates through a potential difference of 1 volt. Assume the speeds are low enough to neglect special relativity. Solve: (a) Converting electron volts to joules,

19171 60 10 J300 eV 300 eV 4 80 10 J

1 00 eV

−−. ×= × = . ×

.

Using the definition of kinetic energy 212 ,K mv=

177

312 2(4 80 10 J) 1 03 10 m/s

9 11 10 kgKvm

−

−. ×= = = . ×

. ×

(b) Likewise, the speed of the H+ ion is 19

6 13

137

27

1 60 10 J3 5 MeV 3 5 10 eV 5 607 10 J1 00 eV

2(5 607 10 J) 2 6 10 m/s1 67 10 kg

v

−−

−

−

. ×. = . × × = . ×.

. ×= = . ×. ×

(c) The mass of the particle is 13

272 7 2

2 2(2 09 MeV) 1 60 10 J 6 69 10 kg1 00 MeV(1 0 10 m/s)

Kmv

−−. . ×= = = = . ×

.. ×

The mass of the particle is the same as the mass of four protons (or two protons and two neutrons). It is probably an alpha particle.

37.16. Model: The electron volt is a unit of energy. It is defined as the energy gained by an electron if it accelerates through a potential difference of 1 volt. Visualize:

Solve: (a) The figure for part (a) shows a Li++ ion accelerating from rest across a parallel-plate capacitor. The energy conservation equation f f i i givesK qV K qV+ = +

f i i f( ) 0 (2 ) 2 (5000 V) 10,000 eV 10 keVK K q V V e V e= + − = + Δ = = =



(b) The potential energy is 9 2 2 19 2

1415 19

0

1 ( )( ) (9 0 10 N m /C )(1 60 10 C) 1 0 eV(2 3 10 J) 0 14 MeV4 10 fm 10 10 m 1 60 10 J

e eUπε

−−

− −. × . × .⎛ ⎞= = = . × = .⎜ ⎟× . ×⎝ ⎠

(c) The energy-conservation equation for the figure for part (c) f f i i isK U K U+ = + 2

f i f i f

1919

0 J ( ) ( ) (0 200 kg)(9 8 m/s )(1 0 m 0 m)1 0 eV(1 96 J) 1 2 10 eV

1 60 10 J

K U U mg y y

−

= + − = − = . . . −

.⎛ ⎞= . = . ×⎜ ⎟. ×⎝ ⎠

37.17. Model: The electron volt is a unit of energy. It is defined as the energy gained by an electron if it accelerates through a potential difference of 1 volt. Solve: (a) The kinetic energy is

2 31 6 2 171 12 2 19

1 0 eV(9 11 10 kg)(5 0 10 m/s) (1 139 10 J) 71 eV1 60 10 J

K mv − −−

.⎛ ⎞= = . × . × = . × =⎜ ⎟. ×⎝ ⎠

(b) The potential energy is 9 2 2 19 2

189 19

0

1 ( )( ) (9 0 10 N m /C )(1 60 10 ) 1 0 eV( 2 30 10 J) 14 eV4 0 10 nm 0 10 10 m 1 60 10 J

e eUπε

−−

− −− − . × . × .⎛ ⎞= = = − . × = −⎜ ⎟. . × . ×⎝ ⎠

(c)

The figure shows a proton accelerating from rest across a parallel-plate capacitor with a potential difference of ΔV = 5000 V. The energy conservation equation f f i i isK qV K qV+ = +

f i i f( ) 0 J (5000 V) 5000 eV 5 0 keVK K q V V e V e= + − = + Δ = = = .

37.18. Visualize: The work-kinetic energy theorem is net .K WΔ = Work done on the proton in slowing it down will be .W q V= − Δ

f iK K q V− = − Δ

Solve: But since f 0,K = then

i (1 0 )(75 V) 75 eVK q V e= Δ = . =

Assess: The plate separation doesn’t matter.

37.19. Model: For a neutral atom, the number of electrons is the same as the number of protons. Solve: (a) Since 105, the BZ = atom has 5 electrons, 5 protons, and 10 – 5 = 5 neutrons.

(b) Since 137, NZ += has 6 electrons, 7 protons, and 13 – 7 = 6 neutrons.

(c) Since Z = 8, the triply charged 17O+++ ion has 5 electrons, 8 protons, and 17 – 8 = 9 neutrons.

