Problems Section 27.1 The Principle of Linear Superposition, Section 27.2 Young's Double-Slit Experiment 1. In a Young's double-slit experiment, the wavelength of the light used is 520 nm (in vacuum), and the separation between the slits is . Determine the angle that locates (a) the dark fringe for which , Answer: REASONING The angles that determine the locations of the dark and bright fringes in a Young's double-slit experiment are related to the integers m that identify the fringes, the wavelength of the light, and the separation d between the slits. Since values are given for m, , and d, the angles can be calculated. SOLUTION The expressions that specify in terms of m, , and d are as follows: (27.1) (27.2) Applying these expressions gives the answers that we seek. (a) (b) (c) (d) (b) the bright fringe for which , Answer: REASONING The angles that determine the locations of the dark and bright fringes in a Young's double-slit experiment are related to the integers m that identify the fringes, the wavelength of the light, and the separation d between the slits. Since values are given for m, , and d, the angles can be calculated. SOLUTION The expressions that specify in terms of m, , and d are as follows: (27.1) (27.2) Applying these expressions gives the answers that we seek. (a) (b)
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Problems
Section 27.1 The Principle of Linear Superposition, Section 27.2 Young's Double-Slit Experiment
1. In a Young's double-slit experiment, the wavelength of the light used is 520 nm (in vacuum), and the separation
between the slits is . Determine the angle that locates
(a) the dark fringe for which ,
Answer:
REASONING The angles that determine the locations of the dark and bright fringes in a Young's
double-slit experiment are related to the integers m that identify the fringes, the wavelength of the light,
and the separation d between the slits. Since values are given for m, , and d, the angles can be calculated.
SOLUTION The expressions that specify in terms of m, , and d are as follows:
(27.1)
(27.2)
Applying these expressions gives the answers that we seek.
(a)
(b)
(c)
(d)
(b) the bright fringe for which ,
Answer:
REASONING The angles that determine the locations of the dark and bright fringes in a Young's
double-slit experiment are related to the integers m that identify the fringes, the wavelength of the light,
and the separation d between the slits. Since values are given for m, , and d, the angles can be calculated.
SOLUTION The expressions that specify in terms of m, , and d are as follows:
(27.1)
(27.2)
Applying these expressions gives the answers that we seek.
(a)
(b)
(c)
(d)
(c) the dark fringe for which , and
Answer:
REASONING The angles that determine the locations of the dark and bright fringes in a Young's
double-slit experiment are related to the integers m that identify the fringes, the wavelength of the light,
and the separation d between the slits. Since values are given for m, , and d, the angles can be calculated.
SOLUTION The expressions that specify in terms of m, , and d are as follows:
(27.1)
(27.2)
Applying these expressions gives the answers that we seek.
(a)
(b)
(c)
(d)
(d) the bright fringe for which .
Answer:
REASONING The angles that determine the locations of the dark and bright fringes in a Young's
double-slit experiment are related to the integers m that identify the fringes, the wavelength of the light,
and the separation d between the slits. Since values are given for m, , and d, the angles can be calculated.
SOLUTION The expressions that specify in terms of m, , and d are as follows:
(27.1)
(27.2)
Applying these expressions gives the answers that we seek.
(a)
(b)
(c)
(d)
REASONING The angles that determine the locations of the dark and bright fringes in a Young's double-slit
experiment are related to the integers m that identify the fringes, the wavelength of the light, and the separation d
between the slits. Since values are given for m, , and d, the angles can be calculated.
SOLUTION The expressions that specify in terms of m, , and d are as follows:
(27.1)
(27.2)
Applying these expressions gives the answers that we seek.
(a)
(b)
(c)
(d)
2. In a Young's double-slit experiment, the angle that locates the second dark fringe on either side of the central bright
fringe is . Find the ratio of the slit separation d to the wavelength of the light.
3. Two in-phase sources of waves are separated by a distance of 4.00 m. These sources produce identical waves that
have a wavelength of 5.00 m. On the line between them, there are two places at which the same type of interference
occurs.
(a) Is it constructive or destructive interference, and
Answer: Destructive interference occurs.
(b) where are the places located?
Answer: 3.25 m and 0.75 m from one of the sources
4. The dark fringe for in a Young's double-slit experiment is located at an angle of . What is the angle
that locates the dark fringe for ?
5. In a Young's double-slit experiment, the seventh dark fringe is located 0.025 m to the side of the central bright
fringe on a flat screen, which is 1.1 m away from the slits. The separation between the slits is . What is
the wavelength of the light being used?
Answer:
6.
Two parallel slits are illuminated by light composed of two wavelengths. One wavelength is
. The other wavelength is and is unknown. On a viewing screen, the light with wavelength
produces its third-order bright fringe at the same place where the light with wavelength produces its
fourth dark fringe. The fringes are counted relative to the central or zeroth-order bright fringe. What is the unknown
wavelength?
7. In a setup like that in Figure 27.7, a wavelength of 625 nm is used in a Young's double-slit experiment. The
separation between the slits is . The total width of the screen is 0.20 m. In one version of the setup,
the separation between the double slit and the screen is , whereas in another version it is .
On one side of the central bright fringe, how many bright fringes lie on the screen in the two versions of the setup? Do
not include the central bright fringe in your counting.
Answer: 6 (version A), 4 (version B)
*8. At most, how many bright fringes can be formed on either side of the central bright fringe when light of wavelength
625 nm falls on a double slit whose slit separation is ?
*9. In a Young's double-slit experiment the separation y between the second-order bright fringe and the central
bright fringe on a flat screen is 0.0180 m when the light has a wavelength of 425 nm. Assume that the angles that
locate the fringes on the screen are small enough so that . Find the separation y when the light has a
wavelength of 585 nm.
Answer: 0.0248 m
**10. In Young's experiment a mixture of orange light (611 nm) and blue light (471 nm) shines on the double slit. The
centers of the first-order bright blue fringes lie at the outer edges of a screen that is located 0.500 m away from the
slits. However, the first-order bright orange fringes fall off the screen. By how much and in which direction (toward or
away from the slits) should the screen be moved so that the centers of the first-order bright orange fringes will just
appear on the screen? It may be assumed that is small, so that sin .
**11. A sheet that is made of plastic covers one slit of a double slit (see the drawing). When the
double slit is illuminated by monochromatic light , the center of the screen appears
dark rather than bright. What is the minimum thickness of the plastic?
Problem 11
Answer: 487 nm
REASONING The light that travels through the plastic has a different path length than the light that
passes through the unobstructed slit. Since the center of the screen now appears dark, rather than bright,
destructive interference, rather than constructive interference occurs there. This means that the difference
between the number of wavelengths in the plastic sheet and that in a comparable thickness of air is .
SOLUTION The wavelength of the light in the plastic sheet is given by Equation 27.3 as
The number of wavelengths contained in a plastic sheet of thickness t is
The number of wavelengths contained in an equal thickness of air is
where we have used the fact that . Destructive interference occurs when the difference,
, in the number of wavelengths is :
Solving this equation for t yields .
Section 27.3 Thin-Film Interference
12. You are standing in air and are looking at a flat piece of glass on which there is a layer of transparent
plastic . Light whose wavelength is 589 nm is vacuum is incident nearly perpendicularly on the coated
glass and reflects into your eyes. The layer of plastic looks dark. Find the two smallest possible nonzero values for the
thickness of the layer.
13. A nonreflective coating of magnesium fluoride covers the glass of a camera lens.
Assuming that the coating prevents reflection of yellow-green light , determine
the minimum nonzero thickness that the coating can have.
Answer: 102 nm
REASONING To solve this problem, we must express the condition for destructive interference in terms of
the film thickness t and the wavelength of the light as it passes through the magnesium fluoride coating. We must
also take into account any phase changes that occur upon reflection.
SOLUTION Since the coating is intended to be nonreflective, its thickness must be chosen so that destructive
interference occurs between waves 1 and 2 in the drawing. For destructive interference, the combined phase difference
between the two waves must be an odd integer number of half wavelengths. The phase change for wave 1 is
equivalent to one-half of a wavelength, since this light travels from a smaller refractive index toward a
larger refractive index .
