SM chapter 43 - Solucionario capitulo 43 Serway 7ma edición PDF

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Solucionario capitulo 43 Serway 7ma edición ...

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43 Molecules and Solids ANSWERS TO QUESTIONS

CHAPTER OUTLINE 43.1 43.2 43.3 43.4 43.5 43.6 43.7 43.8

Molecular Bonds Energy States and Spectra of Molecules Bonding in Solids Free-Electron Theory of Metals Band Theory of Solids Electrical Conduction in Metals, Insulators, and Semiconductors Semiconductor Devices Superconductivity

Q43.2

Q43.1

Ionic bonds are ones between oppositely charged ions. A simple model of an ionic bond is the electrostatic attraction of a negatively charged latex balloon to a positively charged Mylar balloon. Covalent bonds are ones in which atoms share electrons. Classically, two children playing a short-range game of catch with a ball models a covalent bond. On a quantum scale, the two atoms are sharing a wave function, so perhaps a better model would be two children using a single hula hoop. Van der Waals bonds are weak electrostatic forces: the dipole-dipole force is analogous to the attraction between the opposite poles of two bar magnets, the dipole-induced dipole force is similar to a bar magnet attracting an iron nail or paper clip, and the dispersion force is analogous to an alternating-current electromagnet attracting a paper clip. A hydrogen atom in a molecule is not ionized, but its electron can spend more time elsewhere than it does in the hydrogen atom. The hydrogen atom can be a location of net positive charge, and can weakly attract a zone of negative charge in another molecule.

Rotational, vibrational, and electronic (as discussed in Chapter 42) are the three major forms ℏ2 of excitation. Rotational energy for a diatomic molecule is on the order of , where I is 2I the moment of inertia of the molecule. A typical value for a small molecule is on the order of 1 meV = 10 −3 eV. Vibrational energy is on the order of hf, where f is the vibration frequency of the molecule. A typical value is on the order of 0.1 eV. Electronic energy depends on the state of an electron in the molecule and is on the order of a few eV. The rotational energy can be zero, but neither the vibrational nor the electronic energy can be zero.

*Q43.3 If you start with a solid sample and raise its temperature, it will typically melt first, then start emitting lots of far infrared light, then emit light with a spectrum peaking in the near infrared, and later have its molecules dissociate into atoms. Rotation of a diatomic molecule involves less 11 energy than vibration. Absorption and emission of microwave photons, of frequency ~10 Hz, accompany excitation and de-excitation of rotational motion, while infrared photons, of 13 frequency ~10 Hz, accompany changes in the vibration state of typical simple molecules. The ranking is then b > d > c > a.

519

520

Q43.4

Chapter 43

From the rotational spectrum of a molecule, one can easily calculate the moment of inertia of the molecule using Equation 43.7 in the text. Note that with this method, only the spacing between adjacent energy levels needs to be measured. From the moment of inertia, the size of the molecule can be calculated, provided that the structure of the molecule is known.

*Q43.5 Answer (b). At higher temperature, molecules are typically in higher rotational energy levels before as well as after infrared absorption. *Q43.6 (i) Answer (a). An example is NaCl, table salt. (ii) Answer (b). Examples are elemental silicon and carborundum (silicon carbide). (iii) Answer (c). Think of aluminum foil. *Q43.7 (i) Answer (b). The density of states is proportional to the energy to the one-half power. (ii) Answer (a). Most states well above the Fermi energy are unoccupied. Q43.8

In a metal, there is no energy gap between the valence and conduction bands, or the conduction band is partly full even at absolute zero in temperature. Thus an applied electric field is able to inject a tiny bit of energy into an electron to promote it to a state in which it is moving through the metal as part of an electric current. In an insulator, there is a large energy gap between a full valence band and an empty conduction band. An applied electric field is unable to give electrons in the valence band enough energy to jump across the gap into the higher energy conduction band. In a semiconductor, the energy gap between valence and conduction bands is smaller than in an insulator. At absolute zero the valence band is full and the conduction band is empty, but at room temperature thermal energy has promoted some electrons across the gap. Then there are some mobile holes in the valence band as well as some mobile electrons in the conduction band.

