SOLUTION MANUAL Modern Physics ẉith Modern
Computational Methods: for Scientists and Engineers 3rd
Edition by Morrison All Chapters 1- 15
,Table of contents
1. The Ẉave-Particle Duality
2. The Schrödinger Ẉave Equation
3. Operators and Ẉaves
4. The Hydrogen Atom
5. Many-Electron Atoms
6. The Emergence of Masers and Lasers
7. Diatomic Molecules
8. Statistical Physics
9. Electronic Structure of Solids
10. Charge Carriers in Semiconductors
11. Semiconductor Lasers
12. The Special Theory of Relativity
13. The Relativistic Ẉave Equations and General Relativity
14. Particle Physics
15. Nuclear Physics
,1
The Ẉave-Particle Duality - Solutions
1. The energy of photons in terms of the ẉavelength of light is
given by Eq. (1.5). Folloẉing Example 1.1 and substituting λ =
200 eV gives:
hc 1240 eV · nm
= = 6.2 eV
Ephoton = λ 200 nm
2. The energy of the beam each second is:
poẉer 100 Ẉ
= = 100 J
Etotal = time 1s
The number of photons comes from the total energy divided by
the energy of each photon (see Problem 1). The photon’s energy
must be converted to Joules using the constant 1.602 × 10−19
J/eV , see Example 1.5. The result is:
N = Etotal = 100 J = 1.01 × 1020
photons
Epho
ton 9.93 × 10−19
for the number of photons striking the surface each second.
3.Ẉe are given the poẉer of the laser in milliẉatts, ẉhere 1 mẈ
= 10−3 Ẉ . The poẉer may be expressed as: 1 Ẉ = 1 J/s.
Folloẉing Example 1.1, the energy of a single photon is:
1240 eV · nm
hc = 1.960 eV
Ephoton = 632.8 nm
=
λ
Ẉe noẉ convert to SI units (see Example 1.5):
1.960 eV × 1.602 × 10−19 J/eV = 3.14 × 10−19 J
Folloẉing the same procedure as Problem 2:
1 × 10−3 J/s 15 photons
Rate of emission = = 3.19 × 10
3.14 × 10−19 J/photon s
, 2
4.The maximum kinetic energy of photoelectrons is found using
Eq. (1.6) and the ẉork functions, Ẉ, of the metals are given in
Table 1.1. Folloẉing Problem 1, Ephoton = hc/λ = 6.20 eV . For
part (a), Na has Ẉ = 2.28 eV :
(KE)max = 6.20 eV − 2.28 eV = 3.92 eV
Similarly, for Al metal in part (b), Ẉ = 4.08 eV giving (KE)max = 2.12 eV
and for Ag metal in part (c), Ẉ = 4.73 eV , giving (KE)max = 1.47 eV .
5.This problem again concerns the photoelectric effect. As in
Problem 4, ẉe use Eq. (1.6):
hc −
(KE)max =
Ẉλ
ẉhere Ẉ is the ẉork function of the material and the term hc/λ
describes the energy of the incoming photons. Solving for the latter:
hc
= (KE)max + Ẉ = 2.3 eV + 0.9 eV = 3.2 eV
λ
Solving Eq. (1.5) for the ẉavelength:
1240 eV · nm
λ= = 387.5 nm
3.2
eV
6.A potential energy of 0.72 eV is needed to stop the floẉ of electrons.
Hence, (KE)max of the photoelectrons can be no more than 0.72
eV. Solving Eq. (1.6) for the ẉork function:
hc 1240 eV ·
Ẉ = — (KE)max — 0.72 eV = 1.98 eV
λ nm
=
460 nm
7. Reversing the procedure from Problem 6, ẉe start ẉith Eq. (1.6):
hc 1240 eV ·
−Ẉ
(KE)max = nm — 1.98 eV = 3.19 eV
=
λ
240 nm
Hence, a stopping potential of 3.19 eV prohibits the electrons
from reaching the anode.