Similarly, there is a phase change when wave 2 reflects from the right surface of the film, since this light also travels
from a smaller refractive index toward a larger one . Therefore, a phase change of
one-half wavelength occurs at both boundaries, so the net phase change between waves 1 and 2 due to reflection is
zero. Since wave 2 travels back and forth through the film and, and since the light is assumed to be at nearly normal
incidence, the extra distance traveled by wave 2 compared to wave 1 is twice the film thickness, or . Thus, in this
case, the minimum condition for destructive interference is
The wavelength of light in the coating is
(27.3)
Solving the above expression for t, we find that the minimum thickness that the coating can have is
14. When monochromatic light shines perpendicularly on a soap film with air on each side, the second
smallest nonzero film thickness for which destructive interference of reflected light is observed is 296 nm. What is the
vacuum wavelength of the light in nm?
15. A transparent film is deposited on a glass plate to form a nonreflecting coating. The
film has a thickness that is . What is the longest possible wavelength (in vacuum) of light for which
this film has been designed?
Answer:
16.
A tank of gasoline is open to the air . A thin film of liquid floats on the gasoline and has
a refractive index that is between 1.00 and 1.40. Light that has a wavelength of 625 nm (in vacuum) shines
perpendicularly down through the air onto this film, and in this light the film looks bright due to constructive
interference. The thickness of the film is 242 nm and is the minimum nonzero thickness for which constructive
interference can occur. What is the refractive index of the film?
17. Review Conceptual Example 4 before beginning this problem. A soap film with different thicknesses at
different places has an unknown refractive index n and air on both sides. In reflected light it looks multicolored. One
region looks yellow because destructive interference has removed blue from the reflected light,
while another looks magenta because destructive interference has removed green . In these
regions the film has the minimum nonzero thickness t required for the destructive interference to occur. Find the ratio
.
Answer: 1.18
*18. A film of oil lies on wet pavement. The refractive index of the oil exceeds that of the water. The film has the
minimum nonzero thickness such that it appears dark due to destructive interference when viewed in red light
. Assuming that the visible spectrum extends from 380 to 750 nm, for which
visible wavelength(s) in vacuum will the film appear bright due to constructive interference?
*19. Orange light shines on a soap film that has air on either side of it. The light
strikes the film perpendicularly. What is the minimum thickness of the film for which constructive interference causes
it to look bright in reflected light?
Answer: 115 nm
REASONING To solve this problem, we must express the condition for constructive interference in terms of
the film thickness t and the wavelength of the light in the soap film. We must also take into account any phase
changes that occur upon reflection.
SOLUTION For the reflection at the top film surface, the light travels from air, where the refractive index is smaller
, toward the film, where the refractive index is larger . Associated with this reflection there is
a phase change that is equivalent to one-half of a wavelength. For the reflection at the bottom film surface, the light
travels from the film, where the refractive index is larger , toward air, where the refractive index is
smaller . Associated with this reflection, there is no phase change. As a result of these two reflections,
there is a net phase change that is equivalent to one-half of a wavelength. To obtain the condition for constructive
interference, this net phase change must be added to the phase change that arises because of the film thickness t, which is traversed twice by the light that penetrates it. For constructive interference we find that
or
Equation 27.3 indicates that . Using this expression and the fact that for the minimum
thickness t, we find that the condition for constructive interference becomes
or
*20. The drawing shows a cross section of a planoconcave lens resting on a flat glass plate. (A planoconcave lens
has one surface that is a plane and the other that is concave spherical.) The thickness t is . The lens is
illuminated with monochromatic light , and a series of concentric bright and dark rings is
formed, much like Newton's rings. How many bright rings are there?
(Hint: The cross section shown in the drawing reveals that a kind of air wedge exists between the place where the two
pieces of glass touch and the top of the curved surface where the distance t is marked.) **21. A piece of curved glass has a radius of curvature of and is used to form Newton's rings, as in Figure
27.13. Not counting the dark spot at the center of the pattern, there are one hundred dark fringes, the last one being at
the outer edge of the curved piece of glass. The light being used has a wavelength of 654 nm in vacuum. What is the
radius R of the outermost dark ring in the pattern?
(Hint: Note that r is much greater than R, and you may assume that tan for small angles, where must be expressed in radians.)
Answer: 0.0256 m
**22. A uniform layer of water lies on a glass plate . Light shines perpendicularly on
the layer. Because of constructive interference, the layer looks maximally bright when the wavelength of the
light is 432 nm in vacuum and also when it is 648 nm in vacuum.
(a) Obtain the minimum thickness of the film.
(b) Assuming that the film has the minimum thickness and that the visible spectrum extends from 380 to
750 nm, determine the visible wavelength(s) in vacuum for which the film appears completely dark.
Section 27.5 Diffraction
23. (a) As Section 17.3 discusses, high-frequency sound waves exhibit less diffraction than low-frequency sound
waves do. However, even high-frequency sound waves exhibit much more diffraction under normal
circumstances than do light waves that pass through the same opening. The highest frequency that a healthy
ear can typically hear is . Assume that a sound wave with this frequency travels at 343 m/s
and passes through a doorway that has a width of 0.91 m. Determine the angle that locates the first
minimum to either side of the central maximum in the diffraction pattern for the sound. This minimum is
equivalent to the first dark fringe in a single-slit diffraction pattern for light.
Answer:
(b)
Suppose that yellow light passes through a doorway and that the first
dark fringe in its diffraction pattern is located at the angle determined in part (a). How wide would this
hypothetical doorway have to be?
Answer:
24. A dark fringe in the diffraction pattern of a single slit is located at an angle of . With the same light, the
same dark fringe formed with another single slit is at an angle of . Find the ratio of the widths of
the two slits.
25. A diffraction pattern forms when light passes through a single slit. The wavelength of the light is 675 nm.
Determine the angle that locates the first dark fringe when the width of the slit is
(a) and
Answer:
REASONING This problem can be solved by using Equation 27.4 for the value of the angle when
(first dark fringe).
SOLUTION
(a) When the slit width is and , we find, according to
Equation 27.4,
(b) Similarly, when the slit width is and , we find
(b) .
Answer:
REASONING This problem can be solved by using Equation 27.4 for the value of the angle when
(first dark fringe).
SOLUTION
(a) When the slit width is and , we find, according to
Equation 27.4,
(b) Similarly, when the slit width is and , we find
REASONING This problem can be solved by using Equation 27.4 for the value of the angle when
(first dark fringe).
SOLUTION
(a) When the slit width is and , we find, according to Equation
27.4,
(b) Similarly, when the slit width is and , we find
26. A slit has a width of . When light with a wavelength of passes through this
slit, the width of the central bright fringe on a flat observation screen has a certain value. With the screen kept in the
same place, this slit is replaced with a second slit (width ), and a wavelength of is used. The width of
the central bright fringe on the screen is observed to be unchanged. Find .
27. Light that has a wavelength of 668 nm passes through a slit wide and falls on a screen that is 1.85
m away. What is the distance on the screen from the center of the central bright fringe to the third dark fringe on either
side?
Answer: 0.576 m
REASONING The drawing shows a top view of the slit and screen, as well as the position of the central
bright fringe and the third dark fringe. The distance y can be obtained from the tangent function as tan . Since
L is given, we need to find the angle before y can be determined. According to Equation 27.4, the angle is related to
the wavelength of the light and the width W of the slit by , where since we are interested in
the angle for the third dark fringe.
SOLUTION We will first compute the angle between the central bright fringe and the third dark fringe using Equation
27.4 (with ):
The vertical distance is
28. Light shines through a single slit whose width is . A diffraction pattern is formed on a flat screen
located 4.0 m away. The distance between the middle of the central bright fringe and the first dark fringe is 3.5 mm.
What is the wavelength of the light?
