Download Chapter 30: Quantum Physics

Chapter 30: Quantum Physics
8. Predict/Calculate A Famous Double Star Albireo in the constellation Cygnus, which appears as a single star to the naked eye, is actually a beautiful
double-star system, as shown in FIGURE 30-24. The brighter of the two stars is referred to as A (or Beta-01 Cygni), with a surface temperature of TA =
4700 K; its companion is B (or Beta-02 Cygni), with a surface temperature of TB = 13,000 K. (a) When viewed through a telescope, one star is a brilliant
blue color, and the other has a warm golden color, as shown in the photo. Is the blue star A or B? Explain. (b) What is the ratio of the peak frequencies
emitted by the two stars, (ƒA / ƒB)?
8.
Picture the Problem: A negative image of the double star Albireo is shown at right. When
viewed through a telescope, the upper-left star A is a warm golden color and the lower-right
star B is a brilliant blue.
Strategy: According to Wien’s displacement law, a hotter star will emit light with a higher
peak frequency. Because blue has a higher frequency than yellow, the blue star will be the
hotter star. Use Wien’s displacement law (Equation 30-1) to calculate the ratio of the peak
frequencies from these two stars.
Solution: 1. (a) Star B is the blue star. A higher temperature corresponds to a higher frequency, and f blue  f yellow .
10 1
1
f A  5.88 10 s  K  TA
4700 K


 0.36
10 1
1
f B  5.88 10 s  K  TB 13, 000 K
2. (b) Calculate the ratio of the peak frequencies:
Insight: Star A’s peak frequency is in the infrared spectrum (see Chapter 25), hence it has a higher intensity in the redyellow part of the visible spectrum and it appears golden. Star B’s peak frequency is in the ultraviolet spectrum, hence it
has a higher intensity in the blue end of the spectrum and it appears blue.
29. Predict/Calculate Aluminum and calcium have photoelectric work functions of WAl = 4.08 eV and WCa = 2.87 eV, respectively. (a) Which metal
requires higher-frequency light to produce photoelectrons? Explain. (b) Calculate the minimum frequency that will produce photoelectrons from each
surface.
29. Picture the Problem: Photons with energies greater than the work function of a metal can eject electrons from that
metal.
Strategy: Use Equation 30-6 to calculate the cut-off frequencies for the two metals.
Solution: 1. (a) Because higher-frequency photons have higher energies, and since WAl  WCa , aluminum requires
higher-frequency light to produce photoelectrons.
2. (b) Calculate the cutoff frequency for aluminum:
f Al 
3. Calculate the cutoff frequency for calcium:
f Ca 
19
WAl  4.08 eV  1.60 10 J/eV 

 9.85 1014 Hz
h
6.63 1034 J  s
19
WCa  2.87 eV  1.60 10 J/eV 

 6.93 1014 Hz
h
6.63 1034 J  s
Insight: A photon with frequency 9.851014 Hz will eject electrons from either surface. Such a photon has a
wavelength of 304 nm and is in the ultraviolet region of the electromagnetic spectrum (see Chapter 25).
Copyright © 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
30 – 1
James S. Walker, Physics, 5th Edition
Chapter 30: Quantum Physics
36. BIO Owl Vision Owls have large, sensitive eyes for good night vision. Typically, the pupil of an owl’s eye can have a diameter of 8.5 mm (as compared
with a maximum diameter of about 7.0 mm for humans). In addition, an owl’s eye is about 100 times more sensitive to light of low intensity than a human
eye, allowing owls to detect light with an intensity as small as 5.0×10−13 W/m2. Find the minimum number of photons per second an owl can detect,
assuming a frequency of 7.0×1014 Hz for the light.
36. Picture the Problem: The large pupil in an owl’s eye allows for more photons per second to enter the eye than the
number of photons that would enter the human eye. In addition, the retina of the owl is more sensitive to fewer photons
than the human eye.
Strategy: Multiply the minimum intensity of light by the area of the owl’s pupil to determine the power absorbed by the
owl’s eye. Divide this power by the energy per photon to determine the minimum photon flux that the owl can detect.
Solution: 1. Calculate the power absorbed
in the owl’s eye at minimum intensity:
P  I A  I  r 2   5.0 1013 W/m2    0.00425 m  2.837 1017 W
2. Calculate the energy per photon:
E  h f   6.63 1034 J  s  7.0 1014 Hz   4.64 1019 J/photon
3. Divide the power by the energy:
n
2
 2.837 1017 W   61 photons/s
P