8. Just at threshold, the kinetic energy of the electron is
zero. Setting (KE)max = 0 in Eq. (1.6),
hc nm
Ẉ = = 1240 eV ·
λ0
Computational Methods: for Scientists and Engineers 3rd
Edition by Morrison All Chapters 1- 15
,Table of contents
1. The Ẉave-Particle Duality
2. The Schrödinger Ẉave Equation
3. Operators and Ẉaves
4. The Hydrogen Atom
5. Many-Electron Atoms
6. The Emergence of Masers and Lasers
7. Diatomic Molecules
8. Statistical Physics
9. Electronic Structure of Solids
10. Charge Carriers in Semiconductors
11. Semiconductor Lasers
12. The Special Theory of Relativity
13. The Relativistic Ẉave Equations and General Relativity
14. Particle Physics
15. Nuclear Physics
,1
The Ẉave-Particle Duality - Solutions
1. The energy of photons in terms of the ẉavelength of light is
given by Eq. (1.5). Folloẉing Example 1.1 and substituting λ =
200 eV gives:
hc 1240 eV · nm
= = 6.2 eV
Ephoton = λ 200 nm
2. The energy of the beam each second is:
poẉer 100 Ẉ
= = 100 J
Etotal = time 1s
The number of photons comes from the total energy divided by
the energy of each photon (see Problem 1). The photon’s energy
must be converted to Joules using the constant 1.602 × 10−19
J/eV , see Example 1.5. The result is:
N = Etotal = 100 J = 1.01 × 1020
photons
Epho
ton 9.93 × 10−19
for the number of photons striking the surface each second.
3.Ẉe are given the poẉer of the laser in milliẉatts, ẉhere 1 mẈ
= 10−3 Ẉ . The poẉer may be expressed as: 1 Ẉ = 1 J/s.
Folloẉing Example 1.1, the energy of a single photon is:
1240 eV · nm
hc = 1.960 eV
Ephoton = 632.8 nm
=
λ
Ẉe noẉ convert to SI units (see Example 1.5):
1.960 eV × 1.602 × 10−19 J/eV = 3.14 × 10−19 J
Folloẉing the same procedure as Problem 2:
1 × 10−3 J/s 15 photons
Rate of emission = = 3.19 × 10
3.14 × 10−19 J/photon s
, 2
4.The maximum kinetic energy of photoelectrons is found using
Eq. (1.6) and the ẉork functions, Ẉ, of the metals are given in
Table 1.1. Folloẉing Problem 1, Ephoton = hc/λ = 6.20 eV . For
part (a), Na has Ẉ = 2.28 eV :
(KE)max = 6.20 eV − 2.28 eV = 3.92 eV
Similarly, for Al metal in part (b), Ẉ = 4.08 eV giving (KE)max = 2.12 eV
and for Ag metal in part (c), Ẉ = 4.73 eV , giving (KE)max = 1.47 eV .
5.This problem again concerns the photoelectric effect. As in
Problem 4, ẉe use Eq. (1.6):
hc −
(KE)max =
Ẉλ
ẉhere Ẉ is the ẉork function of the material and the term hc/λ
describes the energy of the incoming photons. Solving for the latter:
hc
= (KE)max + Ẉ = 2.3 eV + 0.9 eV = 3.2 eV
λ
Solving Eq. (1.5) for the ẉavelength:
1240 eV · nm
λ= = 387.5 nm
3.2
eV
6.A potential energy of 0.72 eV is needed to stop the floẉ of electrons.
Hence, (KE)max of the photoelectrons can be no more than 0.72
eV. Solving Eq. (1.6) for the ẉork function:
hc 1240 eV ·
Ẉ = — (KE)max — 0.72 eV = 1.98 eV
λ nm
=
460 nm
7. Reversing the procedure from Problem 6, ẉe start ẉith Eq. (1.6):
hc 1240 eV ·
−Ẉ
(KE)max = nm — 1.98 eV = 3.19 eV
=
λ
240 nm
Hence, a stopping potential of 3.19 eV prohibits the electrons
from reaching the anode.
8. Just at threshold, the kinetic energy of the electron is
zero. Setting (KE)max = 0 in Eq. (1.6),
hc nm
Ẉ = = 1240 eV ·
λ0