*29. Light waves with two different wavelengths, 632 nm and 474 nm, pass simultaneously through a single slit
whose width is and strike a screen 1.20 m from the slit. Two diffraction patterns are formed on the
screen. What is the distance (in cm) between the common center of the diffraction patterns and the first occurrence of
a dark fringe from one pattern falling on top of a dark fringe from the other pattern?
Answer: 3.18 cm
*30. The central bright fringe in a single-slit diffraction pattern has a width that equals the distance between the screen
and the slit. Find the ratio of the wavelength of the light to the width W of the slit.
*31. How many dark fringes will be produced on either side of the central maximum if light is
incident on a single slit that is wide?
Answer: 8
REASONING The angle that specifies the location of the dark fringe is given by
(Equation 27.4), where is the wavelength of the light and W is the width of the slit. When has its maximum value of
, the number of dark fringes that can be produced is a maximum. We will use this information to obtain a value
for this number.
SOLUTION Solving Equation 27.4 for m, and setting , we have
Therefore, the number of dark fringes is .
**32. In a single-slit diffraction pattern, the central fringe is 450 times as wide as the slit. The screen is 18 000 times farther
from the slit than the slit is wide. What is the ratio , where is the wavelength of the light shining through the
slit and W is the width of the slit? Assume that the angle that locates a dark fringe on the screen is small, so that
.
Section 27.6 Resolving Power
33. Two stars are apart and are equally distant from the earth. A telescope has an objective lens with a
diameter of 1.02 m and just detects these stars as separate objects. Assume that light of wavelength 550 nm is being
observed. Also assume that diffraction effects, rather than atmospheric turbulence, limit the resolving power of the
telescope. Find the maximum distance that these stars could be from the earth.
Answer:
34.
It is claimed that some professional baseball players can see which way the ball is spinning as it travels toward
home plate. One way to judge this claim is to estimate the distance at which a batter can first hope to resolve two
points on opposite sides of a baseball, which has a diameter of 0.0738 m.
(a) Estimate this distance, assuming that the pupil of the eye has a diameter of 2.0 mm and the wavelength of
the light is 550 nm in vacuum.
(b) Considering that the distance between the pitcher's mound and home plate is 18.4 m, can you rule out the
claim based on your answer to part (a)?
35.
Late one night on a highway, a car speeds by you and fades into the distance. Under these conditions the
pupils of your eyes have diameters of about 7.0 mm. The taillightsof this car are separated by a distance of 1.2 m and
emit red light . How far away from you is this car when its taillights appear to
merge into a single spot of light because of the effects of diffraction?
Answer:
REASONING According to Rayleigh's criterion, the two taillights must be separated by a distance s sufficient
to subtend an angle at the pupil of the observer's eye. Recalling that this angle must be expressed in
radians, we relate to the distances s and L.
SOLUTION The wavelength is 660 nm. Therefore, we have from Equation 27.6
According to Equation 8.1, the distance L between the observer and the taillights is
36. An inkjet printer uses tiny dots of red, green, and blue ink to produce an image. Assume that the dot separation on
the printed page is the same for all colors. At normal viewing distances, the eye does not resolve the individual dots,
regardless of color, so that the image has a normal look. The wavelengths for red, green, and blue are
, , and . The diameter of the pupil through which light enters the eye is 2.0 mm.
For a viewing distance of 0.40 m, what is the maximum allowable dot separation?
37.
A hunter who is a bit of a braggart claims that from a distance of 1.6 km he can selectively shoot either of
two squirrels who are sitting ten centimeters apart on the same branch of a tree. What's more, he claims that he can do
this without the aid of a telescopic sight on his rifle.
(a) Determine the diameter of the pupils of his eyes that would be required for him to be able to resolve the
squirrels as separate objects. In this calculation use a wavelength of 498 nm (in vacuum) for the light.
Answer: 9.7 mm
(b) State whether his claim is reasonable, and provide a reason for your answer. In evaluating his claim,
consider that the human eye automatically adjusts the diameter of its pupil over a typical range of 2 to 8
mm, the larger values coming into play as the lighting becomes darker. Note also that under dark conditions,
the eye is most sensitive to a wavelength of 498 nm.
Answer: The hunter's claim is not reasonable.
38.
Review Conceptual Example 8 as background for this problem. In addition to the data given there, assume that
the dots in the painting are separated by 1.5 mm and that the wavelength of the light is . Find the
distance at which the dots can just be resolved by
(a) the eye and
(b) the camera.
39. Astronomers have discovered a planetary system orbiting the star Upsilon Andromedae, which is at a distance
of from the earth. One planet is believed to be located at a distance of from the star. Using
visible light with a vacuum wavelength of 550 nm, what is the minimum necessary aperture diameter that a telescope
must have so that it can resolve the planet and the star?
Answer: 2.3 m
*40.
The pupil of an eagle's eye has a diameter of 6.0 mm. Two field mice are separated by 0.010 m. From a
distance of 176 m, the eagle sees them as one unresolved object and dives toward them at a speed of 17 m/s. Assume
that the eagle's eye detects light that has a wavelength of 550 nm in vacuum. How much time passes until the eagle
sees the mice as separate objects?
*41. Consult Multiple-Concept Example 7 to see a model for solving this kind of problem. You are
using a microscope to examine a blood sample. Recall from Section 26.12 that the sample should be placed just
outside the focal point of the objective lens of the microscope.
(a) If the specimen is being illuminated with light of wavelength and the diameter of the objective
equals its focal length, determine the closest distance between two blood cells that can just be
resolved. Express your answer in terms of .
Answer:
REASONING Assuming that the angle is small, the distance y between the blood cells
is given by
(8.1)
where f is the distance between the microscope objective and the cells (which is given as the focal
length of the objective). However, the minimum angular separation of the cells is given by the
Rayleigh criterion as (Equation 27.6), where is the wavelength of the light and D
is the diameter of the objective. These two relations can be used to find an expression for y in terms
of .
SOLUTION
(a) Substituting Equation 27.6 into Equation 8.1 yields
Since it is given that , we see that .
(b) Because y is proportional to , the wavelength must be to resolve cells that are closer
together.
(b) Based on your answer to (a), should you use light with a longer wavelength or a shorter wavelength if
you wish to resolve two blood cells that are even closer together?
Answer: shorter wavelength
REASONING Assuming that the angle is small, the distance y between the blood cells
is given by
(8.1)
where f is the distance between the microscope objective and the cells (which is given as the focal
length of the objective). However, the minimum angular separation of the cells is given by the
Rayleigh criterion as (Equation 27.6), where is the wavelength of the light and D
is the diameter of the objective. These two relations can be used to find an expression for y in terms
of .
SOLUTION
(a) Substituting Equation 27.6 into Equation 8.1 yields
Since it is given that , we see that .
(b) Because y is proportional to , the wavelength must be to resolve cells that are closer
together.
REASONING Assuming that the angle is small, the distance y between the blood cells is given by
(8.1)
where f is the distance between the microscope objective and the cells (which is given as the focal length of the
objective). However, the minimum angular separation of the cells is given by the Rayleigh criterion as
(Equation 27.6), where is the wavelength of the light and D is the diameter of the
objective. These two relations can be used to find an expression for y in terms of .
SOLUTION
(a) Substituting Equation 27.6 into Equation 8.1 yields
Since it is given that , we see that .
(b) Because y is proportional to , the wavelength must be to resolve cells that are closer together.
**42. Two concentric circles of light emit light whose wavelength is 555 nm. The larger circle has a radius of 4.0 cm,
and the smaller circle has a radius of 1.0 cm. When taking a picture of these lighted circles, a camera admits
light through an aperture whose diameter is 12.5 mm. What is the maximum distance at which the camera can
(a) distinguish one circle from the other and
(b) reveal that the inner circle is a circle of light rather than a solid disk of light?
Section 27.7 The Diffraction Grating,
Section 27.8 Compact Discs, Digital Video Discs, and the Use of Interference
43. A diffraction grating is 1.50 cm wide and contains 2400 lines. When used with light of a certain wavelength, a
third-order maximum is formed at an angle of . What is the wavelength (in nm)?