E 4.64 1019 J/photon
Insight: If the human eye were sensitive to 61 photons/s, the smaller pupil of the human eye would require a minimum
light intensity of 7.4 1013 W/m2 .
68. Predict/Calculate An electron and a proton have the same de Broglie wavelength. (a) Which has the greater kinetic energy? Explain. (b) Calculate the
ratio of the electron’s kinetic energy to the kinetic energy of the proton.
68. Picture the Problem: An electron and proton have the same de Broglie wavelength, which means that they must also
have the same momentum. However, their kinetic energies will differ because they have different masses.
Strategy: Use Equation 30-16 to write the kinetic energy in terms of the de Broglie wavelength. Then divide the kinetic
energy of the electron by the kinetic energy of the proton to calculate their ratio.
Solution: 1. (a) The proton and electron have the same momentum because they have the same de Broglie wavelength.
The kinetic energy, K  p 2 2m , is inversely proportional to the mass, and me  mp . For identical momenta, an electron
has a greater kinetic energy than a proton.
2. (b) Write the kinetic energy in terms of λ:
K  p 2 2m  h2 2m 2
3. Calculate the ratio of the kinetic energies:
Ke h2 2me  2 mp 1.673 1027 kg



 1836
K p h2 2mp  2 me 9.109 1031 kg
Insight: In this problem we used the classical equation for the kinetic energy. As the speeds approach the speed of light,
keeping the de Broglie wavelengths equal, the ratio of kinetic energies decreases due to relativistic effects.
Copyright © 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
30 – 2
Chapter 31: Atomic Physics
17. How much energy is required to ionize hydrogen when it is in the n = 3 state?
17. Picture the Problem: An amount of energy equal to the binding energy of the electron must be added to a hydrogen
atom in order to ionize it.
Strategy: The Bohr model can be used to determine the binding energy of an electron in any quantum state. The energy
necessary is equal to the energy difference between n = ∞ and the n = 3 Bohr orbits. Use Equation 31-9 to calculate the
binding energy:
Solution: Use Equation 31-9 to find Ebinding :
 13.6 eV 
Ebinding  E  E4  0   
  1.51 eV
32 

Insight: Ionizing a hydrogen atom from the ground state requires 13.6 eV, significantly more than that required to
ionize from the n = 3 state. However, the excitation of the atom to the n = 3 state from the ground state requires
13.6 eV – 1.51 eV = 12.1 eV.
31. Predict/Calculate The potential energy of a hydrogen atom in a particular Bohr orbit is −1.20×10−19 J. (a) Which Bohr orbit does the electron occupy in
this atom? (b) Suppose the electron moves away from the nucleus to the next higher Bohr orbit. Does the potential energy of the atom increase, decrease, or
stay the same? Explain. (c) Calculate the potential energy of the atom for the orbit referred to in part (b).
31. Picture the Problem: The Bohr model of the hydrogen atom can be used to correlate the potential energy of the
electron with the quantum number n.
Strategy: Write the electrostatic potential energy of the atom (Equation 20-8) in terms of the Bohr radius. Solve the
resulting equation for the orbit number. To calculate the potential of the next higher orbit, replace n by n + 1 in the
potential energy equation.
k e2
k e2
 2
rn
n r1
Solution: 1. (a) Substitute the Bohr
radius into Equation 20-8:
Un  
2. Solve for the orbit number:
  8.99 109 N  m2 / C2 1.60 1019 C 
ke2
n