37-8 Chapter 37


37.20. Visualize: Use Appendix C. Solve: Since lithium has 3 protons, the number of protons in the unknown isotope must be 18 3 54,× = which makes the element xenon. The atomic mass number of the isotope must be 11 12 132× = so the answer is 132Xe. Assess: Xenon seems to be the only element in Appendix C with an isotope whose atomic mass number is 132.

37.21. Model: For a neutral atom, the number of electrons is the same as the number of protons, which is the atomic number Z. An atom’s mass number is A = Z + N, where N is the number of neutrons. Solve: (a) This is a neutral atom with A = 1 + 2 = 3. The symbol is 3H. This is also known as tritium. (b) This is a positively charged ion with A = 8 + 10 = 18. The symbol is 18O+.

37.22. Model: For a neutral atom, the number of electrons is the same as the number of protons, which is the atomic number Z. An atom’s mass number A is A = Z + N, where N is the number of neutrons. Solve: (a) This is a neutral atom because it has the same number of protons and electrons. The mass number is A = Z + N = 5 + 6 = 11, so the symbol 11B. (b) This is a singly positively charged ion because it has one more proton than neutron. The atomic number (number of protons) is Z = 6, so the element is C. The mass number is A = Z + N = 6 + 8 = 14. The symbol is 14C+.

37.23. Model: For a neutral atom, the number of electrons is the same as the number of protons, which is the atomic number Z. An atom’s mass number is A = Z + N, where N is the number of neutrons. Solve: (a) For a 197 Au atom, Z = 79, so N = 197 – 79 = 118. A neutral 197 Au atom contains 79 electrons, 79 protons, and 118 neutrons. (b) Assuming that the neutron rest mass is the same as the proton rest mass, the density of the gold nucleus is

( ) ( )27

proton 17 3 nucleus 3 315 154 14.0 4 14.0

3 2 3 2

197 197(1.67 10 kg) 2.29 10 kg/m10 m 10 m

m

π πρ

−

− −

×= = = ×× ×

(c) The nuclear density in part (b) is 132.01 10× times the density of lead. Assess: The mass of the matter is primarily in the nuclei and the volume of the matter is essentially due to the electrons around the nuclei.

37.24. Model: For a neutral atom, the number of electrons is the same as the number of protons, which is the atomic number Z. An atom’s mass number is A = Z + N, where N is the number of neutrons. Solve: (a) For a 207 Pb atom, Z = 82, so N = 207 – 82 = 125. The neutral 207 Pb atom contains 82 electrons, 82 protons, and 125 neutrons. (b) The electric field strength at the surface of a lead nucleus is

9 2 2 1921

2 15 20

1 (8 99 10 N m /C )(82)(1 60 10 C) 2 34 10 V/m4 (7 1 10 m)

QErπε

−

−. × . ×= = = . ×

. ×

37.25. Model: Model the uranium nucleus as a sphere of uniform density. Visualize: The radius of the nucleus is 7.4 fm/2 = 3.7 fm. Solve: (a) The density of the nucleus is

2718 3 18 3

15 343

238 u 1.661 10 kg 1.86 10 kg/m 1.9 10 kg/m1 u(3.7 10 m)

MV

ρπ

−

−

⎛ ⎞×= = = × = ×⎜ ⎟⎜ ⎟× ⎝ ⎠

(b) Solve the density equation for R the radius of the neutron star, and insert the mass of the sun and the density found above.

303 33

17 34 3 3 (1.99 10 kg) 6.3 km3 4 4 (1.86 10 kg/m )

M MV R Rπρ π ρ π

×= = ⇒ = = =×



Assess: We have assumed the density is constant throughout a neutron star, but it is actually less than the density of a nucleus near the surface and greater than the density of a nucleus at the center of the neutron star. This means the neutrons must overlap somewhat in the center.

Problems

37.26. Model: Use the relativistic expression for the total energy. Solve: (a) The energy of the proton is

2 2 2 27 8 2p 2 2 2

9 919

1 1 (7 089)(1 67 10 kg)(3 0 10 m/s)1 / 1 (0 99)

1 eV1 065 10 J 6 66 10 eV 6660 MeV1 6 10 J

E mc mc mcv c

γ −

−−

= = = = . . × . ×− − .

= . × × = . × =. ×

(b) Likewise, the energy of the electron is

31 8 2 13 6192

1 1 eV(9 11 10 kg)(3 0 10 m/s) (5 812 10 J) 3 63 10 eV 3 6 MeV1 6 10 J1 (0 99)

E − −−

⎛ ⎞= . × . × = . × = . × = .⎜ ⎟. ×⎝ ⎠− .