Answer: 644 nm
REASONING The angle that specifies the third-order maximum of a diffraction grating is
(Equation 27.7), where is the wavelength of the light, and d is the separation between the slits of the
grating. The separation is equal to the width of the grating (1.50 cm) divided by the number of lines (2400).
SOLUTION Solving Equation 27.7 for the wavelength, we obtain
44. The light shining on a diffraction grating has a wavelength of 495 nm (in vacuum). The grating produces a second-
order bright fringe whose position is defined by an angle of . How many lines per centimeter does the grating
have?
45. For a wavelength of 420 nm, a diffraction grating produces a bright fringe at an angle of . For an unknown
wavelength, the same grating produces a bright fringe at an angle of . In both cases the bright fringes are of the
same order m. What is the unknown wavelength?
Answer: 630 nm
46. Two diffraction gratings, A and B, are located at the same distance from the observation screens. Light with the
same wavelength is used for each. The separation between adjacent principal maxima for grating A is 2.7 cm, and
for grating B it is 3.2 cm. Grating A has 2000 lines per meter. How many lines per meter does grating B have?
(Hint: The diffraction angles are small enough that the approximation can be used.)
47. The wavelength of the laser beam used in a compact disc player is 780 nm. Suppose that a diffraction grating
produces first-order tracking beams that are 1.2 mm apart at a distance of 3.0 mm from the grating. Estimate the
spacing between the slits of the grating.
Answer:
REASONING AND SOLUTION The geometry of the situation is shown below.
From the geometry, we have
Then, solving Equation 27.7 with for the separation d between the slits, we have
48. The first-order principle maximum produced by a grating is located at an angle of . What is the angle for
the third-order maximum with the same light?
*49. A diffraction grating has 2604 lines per centimeter, and it produces a principal maximum at . The
grating is used with light that contains all wavelengths between 410 and 660 nm. What is (are) the wavelength(s) of
the incident light that could have produced this maximum?
Answer: 640 nm and 480 nm
*50. Light of wavelength 410 nm (in vacuum) is incident on a diffraction grating that has a slit separation of
. The distance between the grating and the viewing screen is 0.15 m. A diffraction pattern is produced
on the screen that consists of a central bright fringe and higher-order bright fringes (see the drawing).
Problem 50
(a) Determine the distance y from the
central bright fringe to the second-
order bright fringe.
(Hint: The diffraction angles are
small enough that the approximation
can be used.)
(b) If the entire apparatus is submerged in water , what is the distance y?
*51. Violet light and red light lie at opposite ends of the visible
spectrum.
(a) For each wavelength, find the angle that locates the first-order maximum produced by a grating with 3300
lines/cm. This grating converts a mixture of all colors between violet and red into a rainbow-like dispersion
between the two angles. Repeat the calculation above for
Answer: violet light:
red light:
REASONING The angle that locates the first-order maximum produced by a grating with 3300
lines/cm is given by Equation 27.7, , with the order of the fringes given by
Any two of the diffraction patterns will overlap when their angular positions are the
same.
SOLUTION Since the grating has 3300 lines/cm, we have
(a) In first order, therefore, for violet light,
Similarly for red light,
(b) Repeating the calculation for the second order maximum , we find that
(c) Repeating the calculation for the third order maximum , we find that
(d) Comparisons of the values for calculated in parts (a), (b) and (c) show that the
.
(b) the second-order maximum and
Answer: violet light:
red light:
REASONING The angle that locates the first-order maximum produced by a grating with 3300
lines/cm is given by Equation 27.7, , with the order of the fringes given by
Any two of the diffraction patterns will overlap when their angular positions are the
same.
SOLUTION Since the grating has 3300 lines/cm, we have
(a) In first order, therefore, for violet light,
Similarly for red light,
(b) Repeating the calculation for the second order maximum , we find that
(c) Repeating the calculation for the third order maximum , we find that
(d) Comparisons of the values for calculated in parts (a), (b) and (c) show that the
.
(c) the third-order maximum.
Answer: violet light:
red light:
REASONING The angle that locates the first-order maximum produced by a grating with 3300
lines/cm is given by Equation 27.7, , with the order of the fringes given by
Any two of the diffraction patterns will overlap when their angular positions are the
same.
SOLUTION Since the grating has 3300 lines/cm, we have
(a) In first order, therefore, for violet light,
Similarly for red light,
(b) Repeating the calculation for the second order maximum , we find that
(c) Repeating the calculation for the third order maximum , we find that
(d) Comparisons of the values for calculated in parts (a), (b) and (c) show that the
.
(d) From your results, decide whether there is an overlap between any of the “rainbows” and, if so, specify
which orders overlap.
Answer: The second and third orders overlap.
REASONING The angle that locates the first-order maximum produced by a grating with 3300
lines/cm is given by Equation 27.7, , with the order of the fringes given by
Any two of the diffraction patterns will overlap when their angular positions are the
same.
SOLUTION Since the grating has 3300 lines/cm, we have
(a) In first order, therefore, for violet light,
Similarly for red light,
(b) Repeating the calculation for the second order maximum , we find that
(c) Repeating the calculation for the third order maximum , we find that
(d) Comparisons of the values for calculated in parts (a), (b) and (c) show that the
.
REASONING The angle that locates the first-order maximum produced by a grating with 3300 lines/cm is
given by Equation 27.7, , with the order of the fringes given by Any two of the
diffraction patterns will overlap when their angular positions are the same.
SOLUTION Since the grating has 3300 lines/cm, we have
(a) In first order, therefore, for violet light,
Similarly for red light,
(b) Repeating the calculation for the second order maximum , we find that
(c) Repeating the calculation for the third order maximum , we find that
(d) Comparisons of the values for calculated in parts (a), (b) and (c) show that the
.
**52. The distance between adjacent slits of a certain diffraction grating is . The grating is illuminated by
monochromatic light with a wavelength of 656.0 nm, and is then heated so that its temperature increases by
. Determine the change in the angle of the seventh-order principal maximum that occurs as a result of the
thermal expansion of the grating. The coefficient of linear expansion for the diffraction grating is
. Be sure to include the proper algebraic sign with your answer: if
the angle decreases.
**53. Two gratings A and B have slit separations and , respectively. They are used with the same light and the same
observation screen. When grating A is replaced with grating B, it is observed that the first-order maximum of A is
exactly replaced by the second-order maximum of B.
(a) Determine the ratio of the spacings between the slits of the gratings.
Answer: 2
(b) Find the next two principal maxima of grating A and the principal maxima of B that exactly replace them
when the gratings are switched. Identify these maxima by their order numbers.
Section 28.3 The Relativity of Time: Time Dilation
1. A particle known as a pion lives for a short time before breaking apart into other particles. Suppose that a pion is
moving at a speed of 0.990c, and an observer who is stationary in a laboratory measures the pion's lifetime to be
.
(a) What is the lifetime according to a hypothetical person who is riding along with the pion?
Answer:
REASONING
(a) The two events in this problem are the creation of the pion and its subsequent decay (or breaking
apart). Imagine a reference frame attached to the pion, so the pion is stationary relative to this
reference frame. To a hypothetical person who is at rest with respect to this reference frame, these
two events occur at the same place, namely, at the place where the pion is located. Thus, this
hypothetical person measures the proper time interval for the decay of the pion. On the other
hand, the person standing in the laboratory sees the two events occurring at different locations, since
the pion is moving relative to that person. The laboratory person, therefore, measures a dilated time
interval . The relation between these two time intervals is given by
(Equation 28.1).
(b) According to the hypothetical person who is at rest in the reference frame attached to the moving
pion, the distance x that the laboratory travels before the pion breaks apart is equal to the speed v of
the laboratory relative to the pion times the proper time interval , or . The speed of the
laboratory relative to the pion is the same as the speed of the pion relative to the laboratory, namely,
.
SOLUTION
(a) The proper time interval is
(28.1)
(b) The distance x that the laboratory travels before the pion breaks apart, as measured by the
hypothetical person, is
(b) According to this hypothetical person, how far does the laboratory move before the pion breaks apart?