*Q43.9 Answer (b). First consider electric conduction in a metal. The number of conduction electrons is essentially fixed. They conduct electricity by having drift motion in an applied electric field superposed on their random thermal motion. At higher temperature, the ion cores vibrate more and scatter more efficiently the conduction electrons flying among them. The mean time between collisions is reduced. The electrons have time to develop only a lower drift speed. The electric current is reduced, so we see the resistivity increasing with temperature. Now consider an intrinsic semiconductor. At absolute zero its valence band is full and its conduction band is empty. It is an insulator, with very high resistivity. As the temperature increases, more electrons are promoted to the conduction band, leaving holes in the valence band. Then both electrons and holes move in response to an applied electric field. Thus we see the resistivity decreasing as temperature goes up. Q43.10 The energy of the photon is given to the electron. The energy of a photon of visible light is sufficient to promote the electron from the lower-energy valence band to the higher-energy conduction band. This results in the additional electron in the conduction band and an additional hole—the energy state that the electron used to occupy—in the valence band. Q43.11 Along with arsenic (As), any other element in group V, such as phosphorus (P), antimony (Sb), and bismuth (Bi), would make good donor atoms. Each has 5 valence electrons. Any element in group III would make good acceptor atoms, such as boron (B), aluminum, (Al), gallium (Ga), and indium (In). They all have only 3 valence electrons. Q43.12 The two assumptions in the free-electron theory are that the conduction electrons are not bound to any particular atom, and that the nuclei of the atoms are fixed in a lattice structure. In this model, it is the “soup” of free electrons that are conducted through metals. The energy band model is more comprehensive than the free-electron theory. The energy band model includes an account of the more tightly bound electrons as well as the conduction electrons. It can be developed into a theory of the structure of the crystal and its mechanical and thermal properties.

Molecules and Solids

521

Q43.13 A molecule containing two atoms of 2 H, deuterium, has twice the mass of a molecule containing two atoms of ordinary hydrogen 1 H. The atoms have the same electronic structure, so the molecules have the same interatomic spacing, and the same spring constant. Then the moment of inertia of the double-deuteron is twice as large and the rotational energies one-half as large 1 as for ordinary hydrogen. Each vibrational energy level for D2 is times that of H 2 . 2 Q43.14 Yes. A material can absorb a photon of energy greater than the energy gap, as an electron jumps into a higher energy state. If the photon does not have enough energy to raise the energy of the electron by the energy gap, then the photon will not be absorbed. *Q43.15 (i) and (ii) Answer (a) for both. Either kind of doping contributes more mobile charge carriers, either holes or electrons. *Q43.16 (a) false

(b) false

(c) true

(d) true

(e) true

SOLUTIONS TO PROBLEMS Section 43.1

Molecular Bonds

(a)

F=

q2 (1.60 × 10− 19 ) (8.99 × 10 9 ) N = 0.921 × 10 −9 N = 2 4π ∈0 r 2 (5.00 × 10 −10 )

(b)

U=

(1.60 × 10−19 ) (8.99 × 109 ) J ≈ −2.88 eV − q2 =− − 4π ∈0 r 5. 00 × 10 1 0

2

P43.1

toward the other ion.

2

P43.2

We are told

K + Cl + 0.7 eV → K+ + Cl−

and

Cl + e− → Cl− + 3.6 eV

or

Cl− → Cl+ e− − 3.6 eV

By substitution,

K + Cl + 0.7 eV → K + + Cl + e− − 3.6 eV K + 4.3 eV → K + + e −

or the ionization energy of potassium is 4.3 eV . P43.3

(a)

Minimum energy of the molecule is found from ⎡2 A⎤1 6 dU = −12 Ar − 13 + 6 Br − 7 = 0 yielding r0 = ⎢ ⎥ ⎣B⎦ dr

(b)

⎡ A B ⎤ ⎡1 E = U r = ⬁ − U r = r0 = 0 − ⎢ 2 2 − ⎥ = −⎢ − ⎣4 ⎣ 4 A B 2A B⎦

1⎤ B 2 B2 = 4A 2⎦⎥ A

This is also the equal to the binding energy, the amount of energy given up by the two atoms as they come together to form a molecule. (c)

⎡ 2 ( 0. 124 × 10 −120 eV ⋅ m 12 )⎤ r0 = ⎢ ⎥ − 60 6 ⎥⎦ ⎣⎢ 1. 488 × 10 eV ⋅ m

(1.488 ×10 4( 0. 124 × 10

−60

E=

eV ⋅m6 )

−120

16

= 7.42 × 10−11 m = 74.2 pm

2

eV⋅ m12 )