Assess: The total energy E for the proton is larger than that for the electron by the factor proton electron/ 1833.m m =

37.27. Model: Use the relativistic expression for the total energy. Solve: (a) The energy of the proton is

27 8 2 192 9

p 2 2

2 2 3

(1 67 10 kg)(3 0 10 m/s) 1 60 10 J500 GeV (500 10 eV)1 eV1 /

1 / 1 879 10 0 999998

E mcv c

v c v c

γ− −

−

⎛ ⎞. × . × . ×= = = = × ⎜ ⎟⎜ ⎟− ⎝ ⎠

− = . × ⇒ = .

(b) Likewise for the electron, 31 8 2

2 2

2 2 4

(9 11 10 kg)(3 0 10 m/s)2 0 GeV1 /

1 / 2 562 10 0 99999997

Ev c

v c v c

−

−

. × . ×= . =−

− = . × ⇒ = .

37.28. Model: The total energy of each proton includes its rest energy plus its kinetic energy. Visualize: Each of the two colliding protons has 6.5 TeV of total energy. Solve: From relativity 2E mc= for the product particle at rest.

12 1923 5

p2 8 22(6.5 10 eV) 1.6 10 J 2.3 10 kg 1.4 10

1 eV(3.0 10 m/s)Em mc

−−⎛ ⎞× ×= = = × = ×⎜ ⎟⎜ ⎟× ⎝ ⎠

Assess: We have assumed the density is constant throughout a neutron star, but it is actually less than the density of a nucleus near the surface and greater than the density of a nucleus at the center of the neutron star. This means the neutrons must overlap somewhat in the center.

37.29. Model: The total energy of the proton includes its rest energy plus its kinetic energy. Visualize: The proton has 6.5 TeV of total energy. Solve: From relativity 2

p .E mcγ=

( )( )( )( )

22 2

p p 2 22

222 27 82

12 19

1 11

1.67 10 kg 3.0 10 m/s1 1 0.9999999896

6.5 10 TeV 1.6 10 J/eV

0.9999999896

E E mcE mcEmc mc

mcE

v c

γ γ ββ

β−

−

= ⇒ = ⇒ = ⇒ − = ⇒−

⎛ ⎞× ×⎛ ⎞ ⎜ ⎟= − = − = ⇒⎜ ⎟ ⎜ ⎟⎝ ⎠ × ×⎜ ⎟⎝ ⎠

=

37-10 Chapter 37


Assess: With p 7000γ ≈ these protons are extremely relativistic.

37.30. Solve: The rest energy of an electron is

2 31 8 2 140 19

1 eV(9 11 10 kg)(3 00 10 m/s) (8 199 10 J) 0 512 MeV1 6 10 J

E mc − −−

⎛ ⎞= = . × . × = . × = .⎜ ⎟. ×⎝ ⎠

The rest energy of a proton is

2 27 8 2 100 19

1 eV(1 67 10 kg)(3 00 10 m/s) 1 503 10 J 939 MeV1 6 10 J

E mc − −−= = . × . × = . × × =

. ×

37.31. Solve: (a) The kinetic energy of the electron is

2 31 8 2 1519

1 eV( 1) (1 01 1)(9 11 10 kg)(3 00 10 m/s) (0 8199 10 J) 0 00512 MeV1 6 10 J

K mcγ − −−

⎛ ⎞= − = . − . × . × = . × = .⎜ ⎟. ×⎝ ⎠

(b) Likewise for the proton, 27 8 2(1 01 1)(1 67 10 kg)(3 00 10 m/s) 9 39 MeVK −= . − . × . × = .

(c) Likewise for the alpha particle, 27 8 2(1 01 1)(4)(1 67 10 kg)(3 00 10 m/s) 37 6 MeVK −= . − . × . × = .

Assess: When the kinetic energy exceeds about 1% of the rest energy, relativistic effects become important.