Answer: 1.5 m
REASONING
(a) The two events in this problem are the creation of the pion and its subsequent decay (or breaking
apart). Imagine a reference frame attached to the pion, so the pion is stationary relative to this
reference frame. To a hypothetical person who is at rest with respect to this reference frame, these
two events occur at the same place, namely, at the place where the pion is located. Thus, this
hypothetical person measures the proper time interval for the decay of the pion. On the other
hand, the person standing in the laboratory sees the two events occurring at different locations, since
the pion is moving relative to that person. The laboratory person, therefore, measures a dilated time
interval . The relation between these two time intervals is given by
(Equation 28.1).
(b) According to the hypothetical person who is at rest in the reference frame attached to the moving
pion, the distance x that the laboratory travels before the pion breaks apart is equal to the speed v of
the laboratory relative to the pion times the proper time interval , or . The speed of the
laboratory relative to the pion is the same as the speed of the pion relative to the laboratory, namely,
.
SOLUTION
(a) The proper time interval is
(28.1)
(b) The distance x that the laboratory travels before the pion breaks apart, as measured by the
hypothetical person, is
REASONING
(a) The two events in this problem are the creation of the pion and its subsequent decay (or breaking apart).
Imagine a reference frame attached to the pion, so the pion is stationary relative to this reference frame. To a
hypothetical person who is at rest with respect to this reference frame, these two events occur at the same place,
namely, at the place where the pion is located. Thus, this hypothetical person measures the proper time interval
for the decay of the pion. On the other hand, the person standing in the laboratory sees the two events
occurring at different locations, since the pion is moving relative to that person. The laboratory person,
therefore, measures a dilated time interval . The relation between these two time intervals is given by
(Equation 28.1).
(b) According to the hypothetical person who is at rest in the reference frame attached to the moving pion, the
distance x that the laboratory travels before the pion breaks apart is equal to the speed v of the laboratory
relative to the pion times the proper time interval , or . The speed of the laboratory relative to
the pion is the same as the speed of the pion relative to the laboratory, namely, .
SOLUTION
(a) The proper time interval is
(28.1)
(b) The distance x that the laboratory travels before the pion breaks apart, as measured by the hypothetical person,
is
2. A radar antenna is rotating and makes one revolution every 25 s, as measured on earth. However, instruments on a
spaceship moving with respect to the earth at a speed v measure that the antenna makes one revolution every 42 s.
What is the ratio of the speed v to the speed c of light in a vacuum?
3. Suppose that you are planning a trip in which a spacecraft is to travel at a constant velocity for exactly six
months, as measured by a clock on board the spacecraft, and then return home at the same speed. Upon your return,
the people on earth will have advanced exactly one hundred years into the future. According to special relativity, how
fast must you travel? Express your answer to five significant figures as a multiple of c—for example, 0.955 85c.
Answer:
REASONING The total time for the trip is one year. This time is the proper time interval , because it is
measured by an observer (the astronaut) who is at rest relative to the beginning and ending events (the times when the
trip started and ended) and who sees them at the same location in spacecraft. On the other hand, the astronaut
measures the clocks on earth to run at the dilated time interval , which is the time interval of one hundred years.
The relation between the two time intervals is given by Equation 28.1, which can be used to find the speed of the
spacecraft.
SOLUTION The dilated time interval is related to the proper time interval by .
Solving this equation for the speed v of the spacecraft yields
(28.1)
4. Suppose that you are traveling on board a spacecraft that is moving with respect to the earth at a speed of 0.975c.
You are breathing at a rate of 8.0 breaths per minute. As monitored on earth, what is your breathing rate?
*5. A 6.00-kg object oscillates back and forth at the end of a spring whose spring constant is 76.0 N/m. An
observer is traveling at a speed of relative to the fixed end of the spring. What does this observer
measure for the period of oscillation?
Answer: 2.28 s
*6. A spaceship travels at a constant speed from earth to a planet orbiting another star. When the spacecraft
arrives, 12 years have elapsed on earth, and 9.2 years have elapsed on board the ship. How far away (in meters) is the
planet, according to observers on earth?
**7. As observed on earth, a certain type of bacterium is known to double in number every 24.0 hours. Two
cultures of these bacteria are prepared, each consisting initially of one bacterium. One culture is left on earth
and the other placed on a rocket that travels at a speed of 0.866c relative to the earth. At a time when the earthbound culture has grown to 256 bacteria, how many bacteria are in the culture on the rocket, according to
an earth-based observer?
Answer: 16
Section
28.4 The Relativity of Length: Length Contraction
8. Suppose the straight-line distance between New York and San Francisco is (neglecting the curvature of
the earth). A UFO is flying between these two cities at a speed of relative to the earth. What do the voyagers
aboard the UFO measure for this distance?
9. How fast must a meter stick be moving if its length is observed to shrink to one-half of a meter?
Answer:
REASONING All standard meter sticks at rest have a length of 1.00 m for observers who are at rest with
respect to them. Thus, 1.00 m is the proper length of the meter stick. When the meter stick moves with speed v
relative to an earth-observer, its length will be a contracted length. Since both and L are known, v can
be found directly from Equation 28.2, .
SOLUTION Solving Equation 28.2 for v, we find that
10. The distance from earth to the center of our galaxy is about 23 000 ly
, as measured by an earth-based observer. A spaceship is to make this
journey at a speed of 0.9990c. According to a clock on board the spaceship, how long will it take to make the trip?
Express your answer in years .
11. A tourist is walking at a speed of along a 9.0-km path that follows an old canal. If the speed of light in
a vacuum were , how long would the path be, according to the tourist?
Answer: 8.1 km
REASONING The tourist is moving at a speed of with respect to the path and, therefore,
measures a contracted length L instead of the proper length of . The contracted length is given by the
length-contraction equation, Equation 28.2.
SOLUTION According to the length-contraction equation, the tourist measures a length that is
12. A Martian leaves Mars in a spaceship that is heading to Venus. On the way, the spaceship passes earth with a
speed relative to it. Assume that the three planets do not move relative to each other during the trip. The
distance between Mars and Venus is , as measured by a person on earth.
(a) What does the Martian measure for the distance between Mars and Venus?
(b) What is the time of the trip (in seconds) as measured by the Martian?
13. Two spaceships A and B are exploring a new planet. Relative to this planet, spaceship A has a speed of 0.60c, and
spaceship B has a speed of 0.80c. What is the ratio of the values for the planet's diameter that each spaceship
measures in a direction that is parallel to its motion?
Answer: 1.3
14. An unstable high-energy particle is created in the laboratory, and it moves at a speed of 0.990c. Relative to a
stationary reference frame fixed to the laboratory, the particle travels a distance of before
disintegrating. What are
(a) the proper distance and
(b) the distance measured by a hypothetical person traveling with the particle? Determine the particle's
(c) proper lifetime and
(d) its dilated lifetime.
*15. As the drawing shows, a carpenter on a space station has constructed a ramp. A rocket moves past the space
station with a relative speed of 0.730c in a direction parallel to side . What does a person aboard the rocket measure
for the angle of the ramp?
Answer:
**16. An object is made of glass and has the shape of a cube 0.11 m on a side, according to an observer at rest relative to it.
However, an observer moving at high speed parallel to one of the object's edges and knowing that the object's mass is
3.2 kg determines its density to be , which is much greater than the density of glass. What is the moving
observer's speed (in units of c) relative to the cube?
**17. A rectangle has the dimensions of when viewed by someone at rest with respect to it. When you move
past the rectangle along one of its sides, the rectangle looks like a square. What dimensions do you observe when you
move at the same speed along the adjacent side of the rectangle?
Answer:
Section 28.5 Relativistic Momentum
18. At what speed is the magnitude of the relativistic momentum of a particle three times the magnitude of the
nonrelativistic momentum?
19. What is the magnitude of the relativistic momentum of a proton with a relativistic total energy of ?
Answer:
20.
