Solution Manual for Physics for Scientists & Engineers with Modern Physics, Global Edition, 5th edition Douglas C. Giancoli

SOLUTION MANUAL Modern Physics with
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Modern Computational Methods: for Scientists
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and Engineers 3rd Edition by Morrison
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Chapters
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1- 15aa aa


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,Table of contents k9 k9 aa

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1. The Wave-Particle Duality aa
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2. The Schrödinger Wave Equation aa
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3. Operators and Waves aa
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4. The Hydrogen Atom aa
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5. Many-Electron Atoms aa
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6. The Emergence of Masers and Lasers aa
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7. Diatomic Molecules aa
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8. Statistical Physics aa
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9. Electronic Structure of Solids aa
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10. Charge Carriers in Semiconductors aa
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11. Semiconductor Lasers aa
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12. The Special Theory of Relativity aa
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13. The Relativistic Wave Equations and General Relativity aa
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14. Particle Physics aa
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15. Nuclear Physics aa
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1 aa


The k aa9 aa Wave-Particle k aa9 aa Duality k aa9 aa - k aa9 aa Solutions aa
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1. The energy of photons in terms of the wavelength of light is
k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k



9 given by Eq. (1.5). Following Example
k9 k9 k9 k9 k9 k aa9 aa 1.1 and substituting λ aa = 200 eV gives:
k9 k9 k9 k 9 k9 k9 k9

aa


aa hc aa1240 aa aaeV aa aa· nm aa k 9 k 9 k9



= aa = 6.2 eV aa k9 k9



Ephoton = aaλ aa 200 nm aa k9 k9




2. The k aa9 aa energy k aa9 aa of k aa9 aa the k aa9 aa beam k aa9 aa each k second
aa9 aa k aa9 aais: aa aa power aa 100 W aa
k9




= aa= 100 J aaEtotal = aatime aa1 s
k9 k9 k9 k9

aa


The number of photons comes from the total energy divided b
k9 k9 k9 k9 k9 k9 k9 k9 k9 k9



aay the energy of each photon (see Problem 1). The photon’s ener
k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9


−1
aagy must be converted to Joules using the constant 1.602 × 10
k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9


9 J/eV , see Example 1.5. The result is: aa
aa k9 k9 k9 k9 k9 k9 k9 k9




N aa= E aatotal k9 k aa9 aa = aa100 J aa= 1.01 × 1020 aaphotons aaEpho aa aa
k9 k9 k9 k9

, −
ton aa 9.93 × 10 19 aa k9 k9




for aa aathe aa
k 9 k 9 aanumber aa aaof aa aaphotons aa aastriking k 9 k 9 k 9 k aa9 aa the k aa9 aa surface k aa9 aa each k aa9

aasecond. aa




3. We are given the power of the laser in milliwatts, where 1 mW
k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9


−3
aak9 = 10 W . The power may be expressed as: 1 W = 1 J/s. Foll
k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9



aa owing Example 1.1, the energy of a single photon is: aa
k9 k9 k9 k9 k9 k9 k9 k9 k9




aa 1240 aa aaeV aa aa· nm aa k 9 k 9 k9



hc aa= 1.960 eV aa k9 k9 k9



Ephoton aa aa= 632.8 nm k 9 k9 aa




aa aa = aaλ aa k 9 k 9



We now aaconvert to SI units (see aaExample aa aa1.5): aa
k9 k9 k9 k9 k9 k9 k9 k 9




1.960 eV × 1.602 × 10−19 J/eV aa aa= 3.14 × 10−19 J aa
k9 k9 k9 k9 k9 k9 k 9 k9 k9 k9 k9




Following the same procedure as Problem 2: aa k9 k9 k9 k9 k9 k9




1 × 10−3 J/s
k9 k9 k9
aa
15 k aa9 aa photons aa
Rate of k9 k aa9 aa emission = k9 k aa9 aa 3.14 k9 k9 −19 k aa9 aa J/photon k aa9 aa = 3.19 × 10 k9 k9 k9 aa s aa

× 10 aa



2
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4. The maximum kinetic energy of photoelectrons is found usi
k9 k9 k9 k9 k9 k9 k9 k9




ng Eq. (1.6) and the work functions, W, of the metals are give
k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9



aan in Table 1.1. Following Problem aa aa1, aa aaEphoton = hc/λ = 6.20 aa aaeV
k9 k9 k9 k9 k9 k 9 k 9 k9 k9 k9 k9 k 9 k9



aa. aa aaFor aa aapart aa aa(a), aa aaNa aa aahas aa aaW aa aa= 2.28 aa aaeV : aa
k 9 k 9 k 9 k 9 k 9 k 9 k 9 k9 k 9 k9