37.32. Model: Mass and energy are equivalent. Solve: The energy released as kinetic energy is

2 27 8 2 1119

1 eV( ) (0 185)(1 66 10 kg)(3 00 10 m/s) (2 76 10 J) 173 MeV1 6 10 J

K E m c − −−

⎛ ⎞= Δ = Δ = . . × . × = . × =⎜ ⎟. ×⎝ ⎠

37.33. Model: Assume the fields between the electrodes are uniform and that they are zero outside the electrodes. Visualize: Please refer to Figures 37.12 and 37.13. Solve: In a crossed-field experiment, the deflection is zero when the magnetic and electric forces exactly balance each other. From Equation 37.8, the speed of the electrons is

37

3/ (150 V)/(5 0 10 m) 3 0 10 m/s

1 0 10 TE V dvB B

−

−Δ . ×= = = = . ×

. ×

When the potential difference across the plates is set equal to zero, the magnetic field deflects the electron into a circular path with radius

31 7

19 3(9 11 10 kg)(3 0 10 m/s) 0 1708 m

(1 6 10 C)(1 0 10 T)mvreB

−

− −. × . ×= = = .

. × . ×

From Figure 37.13, the angle through which the electron is deflected as it passes through the magnetic field is the angle subtended by the arc from where the electron enters the magnetic field to where the electron leaves the magnetic field. Calling this angle θ, we have

2 5 cmsin 0 1463 8 417 08 cm

Lr

θ θ.= = = . ⇒ = . °.

Assess: In view of a rather small magnetic field strength, the small deflection angle is reasonable.

37.34. Model: Assume the fields between the electrodes are uniform and that they are zero outside the electrodes. Solve: In a crossed-field experiment, the deflection is zero when the magnetic and electric forces exactly balance each other. From Equation 37.8,

2 4/ (500 V)/(1 0 10 m) 5 0 10 V/mE V dBv v v v

−Δ . × . ×= = = =



We need to obtain v before we can find B. We know that (i) a potential difference of 500 V across the plates causes the proton to deflect vertically down by 0.50 cm and (ii) the proton travels a horizontal distance of 5.0 cm in the same time (t) as it travels vertically down by 0.50 cm. From kinematics,

2 f i1f i i f i f i2

5 0 cm( ) ( ) ( 0 s)x xx xx x v t t a t t v t v

t t− .− = − + − = − ⇒ = =

The time t can be found from the vertical motion as follows: 2 21 1

f yi f i f i2 2( ) ( ) 0 50 cmi y yy y v t t a t t a t− = − + − ⇒ . =

The force on the electron is

E

19 22 2 8

27 21 1 (1 6 10 C)(500 V)0 50 10 m 4 569 10 s2 2 (1 67 10 kg)(1 0 10 m)

y ye V e VF eE ma a

d mde V tt tmd

−− −

− −

Δ Δ= = = ⇒ =

Δ . ×. × = = ⇒ = . ×. × . ×

The speed of the electron is 2 4

68 6

5 0 cm 5 0 10 m 5 0 10 V/m1 094 10 m/s 46 mT4 569 10 s 1 094 10 m/s

v Bt

−

−. . × . ×= = = . × ⇒ = =

. × . ×

Using the right hand rule we determine that this magnetic field must be directed into the page in order for the magnetic force to point opposite of the electric force. Thus (46 mT, into page)B = .

37.35. Model: Assume the fields are uniform. Visualize:

Solve: Between the plates, the fields are crossed so that the particle passes through the plates without deflection. From Equation 37.8, the speed of the particle is

6187,500 V/m 1 500 10 m/s0 1250 T

EvB

= = = . ×.

After leaving the plates, the particle experiences only a magnetic force and thus follows a circular path of radius r. Thus the charge-to-mass ratio is

67

12

1 500 10 m/s 9 581 10 C/kg(0 1250 T)( 0 2505 m)

mv q vrqB m Br

. ×= ⇒ = = = . ×. × .

This must be a proton because the proton’s charge-to-mass ratio is 19

727

1 602 10 C 9 581 10 C/kg1 672 10 kg

em

−

−. ×= = . ×. ×

37.36. Model: Assume that the electrons transfer all their energy to the foil. Solve: From Chapter 19, the rise in thermal energy of the foil is

6th (10 10 kg)(385 J/kg K)(6 0 C) 0 0231 JE mc T −Δ = Δ = × . ° = .

37-12 Chapter 37


This energy is provided by the electron impacts. Each electron that strikes the foil has been accelerated through a potential difference of 2000 V. Its kinetic energy when it strikes the foil is 162000 eV 10 J.K e V −= Δ = = 3.2× The number N of electrons that strike the foil in 10 seconds is

13th16

0.0231 J 7.2 103.2 10 J

ENK −

Δ= = = ××

The current of the electron beam is 13 19(7 2 10 )(1 6 10 C) 1 2 A

10 sNq

tμ. × . × = = .