A spacecraft has a nonrelativistic (or classical) momentum whose magnitude is . The
spacecraft moves at such a speed that the pilot measures the proper time interval between two events to be one-half
the dilated time interval. Find the relativistic momentum of the spacecraft.
21. A woman is 1.6 m tall and has a mass of 55 kg. She moves past an observer with the direction of the motion
parallel to her height. The observer measures her relativistic momentum to have a magnitude of
. What does the observer measure for her height?
Answer: 1.0 m
REASONING The height of the woman as measured by the observer is given by Equation 28.2 as
, where is her proper height. In order to use this equation, we must determine the speed v of
the woman relative to the observer. We are given the magnitude of her relativistic momentum, so we can determine v
from p.
SOLUTION According to Equation 28.3 , so Squaring both sides,
we have
Solving for v and substituting values, we have
Then, the height that the observer measures for the woman is
22. Three particles are listed in the table. The mass and speed of each particle are given as multiples of the
variables m and v, which have the values and . The speed of light in a
vacuum is . Determine the momentum for each particle according to special relativity.
Particle Mass Speed
a m v
b
c
*23. Starting from rest, two skaters push off against each other on smooth level ice, where friction is
negligible. One is a woman and one is a man. The woman moves away with a velocity of relative
to the ice. The mass of the woman is 54 kg, and the mass of the man is 88 kg. Assuming that the speed of light
is , so that the relativistic momentum must be used, find the recoil velocity of the man relative to the
ice.
(Hint: This problem is similar to Example 6 in Chapter 7.)
Answer:
REASONING The magnitude p of the relativistic momentum of an object is given by
(Equation 28.3), where m is the object's mass, v is the object's speed, and c is the speed
of light in a vacuum. The principle of conservation of linear momentum (see Section 7.2) states that the total
momentum of a system is conserved when no net external force acts on the system. This principle applies at
speeds approaching the speed of light in a vacuum, provided that Equation 28.3 is used for the individual
momenta of the objects that comprise the system.
SOLUTION The total momentum of the man/woman system is conserved, since friction is negligible, so that
no net external force acts on the system. Therefore, the final total momentum must equal the initial total momentum, which is zero. As a result, where Equation 28.3 must be used for the momenta
and . Thus, we find
(1)
We know that , , and . Remember that c has the hypothetical
value of 3.0 m/s. Solving Equation 1 for reveals that . We choose the negative value,
since the man and woman recoil from one another and it is stated that the woman moves away in the positive
direction. Therefore, we find that .
Section 28.6 The Equivalence of Mass and Energy
24. Radium is a radioactive element whose nucleus emits an particle (a helium nucleus) with a kinetic energy of about
(4.9 MeV). To what amount of mass is this energy equivalent?
25. How much work must be done on an electron to accelerate it from rest to a speed of 0.990c?
Answer:
REASONING According to the work-energy theorem, Equation 6.3, the work that must be done on the
electron to accelerate it from rest to a speed of is equal to the kinetic energy of the electron when it is moving
at .
SOLUTION Using Equation 28.6, we find that
26. Review Conceptual Example 9 for background pertinent to this problem. Suppose that the speed of light in a vacuum
were one million times smaller than its actual value, so that . The spring constant of a spring is
850 N/m. Determine how far you would have to compress the spring from its equilibrium length in order to increase
its mass by 0.010 g.
27. Suppose that one gallon of gasoline produces of energy, and this energy is sufficient to operate a car for
twenty miles. An aspirin tablet has a mass of 325 mg. If the aspirin could be converted completely into thermal
energy, how many miles could the car go on a single tablet?
Answer:
28.
Two kilograms of water are changed
(a) from ice at into liquid water at and
(b) from liquid water at into steam at . For each situation, determine the change in mass of
the water.
29.
Determine the ratio of the relativistic kinetic energy to the nonrelativistic kinetic energy when a
particle has a speed of
(a) and
Answer: 1.0
REASONING AND SOLUTION
(a) In Section 28.6 it is shown that when the speed of a particle is (or less), the relativistic kinetic
energy becomes nearly equal to the nonrelativistic kinetic energy. Since the speed of the particle here
is , the ratio of the relativistic kinetic energy to the nonrelativistic kinetic energy is .
(b) Taking the ratio of the relativistic kinetic energy, Equation 28.6, to the nonrelativistic kinetic energy,
, we find that
(b) 0.970c.
Answer: 6.6
REASONING AND SOLUTION
(a) In Section 28.6 it is shown that when the speed of a particle is (or less), the relativistic kinetic
energy becomes nearly equal to the nonrelativistic kinetic energy. Since the speed of the particle here
is , the ratio of the relativistic kinetic energy to the nonrelativistic kinetic energy is .
(b) Taking the ratio of the relativistic kinetic energy, Equation 28.6, to the nonrelativistic kinetic energy,
, we find that
REASONING AND SOLUTION
(a) In Section 28.6 it is shown that when the speed of a particle is (or less), the relativistic kinetic energy
becomes nearly equal to the nonrelativistic kinetic energy. Since the speed of the particle here is , the
ratio of the relativistic kinetic energy to the nonrelativistic kinetic energy is .
(b)
Taking the ratio of the relativistic kinetic energy, Equation 28.6, to the nonrelativistic kinetic energy, ,
we find that
30. Multiple-Concept Example 6 reviews the principles that play a role in this problem. A nuclear power reactor generates
of power. In one year, what is the change in the mass of the nuclear fuel due to the energy being taken
from the reactor?
*31. Multiple-Concept Example 6 explores the approach taken in problems such as this one. Quasars are believed to
be the nuclei of galaxies in the early stages of their formation. Suppose that a quasar radiates electromagnetic energy
at the rate of . At what rate (in kg/s) is the quasar losing mass as a result of this radiation?
Answer:
*32.
An electron is accelerated from rest through a potential difference that has a magnitude of . The
mass of the electron is , and the negative charge of the electron has a magnitude of
.
(a) What is the relativistic kinetic energy (in joules) of the electron?
(b) What is the speed of the electron? Express your answer as a multiple of c, the speed of light in a vacuum.
*33. An object has a total energy of and a kinetic energy of . What is the magnitude of
the object's relativistic momentum?
Answer:
Section 28.7 The Relativistic Addition of Velocities
34. You are driving down a two-lane country road, and a truck in the opposite lane is traveling toward you. Suppose
that the speed of light in a vacuum is . Determine the speed of the truck relative to you when
(a) your speed is and the truck's speed is and
(b) your speed is and the truck's speed is . The speeds given in parts (a) and (b) are relative
to the ground.
35. A spacecraft approaching the earth launches an exploration vehicle. After the launch, an observer on earth sees
the spacecraft approaching at a speed of 0.50c and the exploration vehicle approaching at a speed of . What is
the speed of the exploration vehicle relative to the spaceship?
Answer:
REASONING Let's define the following relative velocities, assuming that the spaceship and exploration
vehicle are moving in the positive direction.
of Exploration vehicle relative to the Spaceship.
of Exploration vehicle relative to an Observer on
of Spaceship relative to an Observer on
The velocity can be determined from the velocity-addition formula, Equation 28.8:
The velocity of the observer on earth relative to the spaceship is not given. However, we know that is the
negative of , so .
SOLUTION The velocity of the exploration vehicle relative to the spaceship is
The speed of the exploration vehicle relative to the spaceship is the magnitude of this result or .
36. Spaceships of the future may be powered by ion-propulsion engines in which ions are ejected from the back of
the ship to drive it forward. In one such engine the ions are to be ejected with a speed of 0.80c relative to the
spaceship. The spaceship is traveling away from the earth at a speed of relative to the earth. What is the velocity
of the ions relative to the earth? Assume that the direction in which the spaceship is traveling is the positive direction,
and be sure to assign the correct plus or minus signs to the velocities.
37. The spaceship Enterprise 1 is moving directly away from earth at a velocity that an earth-based observer measures to
be . A sister ship, Enterprise 2, is ahead of Enterprise 1 and is also moving directly away from earth along
the same line. The velocity of Enterprise 2 relative to Enterprise 1 is . What is the velocity of Enterprise 2,
as measured by the earth-based observer?