 6
r1U n
5.29 1011 m  1.20 1019 J 
2
3. (b) If n increases, the potential energy will become a smaller negative number and thus increase. This is because
positive work is done on the electron to move it farther away from the nucleus.
4. (c) Set n = 7 in the
potential energy equation:
8.99 109 N  m2 / C2 1.60 1019 C 
k e2
U7  

r7
7 2  5.29  1011 m 
2
  0.888 1019 J   0.555 eV
Insight: The potential energy is equal to twice the total energy, so the orbit number could also have been found by
using Equation 31-9.
See class notes for a hint on homework problem 31.75
Copyright © 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
31 – 1
James S. Walker, Physics, 5th Edition
Chapter 31: Atomic Physics
38. Find an expression, in terms of n and the fundamental constants, for the de Broglie wavelength of an electron in the nth state of the hydrogen atom.
38. Picture the Problem: An electron in the nth state of a hydrogen atom has a corresponding de Broglie wavelength.
Strategy: The de Broglie wavelength is given in Equation 30-16 in terms of the electron momentum. Write the
momentum as the product of mass and velocity, with the velocity given by Equation 31-6.
Solution: 1. Write the de Broglie wavelength
in terms of the velocity in the nth state:
n 
2. Substitute the expression for the
velocity using Equation 31-6:
n 
h
h

pn mvn
h
m  2 ke nh 
2

nh 2
2 mke2
Insight: In general, the de Broglie wavelength is n times the ground state wavelength, or  n  n1  n  0.332 nm  .
39. What is the radius of the hydrogen-atom Bohr orbit shown in FIGURE 31-40?
39. Picture the Problem: The image shows the de Broglie wavelength associated with a certain
Bohr orbit for a hydrogen atom.
Strategy: The circumference of each Bohr orbit accommodates an integer multiple of the de
Broglie wavelength. There are five wavelengths contained in the orbit shown in the figure,
and we conclude the electron is in the n = 5 state. Use Equation 31-5 to calculate the radius
of the n = 5 orbit.
Solution: Calculate the radius for n = 5:
rn  n 2 r1
r5  52  5.29 10 –11 m   1.32 nm
Insight: The de Broglie wavelength is proportional to the orbit number (see Problem 38), and the number of
wavelengths contained in the circumference is equal to the orbit number. These two facts require the radius to be
proportional to n2 (Equation 31-5).
41. (a) Find the kinetic energy (in eV) of an electron whose de Broglie wavelength is equal to 0.5 Å = 0.05 nm, a typical atomic size. (b) If the wavelength
of the electron is reduced to 10−15 m, a typical nuclear size, is its kinetic energy greater than, less than, or equal to the value found in part (a)? Explain.
41. Picture the Problem: An electron that has a de Broglie wavelength equal to the size of an atom (5×10−11 m = 0.5 Å) or
the size of a nucleus (10−15 m = 1 fm) has a definite momentum and a corresponding kinetic energy.
Strategy: Write the kinetic energy as K  p 2 2m and use Equation 30-16 to write the momentum in terms of the de
Broglie wavelength. For part (b) the energy of an electron with a smaller wavelength is very much larger than it was in
part (a). [In fact, the electron’s kinetic energy would be larger than its 511-keV rest energy, so that relativistic
calculations (Equation 29-9 and others) must be employed.]
2
p2
1 h
h2


 
2m 2m   
2m 2
Solution: 1. (a) Solve for K in terms
of the de Broglie wavelength:
K
2. Substitute a wavelength of 0.5 Å:
 6.63 10 J  s 
K
2  9.1110 kg  0.5 10
–34
–31
2
–10


1 eV
 600 eV

–19 
1.60

10
J

m 
2
3. (b) The expression K  h2 2m 2 indicates that if the de Broglie wavelength of the electron is reduced, its kinetic
energy will be greater than that calculated in part (a).
Insight: If we use the relativistic expression K 
 hc /  
2
  m0 c 2   m0 c 2 , we find that the kinetic energy of the
2
electron with a de Broglie wavelength of 1.0 fm is 1240 MeV. If we instead employed the classical approach of part
(a), we would have calculated K = 2×1012 eV. Notice that the energies at the tiny scale of a nucleus are much greater
than those at the atomic scale.
Copyright © 2017 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
31 – 2