Δ

2

37.37. Model: Model the Li atom as a single valence electron orbiting a sphere with net charge q = +1e due to the 3 protons and 2 inner electrons. A sphere of charge acts like a point charge with the total charge concentrated at the center of the sphere. Visualize:

Solve: The electron’s energy is both kinetic and potential: 212 ( )/ .E mv Kq e r= + − To say that the energy needed to

ionize the atom is 5.14 eV means that you would need to increase the electron’s energy by 5.14 eV to remove it from the atom, taking it to r ≈ ∞. Since we define E = 0 for charged particles that are infinitely separated, the energy of the atom must be 195.14 eV 8.224 10 J.E −= − = − × Negative energy indicates that the system is bound, and the absolute value of this energy is the binding energy. Thus, from energy considerations we learn that

22 191

2 8 224 10 JKeE mvr

−= − = − . ×

The Coulomb force on the electron provides the centripetal acceleration of circular motion. The force equation is 2 2 2

212 2 2r r

Ke mv KeF ma mvr rr

= = = ⇒ =

Substitute this expression for the kinetic energy into the energy equation: 2 2 2 2

2 1912

9 2 2 19 210

19

8 224 10 J2 2

(8 99 10 N m /C )(1 60 10 C) 1 40 10 m 0 140 nm2(8 224 10 J)

Ke Ke Ke Kemvr r r r

r

−

−−

−

− = − = − = − . ×

. × . ×= = . × = .. ×

With the radius now known, we can use the result of the force equation to find that 2

61 34 10 m/sKevmr

= = . ×

37.38. Model: The kinetic energy is conserved because the collision is elastic. This is a one-dimensional collision in the x-direction. Visualize: Both kinetic energy and momentum are conserved. This problem has essentially been done in Model 11.1 in the chapter on collisions where we are given



1 2 1f 1 i 1 f 2 i 1

1 2 1 2

2( ) ( ) ( ) ( )x x x xm m mv v v vm m m m

−= =+ +

if object 2 is initially at rest. Look up the masses in Appendix C: H 1 1.0078 um m= = and C 2 12.00 u.m m= = Solve: First find the speed of the incoming hydrogen atom.

195

iH 272 2(200 eV)(1.6 10 J/eV) 1.955 10 m/s

(1.0078 u)(1.661 10 kg/u)Kvm

−

−×= = = ×

×

Now use the equations above in the kinetic energy equation.

( ) ( )

( )

2 2275H C

fH H iHH C

17

2 27H

fC C iHH C

1 1 1.661 10 kg 1.0078 u 12.00 u1.0078 u 1.955 10 m/s2 2 1 u 1.0078 u 12.00 u

2.286 10 J 143 eV

2 11 2 1 1.661 10 kg12.00 u2 2 1 u

m mK m vm m

mK m vm m

−

−

−

⎛ ⎞⎛ ⎞− × −⎛ ⎞= = ×⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠ ⎝ ⎠

= × =

⎛ ⎞⎛ ⎞ ×= = ⎜ ⎟⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠

( ) ( )2

5

18

.0078 u1.955 10 m/s

1.0078 u 12.00 u

9.149 10 J 57.2 eV−

⎛ ⎞×⎜ ⎟⎜ ⎟+⎝ ⎠

= × =

Assess: These final kinetic energies add up to the 200 eV initial kinetic energy.

37.39. Model: The mass of an atom is concentrated almost entirely in the nucleus. Solve: The volumes of the nucleus and the atom are

3 314 1042 3 31 3

nucleus atom

42 313nucleus

31 3atom

4 1 0 10 m 4 1 2 10 m0 524 10 m 9 05 10 m3 2 3 2

0 524 10 m 5 8 109 05 10 m

V V

VV

π π− −− −

−−

−

⎛ ⎞ ⎛ ⎞. × . ×= = . × = = . ×⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

. ×= ≈ . ×. ×

Thus 115.8 10 % 0.000000000058−× = % of the atom’s volume contains all the mass. The percent of empty space in an atom is 99.999999999942%.