Answer:
*38.
A person on earth notices a rocket approaching from the right at a speed of 0.75c and another rocket
approaching from the left at 0.65c. What is the relative speed between the two rockets, as measured by a passenger on
one of them?
*39. Refer to Conceptual Example 11 as an aid in solving this problem. An intergalactic cruiser has two types of
guns: a photon cannon that fires a beam of laser light and an ion gun that shoots ions at a velocity of 0.950c relative to
the cruiser. The cruiser closes in on an alien spacecraft at a velocity of 0.800c relative to this spacecraft. The captain
fires both types of guns. At what velocity do the aliens see
(a) the laser light and
Answer: These answers assume that the direction of the cruiser, the ions, and the laser light is the positive direction.
REASONING AND SOLUTION In all parts of this problem, the direction of the intergalactic
cruiser, the ions, and the laser light is taken to be the positive direction.
(a) According to the second postulate of special relativity, all observers measure the speed of light to be
c, regardless of their velocities relative to each other. Therefore, the aliens aboard the hostile
spacecraft see the photons of the laser approach .
(b) To find the velocity of the ions relative to the aliens, we define the relative velocities as follows:
of the Ions relative to the alien Spacecraft
of the Ions relative to the intergalactic
of the intergalactic Cruiser relative to the alien
These velocities are related by the velocity-addition formula, Equation 28.8. The velocity of the ions
relative to the alien spacecraft is:
(c) The aliens see the laser light (photons) moving with respect to the cruiser at a velocity
(d) The aliens see the ions moving away from the cruiser at a velocity
(b) the ions approach them? At what velocity do the aliens see
Answer:
REASONING AND SOLUTION In all parts of this problem, the direction of the intergalactic
cruiser, the ions, and the laser light is taken to be the positive direction.
(a) According to the second postulate of special relativity, all observers measure the speed of light to be
c, regardless of their velocities relative to each other. Therefore, the aliens aboard the hostile
spacecraft see the photons of the laser approach .
(b) To find the velocity of the ions relative to the aliens, we define the relative velocities as follows:
of the Ions relative to the alien Spacecraft
of the Ions relative to the intergalactic
of the intergalactic Cruiser relative to the alien
These velocities are related by the velocity-addition formula, Equation 28.8. The velocity of the ions
relative to the alien spacecraft is:
(c) The aliens see the laser light (photons) moving with respect to the cruiser at a velocity
(d) The aliens see the ions moving away from the cruiser at a velocity
(c) the laser light and
Answer:
REASONING AND SOLUTION In all parts of this problem, the direction of the intergalactic
cruiser, the ions, and the laser light is taken to be the positive direction.
(a) According to the second postulate of special relativity, all observers measure the speed of light to be
c, regardless of their velocities relative to each other. Therefore, the aliens aboard the hostile
spacecraft see the photons of the laser approach .
(b) To find the velocity of the ions relative to the aliens, we define the relative velocities as follows:
of the Ions relative to the alien Spacecraft
of the Ions relative to the intergalactic
of the intergalactic Cruiser relative to the alien
These velocities are related by the velocity-addition formula, Equation 28.8. The velocity of the ions
relative to the alien spacecraft is:
(c) The aliens see the laser light (photons) moving with respect to the cruiser at a velocity
(d) The aliens see the ions moving away from the cruiser at a velocity
(d) the ions move away from the cruiser?
Answer:
REASONING AND SOLUTION In all parts of this problem, the direction of the intergalactic
cruiser, the ions, and the laser light is taken to be the positive direction.
(a) According to the second postulate of special relativity, all observers measure the speed of light to be
c, regardless of their velocities relative to each other. Therefore, the aliens aboard the hostile
spacecraft see the photons of the laser approach .
(b) To find the velocity of the ions relative to the aliens, we define the relative velocities as follows:
of the Ions relative to the alien Spacecraft
of the Ions relative to the intergalactic
of the intergalactic Cruiser relative to the alien
These velocities are related by the velocity-addition formula, Equation 28.8. The velocity of the ions
relative to the alien spacecraft is:
(c) The aliens see the laser light (photons) moving with respect to the cruiser at a velocity
(d) The aliens see the ions moving away from the cruiser at a velocity
REASONING AND SOLUTION In all parts of this problem, the direction of the intergalactic cruiser, the
ions, and the laser light is taken to be the positive direction.
(a) According to the second postulate of special relativity, all observers measure the speed of light to be c,
regardless of their velocities relative to each other. Therefore, the aliens aboard the hostile spacecraft see the
photons of the laser approach .
(b) To find the velocity of the ions relative to the aliens, we define the relative velocities as follows:
of the Ions relative to the alien Spacecraft
of the Ions relative to the intergalactic
of the intergalactic Cruiser relative to the alien
These velocities are related by the velocity-addition formula, Equation 28.8. The velocity of the ions relative to
the alien spacecraft is:
(c) The aliens see the laser light (photons) moving with respect to the cruiser at a velocity
(d) The aliens see the ions moving away from the cruiser at a velocity
*40. Two identical spaceships are under construction. The constructed length of each spaceship is 1.50 km. After
being launched, spaceship A moves away from earth at a constant velocity (speed is 0.850c) with respect to the earth.
Spaceship B follows in the same direction at a different constant velocity (speed is 0.500c) with respect to the earth.
Determine the length that a passenger on one spaceship measures for the other spaceship.
**41. Two atomic particles approach each other in a head-on collision. Each particle has a mass of . The
speed of each particle is when measured by an observer standing in the laboratory.
(a) What is the speed of one particle as seen by the other particle?
Answer:
(b) Determine the magnitude of the relativistic momentum of one particle, as it would be observed by the other.
Section 29.3 Photons and the Photoelectric Effect 1. The dissociation energy of a molecule is the energy required to break the molecule apart into its separate atoms. The
dissociation energy for the cyanogen molecule is . Suppose that this energy is provided by a single
photon. Determine the
(a) wavelength and
Answer:
(b) frequency of the photon.
Answer:
(c) In what region of the electromagnetic spectrum (see Figure 24.9) does this photon lie?
Answer: ultraviolet region
2. An AM radio station broadcasts an electromagnetic wave with a frequency of 665 kHz, whereas an FM station
broadcasts an electromagnetic wave with a frequency of 91.9 MHz. How many AM photons are needed to have a total
energy equal to that of one FM photon?
3. Ultraviolet light with a frequency of strikes a metal surface and ejects electrons that have a
maximum kinetic energy of 6.1 eV. What is the work function (in eV) of the metal?
Answer: 6.3 eV
REASONING According to Equation 29.3, the work function is related to the photon energy and the
maximum kinetic energy by . This expression can be used to find the work function of
the metal.
SOLUTION is 6.1 eV. The photon energy (in eV) is, according to Equation 29.2,
The work function is, therefore,
4. Light is shining perpendicularly on the surface of the earth with an intensity of . Assuming that all
the photons in the light have the same wavelength (in vacuum) of 730 nm, determine the number of photons per
second per square meter that reach the earth.
5. Ultraviolet light is responsible for sun tanning. Find the wavelength (in nm) of an ultraviolet photon whose
energy is .
Answer: 310 nm
REASONING The energy of the photon is related to its frequency by Equation 29.2, . Equation 16.1,
, relates the frequency and the wavelength for any wave.
SOLUTION Combining Equations 29.2 and 16.1, and noting that the speed of a photon is c, the speed of light in a
vacuum, we have
6. The maximum wavelength that an electromagnetic wave can have and still eject electrons from a metal surface is 485
nm. What is the work function of this metal? Express your answer in electron volts.
7. Radiation of a certain wavelength causes electrons with a maximum kinetic energy of 0.68 eV to be ejected from a
metal whose work function is 2.75 eV. What will be the maximum kinetic energy (in eV) with which this same
radiation ejects electrons from another metal whose work function is 2.17 eV?