= 4.46 eV

522

P43.4

Chapter 43

(a)

We add the reactions K +4.34 eV → K+ + e− and

I + e− → I− + 3.06 eV

to obtain

K + I → K+ + I− + (4.34 − 3.06 ) eV

The activation energy is 1.28 eV . (b)

13 7 dU 4 ∈⎡ ⎛s ⎞ ⎤ ⎛s ⎞ = ⎢ −12 ⎜ ⎟ +6 ⎜ ⎟ ⎥ dr s ⎣ ⎝r ⎠ ⎝r⎠ ⎦

At r = r0 we have s = 2 −1 6 r0

13 7 dU ⎞ ⎛ ⎞ = 0. Here ⎛⎜s ⎟ = 1 ⎜s ⎟ dr 2 ⎝ r0 ⎠ ⎝ r0 ⎠ −1 6

s =2

( 0.305) nm =

0.272 nm = s

Then also ⎡⎛ − 1 6 ⎞ 12 ⎛ − 1 6 ⎞ 6 ⎤ ⎡ 1 1⎤ 2 r0 2 r0 ⎥ U (r 0 ) = 4 ∈ ⎢⎜ ⎟ −⎜ ⎟ + Ea = 4 ∈ ⎢ − ⎥ + Ea = − ∈ + E a ⎣ 4 2⎦ ⎢⎣⎝ r0 ⎠ ⎝ r0 ⎠ ⎥⎦ ∈= Ea − U ( r0 ) = 1.28 eV + 3.37 eV= 4.65 eV = ∈ (c)

F ( r) = −

13 7 dU 4 ∈⎡ ⎛s ⎞ ⎛⎜s ⎞ ⎤ ⎜ ⎟ − ⎟ = 12 6 ⎥ ⎢ dr s ⎣ ⎝r ⎠ ⎝r ⎠ ⎦

To find the maximum force we calculate 42 ⎞ ⎟ = ⎜⎛ ⎝ rrupture 156 ⎠ s

Fmax =

14 8 dF 4 ∈ ⎡ ⎛s ⎞ ⎤ ⎛s ⎞ = 2 ⎢ − 156 ⎜ ⎟ + 42 ⎜ ⎟ ⎥ = 0 when ⎝r⎠ ⎝r ⎠ ⎦ dr s ⎣

16

76 13 6 1. 6 × 10−1 9 Nm 4 (4 .65 eV) ⎡ ⎛ 42 ⎞ ⎛ 42 ⎞ ⎤ − 12 6 ⎟⎠ ⎜⎝ ⎟⎠ ⎥ = −41. 0 eV nm = − 41. 0 ⎢ ⎜⎝ 156 0. 272 nm ⎣ 156 10−9 m ⎦

= −6 .55 nN Therefore the applied force required to rupture the molecule is +6.55 nN away from the center. (d)

⎡ ⎛2 −1 6 r0 ⎞12 ⎛ 2 −1 6 ⎞6 ⎤ ⎡ ⎛ s ⎞ 12 ⎛ s ⎞ 6 ⎤ U ( r0 + s) = 4 ∈⎢ ⎜ ⎟ −⎜ ⎟ −⎜ ⎟ ⎥ + Ea ⎟ ⎥ + Ea = 4 ∈ ⎢ ⎜ + ⎢⎣ ⎝ r0 + s ⎠ ⎝ r0 + s ⎠ ⎥⎦ ⎝ r0 + s⎠ ⎥⎦ ⎣⎢ ⎝ r0 s ⎠ − 12 −6 ⎤ ⎡1 ⎛ 1⎛ s⎞ s⎞ = 4 ∈ ⎢ ⎜1 + ⎟ − ⎜1 + ⎟ ⎥ + Ea ⎢⎣4 ⎝ r0 ⎠ 2 ⎝ r0 ⎠ ⎥⎦ ⎞⎤ ⎞ 1⎛ ⎡1 ⎛ s2 s s2 s = 4 ∈ ⎢ ⎜1 − 12 + 78 2 − ⋯⎟ − ⎜1− 6 + 21 2 −⋯⎟⎥+ Ea r0 r0 r0 r0 ⎠⎦ ⎠ 2⎝ ⎣4 ⎝ = ∈ − 12 ∈

s s2 s s2 + 78 ∈ 2 − 2 ∈ + 12 ∈ − 42 ∈ 2 + E a + ⋯ r0 r0 r0 r0

⎛ s⎞ s2 = − ∈ + Ea + 0 ⎜ ⎟ +36 ∈ 2 + ⋯ r0 ⎝ r0 ⎠ 1 U ( r 0 + s) ≈ U ( r 0) + ks 2 2 ∈ 72 (4.65 eV) where k = 72 2 = = 3 599 eV nm 2 = 576 N m 2 r0 (0.305 nm )