37.40. Model: (a) 3 31 3atom atom

4 9 05 10 m .3

V rπ −= = . × Since the atomic mass of aluminum is 27 u, the mass per

atom is 27 26(27)(1 661 10 kg) 4 48 10 kgm − −= . × = . ×

Therefore, the average density of an aluminum atom is 26

4 3 4 3atom 31 3

4 48 10 kg 4 96 10 kg/m 5 0 10 kg/m9 05 10 m

ρ−

−. ×= = . × ≈ . ×. ×

(b) The average density of an aluminum atom is found in part (a) to be larger than the density of solid aluminum 3

1( 2700 kg/m ).Aρ = The volume per atom in the solid is

2629 3 29 3

34 48 10 kg 1 66 10 m 1 7 10 m

2700 kg/m

−− −. × = . × ≈ . ×

So, using 3 29 34sphere3 1 66 10 m ,r −π = . × we get 10

sphere 1 6 10 m.r −= . ×

(c) The atomic mass is almost entirely in the nucleus, so the density of the nucleus is 26

17 3nucleus 15 34

3

4 48 10 kg 1 7 10 kg/m(4 10 m)π

ρ−

−. ×= = . ×

×

37-14 Chapter 37


Compared to the density of solid aluminum, which is 32700 kg/m , the nuclear density is approximately

13 146.2 10 10× ≈ times larger.

37.41. Model: The alpha particle is a doubly charged He nucleus having a mass of 4 times that of the proton. Solve: The force required for the particle to execute circular motion is (see Equation 8.6)

2

cmvF

r=

The force exerted on the alpha particle by the magnetic field is

MF qvB=

Equating the two, and using the expression 212K mv= for the kinetic energy of the alpha particle gives

2 2 2 2 19 2 2 213

272 2 2 (2 1.60 10 C) (0.75 T) (0.45 m) 8 8 10 J 5.5 MeV

2 2 2(4 1 67 10 kg)

K

mv K q B rqvB K qB Kr r m m

−−

−× ×= ⇒ = ⇒ = = = . × =

× . ×

37.42. Model: Since the 211Po nucleus was at rest the magnitudes of the momenta of the products are equal. Visualize: Use 2 /2 .K p m= Look up the masses of 207Pb and 4He in Appendix C: Pb 206.976 um = and 4.002 u.mα = Solve: The magnitude of the momentum of the 207Pb nucleus is

2Pb Pb Pb Pb Pb Pb2 ( ) 2p K m p K m= ⇒ =

Now the kinetic energy of the alpha particle is 2

Pb Pb Pb Pb2 (0.14 MeV)(206.976 u) 7.24 MeV2 2 4.002 up K m K mKm m mα

α α α= = = = =

Add the two kinetic energies together to get the total energy released in the decay:

tot Pb 7.24 MeV 1.4 Mev 7.4 MeVK K Kα= + = + =

Assess: It is expected that the lighter particle would have most of the kinetic energy.

37.43. Model: Assume the nucleus is at rest. Use conservation of energy. Visualize:

Solve: The energy-conservation equation f f i iK U K U+ = + gives

09 2 2 19 2 19

615

1 (2 )( )0 J 6 24 MeV 0 J4 (6 0 fm)

(9 0 10 N m /kg )(2)(1 60 10 C) 1 60 10 J(6 24 10 eV)1 0 eV6 0 10 m

e Ze

Z

π ε− −

−

+ = . +.

⎛ ⎞. × . × . ×= . × ⎜ ⎟⎜ ⎟.. × ⎝ ⎠



Solving for Z gives Z = 13. The element is aluminum.

37.44. Model: Assume the 238U nucleus is at rest. Energy is conserved. Visualize:

Solve: The energy conservation equation f f i iK U K U+ = + is

i150

9 2 2 19 212

i 15 19

1 (2 )(92 )0 J 0 J4 7.5 10 m

(9 0 10 N m /C )(184)(1 60 10 C) 1 0 eV(5 65 10 J) 35.33 MeV7 5 10 m 1 60 10 J

e e K

K

πε −

−−

− −

+ = +×

. × . × .⎛ ⎞= = . × =⎜ ⎟. × . ×⎝ ⎠

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2e gains kinetic energy K eV by accelerating through a potential difference 1

2 V.V KΔ = Thus, the alpha particle will just

reach the 238U nucleus after being accelerated through a potential difference

735 33 MV 1 8 10 V2

V . Δ = = . ×

37.45. Model: Assume the 16O nucleus is at rest. Energy is conserved. Visualize:

A proton with initial velocity iv and zero electric potential energy (due to its infinite separation from an 16O nucleus) moves toward 16O and reverses direction when it is 1.0 fm from the surface. Since r is measured from the center of the nucleus, f 4 0 fmr = . . Solve: (a) The energy conservation equation f f i iK U K U+ = + gives

21i2

09 2 2 19 2

2 14 2 2 7i i15 27

1 (8 )0 J 0 J4 (3 0 fm 1 0 fm)

(2)(9 0 10 N m /C )(8)(1 6 10 C) 5 52 10 m /s 2 3 10 m/s(4 0 10 m)(1 67 10 kg)

e e mv

v v

πε−

− −

+ = +. + .