(KE)max = 6.20 eV − 2.28 eV aa aa= 3.92 eV aa
k9 k9 k9 k9 k9 k9 k 9 k9 k9




Similarly, for Al metal in part (b), W aa aa= 4.08 eV aa aagiving (KE)max = 2.12 e aaV aa
k9 k9 k9 k9 k9 k9 k9 k 9 k9 k9 k 9 k9 k9 k9 k9



and for Ag metal in part (c), W = 4.73 eV , giving (KE)max = 1.47 eV . aa
k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9




5. This problem again concerns the photoelectric effect. As in Pr
k9 k9 k9 k9 k9 k9 k9 k9 k9



aa oblem 4, we use Eq. (1.6): aa k9 k9 k9 k9 k9




aa (KE)max aa= aa hc − k9 k9 k9



λ
aaW k9




where aa aaW aa aais aa aathe aa aawork aa aafunction aa aaof aa aathe aa aamaterial aa aaand
k 9 k 9 k 9 k 9 k 9 k 9 k 9 k 9 k 9 k the
aa9 aa k aa9

aaterm aa aahc k 9




/λ aa aadescribes the energy of the incoming photons. Solving for the la
k 9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9



aatter: aahc aa λ aa aa= (KE) + W aa aa= 2.3 eV aa aa+ 0.9 eV aa aa= 3.2 eV aa k 9 k9 maxk9 k9 k 9 k9 k9 k 9 k9 k9 k 9 k9 k9




Solving Eq. (1.5) for the wavelength: aa k9 k9 k9 k9 k9




1240 k aa9 aa eV aa aa· nm aa k 9 k9

, λ = aa
k9
= 387.5 nm aa k9 k9



3.2 k aa9

aae



V aa
6. A potential energy of 0.72 eV is needed to stop the flow of electro
k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9



aans. Hence, (KE)max of the photoelectrons can be no more than 0.7
k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9



aa2 eV. Solving Eq. (1.6) for the work function: aa
k9 k9 k9 k9 k9 k9 k9 k9




hck9 aa k aa9 aa — aa 1240 k aa9 aa eV aa aa· aa— 0.72
k 9 k9 k9 k aa9 aa eV aa aa= 1.98
k 9 k9 k aa9 aa eV aa
W aa=
λ nm aa
(KE)max k



9 = aa aa
aa

460 nm aa k9




7. Reversing aa aathe k 9 k aa9 aa procedure k aa9 aa from k aa9 aa Problem k aa9 aa 6, k aa9 aa we k aa9 aa start k with
aa9 aa k



aa aaEq. aa aa(1.6): aa
9 k 9




−W
aa k9 hc k aa9 aa k9 k aa9 1240 k aa9 aa eV aa aa· aa— 1.98 k 9 k9 k9 k aa9 aa eV aa aa= 3.19
k 9 k9 k eV aa
aa9 aa


(KE)max aa= aa
= aa nm aa
λ aa aa aa


240 nm aa k9




Hence, a stopping potential of 3.19 eV prohibits the electrons fro
k9 k9 k9 k9 k9 k9 k9 k9 k9 k9



aam reaching the anode. aa
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8. Just aa aaat aa aathreshold, aa aathe aa aakinetic aa aaenergy aa aaof
k 9 k 9 k 9 k 9 k 9 k 9 k aa9 aa the k aa9 aa electron k i
aa9 aa

aas aa aazero. aa aaSetting (KE)max = 0 aa aain aa aaEq. aa aa(1.6), aa
k 9 k 9 k9 k9 k9 k 9 k 9 k 9




hc aa1240 aa aaeV aa aa· aa
k9 k 9 k 9 k9




W aa= aa λ0 aa= aa nm aa = 3.44 k9 k aa9 aa eV aa

aa aa

360 nm aa k9



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9. A frequency of 1200 THz is equal to 1200 × 1012 Hz. Using Eq. (1.10), aa
k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9 k9




Ephoton = hf = 4.136 × 10−15 eV aa aa· s × 1.2 × 1015 Hz = 4.96 eV aa
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