Molecules and Solids

P43.5

kB T ≈ 10−3 eV = 10−3 (1. 6 × 10−19 J )

At the boiling or condensation temperature,

T≈

Section 43.2 *P43.6

(a)

523

1.6 × 10− 22 J ~ 10 K 1.38 × 10− 23 J K

Energy States and Spectra of Molecules With r representing the distance of each atom from the center of mass, the moment of inertia is 2

⎛ 0 .75 × 10 −10 m ⎞ − −48 2 mr2 + mr2 = 2 (1 .67 × 10 27 kg ) ⎜ ⎟⎠ = 4.70 × 10 kg ⋅ m ⎝ 2 The rotational energy is Erot =

ℏ2 J ( J + 1) or it is zero for J = 0 and for J = 1 it is 2I

( 6. 626× 10 34 J⋅ s) ⎛⎜ 1 eV ⎟⎞ = 0 .0148 eV h 21(2 ) = 2 4 π 2 I 4 π2 ( 4.70 × 10−48 kg ⋅ m 2 ) ⎝1.6 × 10− 19 J ⎠ −

(b)

P43.7

µ=

λ=

2

c ch (2.998× 108 m/s)(6.626 × 10−34 J ⋅ s) 1 eV 1240 eV ⋅ nm = = = = 83.8 µ m −19 × f E 0.0 0148 eV 1.602 10 J 0.0148 eV

m1m 2 132 .9 (126 .9 ) = (1.66 × 10−27 kg ) = 1.08 × 10− 25 kg m 1 + m 2 132.9 + 126.9

I = µ r2 = ( 1.08 × 10−25 kg) ( 0.127 × 10−9 m) = 1.74× 10 −45 kg ⋅ m 2 2

(a)

1

E=

Iω 2 =

( Iω )2

2 2I J = 0 gives E = 0

=

J ( J + 1) ℏ 2I

2

(6 .626 × 10 J ⋅ s ) = 6. 41× 10− 24 J = 40. 0 µeV ℏ2 = I 4π 2 (1 .74 × 10− 45 kg ⋅ m 2 ) −34

J = 1 gives E =

2

hf = 6.41 × 10 −24 J − 0 gives f = 9.66 × 10 9 Hz (b)

*P43.8

f=

∆E vib =

µ= µ=

2 h E1 ℏ = = ∝ r −2 2 h hI 4 π µr 2

h 2π

k 4π f 2

2

k = hf µ =

so

If r is 10% too small, f is 20% too large.

k = 4π 2 f 2µ

1530 N/m = 1 .22 × 10 − 2 6 kg 4 π (56 .3 × 10 12 /s) 2 2

− m1m 2 14.007 u 15.999 u 1.66 × 10 27 kg = = 1. 24 × 10 −26 kg 1u m 1 + m 2 14 .007 u +15.99 99 u

The reduced masses agree, because the small apparent difference can be attributed to uncertainty in the data.

524

P43.9

Chapter 43

For the HCl molecule in the J = 1 rotational energy level, we are given r0 = 0.127 5 nm.

Taking J = 1 , we have

Erot =

ℏ2 J (J + 1) 2I

Erot =

2ℏ2 ℏ2 1 2 ℏ = Iω or ω = = 2 I 2 I2 I

The moment of inertia of the molecule is given by Equation 43.3.