. × . ×= = . × ⇒ = . ×. × . ×

(b) 2 27 14 2 2 131 1i2 2 (1 67 10 kg)(5 52 10 m /s ) 4 60 10 J 2 9 MeVK mv − −= = . × . × = . × = .

37-16 Chapter 37


37.46. Model: Assume the 12C nucleus is at rest. Energy is conserved. Visualize:

Solve: (a) A proton with an initial velocity iv and zero electric potential energy (due to its infinite separation from

the 12C nucleus) moves toward the 12C nucleus. It must impact the 12C nucleus, which has a radius of 2.75 fm, with an energy of 3.00 MeV. The energy conservation equation f f i iK U K U+ = + gives

2proton i15

0

19 9 2 2 19 26 27 2

i15

7i

1 (6 ) 13 00 MeV 0 J4 2(2 75 10 m)

1 60 10 J (8 99 10 N m /C )(6)(1 60 10 C) 1(3 00 10 eV) (1 67 10 kg)1 00 eV 22 75 10 m

3 43 10 m/s

e e m v

v

v

πε −

− −−

−

. + = +. ×

⎛ ⎞. × . × . ×. × + = . ×⎜ ⎟⎜ ⎟. . ×⎝ ⎠

= . ×

(b) The initial kinetic energy is 21i i2 .K mv e V= = Δ Thus,

27 7 26

191 (1 67 10 kg)(3 43 10 m/s) 6 14 10 V2 1 60 10 C

V−

−. × . ×Δ = = . ×

. ×

Challenge Problems

37.47. Model: Momentum is conserved during the interaction. Assume the nucleus is initially at rest. Visualize:

Solve: Momentum is a vector, so both the x- and y-components of p must be conserved. The x-equation is

7 7i f n fn fn

5fn fn

4(1 50 10 m/s) cos cos 4(1 49 10 m/s)cos(49 ) 197 cos

cos ( ) 1 0609 10 m/sx

m v m v m v v

v vα α α α θ φ φ

φ

= . × = + = . × ° +

= = . ×

The y-equation is 7

f n fn fn5

fn fn

0 sin sin 4(1 49 10 m/s)sin(49 ) 197 sin

sin ( ) 2 283 10 m/sy

m v m v v

v vα α θ φ φ

φ

= − = . × ° −

= = . ×



We can work with relative masses (4 and 192) since the conversion to kg occurs in all terms of the equation and cancels out. Dividing the y-equation by the x-equation gives

1fn

fn

sin 2 283tan 2 152 tan (2 152) 65 1cos 1 0609

vv

φφ φφ

−.= = = . ⇒ = . = . °.

The speed of the recoiling nucleus is 2 2 2 2 5 5

fn fn fn( ) ( ) (1 0609) (2 203) 10 m/s 2 52 10 m/sx yv v v= + = . + . × = . ×

Thus the nucleus recoils at 52.52 10 m/s× in a direction 65.1° below the x-axis.

37.48. Solve: (a) The Coulomb force between the electron and the proton causes the centripetal acceleration of the electron in its circular orbit. That is,

2 22

20 0

1 ( ) 14 4r

mv e e eF mvr rrπε πε

= = ⇒ =

The total energy of the electron is a sum of its kinetic energy and electric potential energy: 2 2 2

2

0 0 0 0

1 1 ( ) 1 1 1 ( )2 4 2 4 4 8

e e e e eE mvr r r rπε πε πε πε

⎛ ⎞− −= + = + = −⎜ ⎟⎜ ⎟⎝ ⎠

(b) The ratio of the potential energy to the kinetic energy is 2 2

0 02 21 1

02 2

/4 /4 2( /4 )

U e r e rK mv e r

πε πεπε

− −= = = −

(c) The part (a) energy is negative because the electron is bound. To remove the electron to infinity, where E = 0, you need to give it energy ionization .E E= Thus