FIG. P43.9

⎛ mm ⎞ I = µ r02 = ⎜ 1 2 ⎟ r02 ⎝ m1 + m2 ⎠ ⎡( 1.01 u)( 35.5 u) ⎤ 2 2 −27 −10 65 × 10 −47 kg ⋅ m 2 I =⎢ ⎥ r0 = ( 0.982 u )(1. 66 × 10 kg u ) ( 1. 275 × 10 m) = 2.6 + ⎣ 1.01 u 35.5 u ⎦ Therefore, ω = 2

P43.10

hf = ∆E =

34 ℏ 2 6. 626 × 10− J ⋅ s = 5.63 × 1012 rad s = 47 I 2π (2.65× 10 − kg⋅ m 2)

ℏ2 ℏ2 ℏ2 2 ( 2 + 1)] − [1 (1 + 1)] = (4 ) [ 2I 2I 2I

h 4 ( h 2π ) 6.626 × 10 −34 J ⋅ s = = = 1.46 × 10 −46 kg ⋅ m 2 2 2hf 2 π f 2π 2 ( 2. 30 × 10 11 Hz) 2

I= P43.11

P43.12

I = m1 r12 + m2 r22 mr Then r1 = 2 2 m1 m1 r1 Also, r2 = . m2

where so Thus,

m1r1 = m 2r2 m2 r2 + r2 = r m1 mr r1 + 1 1 = r m2

and and and

I = m1

m1 m2 r2 (m2 + m1 ) m1m 2r 2 m2 m12 r 2 m22 r 2 = + = = µ r2 2 2 2 + m m + + + m m m m m m ( 1 2) ( 1 2) ( 1 2) 1 2

(a)

µ=

r1 + r2 = r m1 r m1 + m2 m2 r r1 = m1 + m 2 r2 =

22.99 ( 35.45) (1.66 × 10 −27 kg) = 2.32 × 10 −26 kg ( 22.99 + 35.45)

I = µr2 = ( 2.32 × 10 −26 kg)( 0.280 × 10 −9 m) = 1.81 × 10 −45 kg ⭈ m 2 2

(b)

ℏ2 hc ℏ 2 3ℏ 2 ℏ 2 2ℏ 2 2 h2 = − = = 2 ( 2 + 1) − 1 (1 + 1 )= I I I 2I 4π 2 I λ 2I

λ=

2 − 45 8 2 c 4 π 2I (3 .00 × 10 m s) 4 π (1 .81 × 10 kg⋅ m ) = = 1. 62 cm 34 2h 2( 6.626 × 10− J ⋅ s )

Molecules and Solids

P43.13

⎛ℏ 2 ⎞ The energy of a rotational transition is ∆E =⎜ ⎟ J where J is the rotational quantum number of ⎝ I⎠ the higher energy state (see Equation 43.7). We do not know J from the data. However, hc ( 6.626 × 10 = λ

− 34

∆E =

J ⋅ s) ( 3.00 × 108 m s ) ⎛ 1 eV ⎞ ⎜⎝ ⎟. λ 1..60 × 10− 19 J ⎠

For each observed wavelength, l (mm)

∆E (eV)

0.120 4

0.010 32

0.096 4

0.012 88

0.080 4

0.015 44

0.069 0

0.018 00

0.060 4

0.020 56 ℏ2 = 0. 002 56 eV I

E1 =

The ∆E⬘s consistently increase by 0.002 56 eV.

2 (1.055 × 10 −34 J ⋅ s) ⎛⎜ 1 eV ⎟⎞ = 2 .72 × 10−47 kg ⋅ m 2 and I = ℏ = E1 ( 0.002 56 eV) ⎝1.660 ×10 −19 J ⎠ 2

*P43.14 (a)

I 2. 72 × 10−47 m = 0.130 nm = 1.62 × 10−27 µ

r=

For the HCl molecule, the internuclear radius is

Minimum amplitude of vibration of HI is characterized by 1 2 1 h kA = ℏ ω = 2 2 4π A=

k µ

1 /4

so

A=

h ⎛ 1 ⎞ 2π ⎜⎝ k µ ⎟⎠

⎞ 6.626 × 10− 34 J ⋅s ⎛ 1 ⎜⎝ 2π (320 N/m)(127/128)(1.66 × 10− 27 kg)⎟⎠

1/ 4

= 12 .1 pm 1/ 4

P43.15

525

⎞ 6.626 ×10 −34 J ⋅s ⎛ 1 ⎜⎝ 2π ( 970 N/m)(19/20)(1.66 × 10− 27 kg) ⎟⎠

(b)

For HF, A =

(c)

Since HI has the smaller k, it is more weakly bound.

µ=

m1m 2 35 27 27 = × 1.66 × 10 − kg = 1.61 ×10 − kg g m 1 + m 2 36

∆Evib = ℏ

k 480 = (1.055 ×10− 34) = 5.74 × 10 −20 J = 0 .358 eV 1.61 × 10 −27 µ

= 9. 23 pm

526

P43.16

Chapter 43

(a)

The reduced mass of the O2 is

µ=

(16 u )(16 u ) = 8 (16 u) + (16 u)

u = 8 (1 .66 ×10 − 27 kg) = 1 .33 ×10 −26 kg

The moment of inertia is then 2 26 10 I = µr = (1 .33 ×10 − kg) (1 .20 × 10 − m)

2

= 1.91× 10 −4 6 kg ⋅ m2 2 ( 1.055× 10−34 J ⋅ s ) J J + 1 The rotational energies are E rot = ℏ J ( J + 1) = ( ) 2 − 2I 2 (1.91 × 10 4 6 kg ⋅ m ) 2

(b)

Thus

Erot = ( 2.91 × 10 −23 J) J ( J + 1)

And for J = 0, 1, 2,

Erot = 0, 3.64 ×10−4 eV, 1.09 ×10−3 eV

⎛ 1⎞ k ⎛ 1⎞ 1177 N m Evib = ⎜v + ⎟ ℏ = ⎜ v + ⎟( 1 .055 × 10− 34 J ⋅ s) 8 1 µ ⎠ ⎝ ⎠ ⎝ 2 2 ( . 66 × 10 −27 kg ) 1 eV 1 ⎞ ⎛ Evib = ⎛⎜ v + ⎞⎟ (3 .14 × 10 −20 J) ⎛⎜ ⎟ = ⎜v + 19 2⎠ ⎝ 1.60 × 10− J⎠ ⎝ ⎝ For v = 0, 1, 2,

P43.17

1⎞ ⎟ ( 0.196 eV) 2⎠

Evib = 0.098 2 eV, 0.295 eV, 0.491 eV .

In Benzene, the carbon atoms are each 0.110 nm from the axis and each hydrogen atom is (0.110 + 0.100 nm ) = 0.210 nm from the axis. Thus, I = Σmr 2 : I = 6 (1.99 ×10 − 26 kg) (0.110 ×10 − 9 m ) +6 (1.67 ×10 − 2

27

kg) ( 0.210 × 10−9 m)

2

= 1.89 × 10−45 kg ⋅ m 2 The allowed rotational energies are then ℏ2 (1.055 ×10 −34 J ⋅s ) J (J + 1) = ( 2.95 × 10 − 24 J ) J ( J + 1) J ( J + 1) = 2I 2 (1.89 ×10 −4 5 kg ⋅m 2 ) 2

Erot =

= ( 18.4 × 10−6 eV )J (J + 1) Erot = (18.4 µ eV) J ( J +1) where J = 0, 1, 2, 3, . . . The first five of these allowed energies are: Erot = 0, 36.9µ eV, 111µ eV, 221µ eV, and 369µ eV.

Molecules and Solids

P43.18

527

We carry extra digits through the solution because part (c) involves the subtraction of two close numbers. The longest wavelength corresponds to the smallest energy difference between the ℏ2 rotational energy levels. It is between J = 0 and J = 1, namely I hc hc 4 π 2 Ic λ= . If m is the reduced mass, then = = ∆E min ℏ 2 I h I = µ r2 = µ (0.127 46 ×10 −9 m ) = (1.624 605 × 10 −20 m 2 ) µ 2

4π 2 (1. 624 605 × 10

− 20

Therefore λ =

µ35 =

(a)

m 2 ) µ ( 2. 997 925 × 10 8 m s)

6. 626 075 × 10 −34 J ⋅ s

(1) = ( 2. 901 830 × 10 23 m kg ) µ

(1. 007 825 u)(34. 968 853 u) = 0 .979 593u = 1 .626 653 × 10−27 kg 1 .007 825 u + 34 .968 853u

From (1): λ 35 = ( 2.901830 × 10 23 m kg ) (1.626 653 × 10 − 27 kg ) = 472 µm

µ37 =

(b)

(1.007 825 u)(36.965 903 u) 1. 007 825u + 36. 965 903 u

=0 .981 077 u =1.629118 ×10 −27 kg

From (1): λ 37 = (2.901830 × 10 23 m kg )(1.629118 × 10− 27 kg ) = 473 µm

λ 37 − λ 35 = 472.742 4 µ m − 472.027 0 µ m = 0.715 µ m

(c)

P43.19

We find an average s...