19 2

ionization0

9 2 2 19 211

19

1 60 10 J(13 6 eV)1 00 eV 8

1 (8 99 10 N m /C )(1 60 10 C) (1 00 eV) 5 29 10 m2 (13 6 eV)(1 60 10 J)

eEr

r

πε

−

−−

−

⎛ ⎞. ×= . =⎜ ⎟⎜ ⎟.⎝ ⎠

. × . × .= = . ×. . ×

To calculate the speed, we have 21ionization 2( ) ( 2 ) .E E U K K K K mv= − = − + = − − + = = Therefore, the speed is

196ionization

312 2(13 6 eV) 1 60 10 J 2 19 10 m/s

1 00 eV9 11 10 kgEv

m

−

−

⎛ ⎞⎡ ⎤. . ×= = = . ×⎜ ⎟⎢ ⎥⎜ ⎟.. ×⎢ ⎥⎣ ⎦⎝ ⎠

37.49. Model: Assume that the electric field between the plates is uniform. Visualize: Please refer to Figure 37.14. Solve: (a) The electric force on a positive charge q in a field E is EF qE= . If a droplet is charged with charge q and is suspended between the plates by applying a field 0,E then the electric force must exactly cancel the weight force:

net E 00

0 N 0 N mgF F w qE mg qE

= + = ⇒ − = ⇒ =

(b) A droplet falling at the terminal speed has no net force acting on it. The velocity is negative for a falling drop, or term.v v= − Thus,

net drag term term0 N 0 N mgF F w bv mg bv mg vb

= + = ⇒ − − = + − = ⇒ =

37-18 Chapter 37


(c) Because drag 6 , 6 .F rv bv b rπη πη= − = − = The result in part (b) for termv thus simplifies to term /(6 ).v mg rπη=

Noting that 343 ,m V rπρ ρ= = the expression for the terminal velocity is

( )34 23 term

term2 9

6 9 2

r g r g vv rr g

πρ ρ ηπη η ρ

= = ⇒ =

(d) We first need to find the mass m of the droplet before we can obtain q from the expression in part (a). Using the expression for r in part (c), the mass of the oil drop is

3/23 air term

oil oil oiloil

4 4 93 3 2

vm V rg

π π ηρ ρ ρρ

⎡ ⎤⎛ ⎞ ⎛ ⎞= = = ⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎣ ⎦

The terminal velocity is 3term 3 00 10 m/7 33 s.v −= . × . The mass of the oil drop is

3/25 33 14

3 2

14 2 218

4 9(1 83 10 kg m/s)(3 00 10 m)/(7 33 s)(860 kg/m ) 2 881 10 kg3 2(860 kg/m )(9 8 m/s )

(2 881 10 kg)(9 8 m/s )(1 0 10 m) 2 4 10 C( / ) 1177 V

m

mgqV d

π − −−

− −−

⎡ ⎤. × . × .= = . ×⎢ ⎥.⎢ ⎥⎣ ⎦

. × . . ×= = = . ×Δ

(e) The number of fundamental electric charges on the oil drop is 18

192 4 10 C 151 6 10 C

qe

−

−. ×= =. ×

37.50. Model: Assume the classical electric force is the centripetal force. 2 2 2

e2e

Ke v KeF m vr m rr

= = ⇒ =

Visualize: We are given 600 nmλ = . We also know that for electromagnetic waves .c fλ= Solve: (a) The speed is the circumference of the orbit divided by the period.

2

e

2 2

2 2 2

Kem r

rv rfT

v vrf c c

π π

λλπ π π

= =

= = =

Square both sides. 2

e

22

2 24

Kem rr

c

λ

π=

Multiply both sides by r. 2 2

32 2

e4Ker

c mλ

π=

Take the cube roots. 2 2

3 2 2

9 2 2 19 2 9 23 2 8 2 31

10

4

(8.99 10 N m /C )(1 60 10 C) (600 10 m)4 (3 00 10 m/s) (9 11 10 kg)

2 95 10 m0 295 nm

Ke

er

c mλ

π

π

− −

−

−

=

× ⋅ . × ×=. × . ×

= . ×= .



(b) The total mechanical energy of the classical atom was found in Problem 37.48. 2 19 2

192 12 2 2 10

0

(1.60 10 C)3.90 10 J 2.44 eV

8 8 (8.85 10 C /N m )(2.95 10 m)e

Erπε π

−−

− −×

= = − = − × = −× ⋅ ×

Assess: While we don’t expect the classical analysis to be perfect, the result is in a reasonable range.

the foundations of modern physics

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