Answer: 1.26 eV
8. Multiple-Concept Example 3 reviews the concepts necessary to solve this problem. Radiation with a wavelength
of 238 nm shines on a metal surface and ejects electrons that have a maximum speed of . Which one
of the following metals is it, the values in parentheses being the work functions: potassium (2.24 eV), calcium (2.71
Section 31.1 Nuclear Structure, Section 31.2 The Strong Nuclear Force and the Stability of the Nucleus
1. For find
(a) the net electrical charge of the nucleus,
Answer:
REASONING For an element whose chemical symbol is X, the symbol for the nucleus is ,
where A represents the total number of protons and neutrons (the nucleon number) and Z represents the
number of protons in the nucleus (the atomic number). The number of neutrons N is related to A and Z by
Equation 31.1: .
SOLUTION For the nucleus , we have and .
(a) The net electrical charge of the nucleus is equal to the total number of protons multiplied by the
charge on a single proton. Since the nucleus contains 82 protons, the net electrical charge of the
nucleus is
(b) The number of neutrons is .
(c) By inspection, the number of nucleons is .
(d) The approximate radius of the nucleus can be found from Equation 31.2, namely
(e) The nuclear density is the mass per unit volume of the nucleus. The total mass of the nucleus can be
found by multiplying the mass of a single nucleon by the total number A of nucleons in the
nucleus. Treating the nucleus as a sphere of radius r, the nuclear density is
Therefore,
(b) the number of neutrons,
Answer: 126
REASONING For an element whose chemical symbol is X, the symbol for the nucleus is ,
where A represents the total number of protons and neutrons (the nucleon number) and Z represents the
number of protons in the nucleus (the atomic number). The number of neutrons N is related to A and Z by
Equation 31.1: .
SOLUTION For the nucleus , we have and .
(a) The net electrical charge of the nucleus is equal to the total number of protons multiplied by the
charge on a single proton. Since the nucleus contains 82 protons, the net electrical charge of the
nucleus is
(b) The number of neutrons is .
(c) By inspection, the number of nucleons is .
(d) The approximate radius of the nucleus can be found from Equation 31.2, namely
(e) The nuclear density is the mass per unit volume of the nucleus. The total mass of the nucleus can be
found by multiplying the mass of a single nucleon by the total number A of nucleons in the
nucleus. Treating the nucleus as a sphere of radius r, the nuclear density is
Therefore,
(c) the number of nucleons,
Answer: 208
REASONING For an element whose chemical symbol is X, the symbol for the nucleus is ,
where A represents the total number of protons and neutrons (the nucleon number) and Z represents the
number of protons in the nucleus (the atomic number). The number of neutrons N is related to A and Z by
Equation 31.1: .
SOLUTION For the nucleus , we have and .
(a) The net electrical charge of the nucleus is equal to the total number of protons multiplied by the
charge on a single proton. Since the nucleus contains 82 protons, the net electrical charge of the
nucleus is
(b) The number of neutrons is .
(c) By inspection, the number of nucleons is .
(d) The approximate radius of the nucleus can be found from Equation 31.2, namely
(e) The nuclear density is the mass per unit volume of the nucleus. The total mass of the nucleus can be
found by multiplying the mass of a single nucleon by the total number A of nucleons in the
nucleus. Treating the nucleus as a sphere of radius r, the nuclear density is
Therefore,
(d) the approximate radius of the nucleus, and
Answer:
REASONING For an element whose chemical symbol is X, the symbol for the nucleus is ,
where A represents the total number of protons and neutrons (the nucleon number) and Z represents the
number of protons in the nucleus (the atomic number). The number of neutrons N is related to A and Z by
Equation 31.1: .
SOLUTION For the nucleus , we have and .
(a) The net electrical charge of the nucleus is equal to the total number of protons multiplied by the
charge on a single proton. Since the nucleus contains 82 protons, the net electrical charge of the
nucleus is
(b) The number of neutrons is .
(c) By inspection, the number of nucleons is .
(d) The approximate radius of the nucleus can be found from Equation 31.2, namely
(e) The nuclear density is the mass per unit volume of the nucleus. The total mass of the nucleus can be
found by multiplying the mass of a single nucleon by the total number A of nucleons in the
nucleus. Treating the nucleus as a sphere of radius r, the nuclear density is
Therefore,
(e) the nuclear density.
Answer:
REASONING For an element whose chemical symbol is X, the symbol for the nucleus is ,
where A represents the total number of protons and neutrons (the nucleon number) and Z represents the
number of protons in the nucleus (the atomic number). The number of neutrons N is related to A and Z by
Equation 31.1: .
SOLUTION For the nucleus , we have and .
(a) The net electrical charge of the nucleus is equal to the total number of protons multiplied by the
charge on a single proton. Since the nucleus contains 82 protons, the net electrical charge of the
nucleus is
(b) The number of neutrons is .
(c) By inspection, the number of nucleons is .
(d) The approximate radius of the nucleus can be found from Equation 31.2, namely
(e) The nuclear density is the mass per unit volume of the nucleus. The total mass of the nucleus can be
found by multiplying the mass of a single nucleon by the total number A of nucleons in the
nucleus. Treating the nucleus as a sphere of radius r, the nuclear density is
Therefore,
REASONING For an element whose chemical symbol is X, the symbol for the nucleus is , where A
represents the total number of protons and neutrons (the nucleon number) and Z represents the number of protons in
the nucleus (the atomic number). The number of neutrons N is related to A and Z by Equation 31.1: .
SOLUTION For the nucleus , we have and .
(a) The net electrical charge of the nucleus is equal to the total number of protons multiplied by the charge on a
single proton. Since the nucleus contains 82 protons, the net electrical charge of the nucleus is
(b) The number of neutrons is .
(c) By inspection, the number of nucleons is .
(d) The approximate radius of the nucleus can be found from Equation 31.2, namely
(e) The nuclear density is the mass per unit volume of the nucleus. The total mass of the nucleus can be found by
multiplying the mass of a single nucleon by the total number A of nucleons in the nucleus. Treating
the nucleus as a sphere of radius r, the nuclear density is
Therefore,
2. A nucleus contains 18 protons and 22 neutrons. What is the radius of this nucleus?
3. In each of the following cases, what element does the symbol X represent and how many neutrons are in the nucleus?
Use the periodic table on the inside of the back cover as needed.
(a)
Answer: 117 neutrons, platinum (Pt)
(b)
Answer: 16 neutrons, sulfur (S)
(c)
Answer: 34 neutrons, copper (Cu)
(d)
Answer: 6 neutrons, boron (B)
(e)
Answer: 145 neutrons, plutonium (Pu)
4. By what factor does the nucleon number of a nucleus have to increase in order for the nuclear radius to double?
5. In electrically neutral atoms, how many
(a) protons are in the uranium nucleus,
Answer: 92 protons
REASONING
(a) The number of protons in a given nucleus is specified by its atomic number Z.
(b) The number N of neutrons in a given nucleus is equal to the nucleon number A (the number of
protons and neutrons) minus the atomic number Z (the number of protons): (Equation
31.1).
(c) In an electrically neutral niobium atom, the number of electrons in orbit about the nucleus is equal to
the number of protons in the nucleus.
SOLUTION
(a) The number of protons in the uranium nucleus is .
(b) The number N of neutrons in the nucleus is .
(c) The number of electrons that orbit the nucleus in the neutral niobium atom is equal to the
number of protons in the nucleus, or .
(b) neutrons are in the mercury nucleus, and
Answer: 122 neutrons
REASONING
(a) The number of protons in a given nucleus is specified by its atomic number Z.
(b) The number N of neutrons in a given nucleus is equal to the nucleon number A (the number of
protons and neutrons) minus the atomic number Z (the number of protons): (Equation
31.1).
(c) In an electrically neutral niobium atom, the number of electrons in orbit about the nucleus is equal to
the number of protons in the nucleus.
SOLUTION
(a) The number of protons in the uranium nucleus is .
(b) The number N of neutrons in the nucleus is .
(c) The number of electrons that orbit the nucleus in the neutral niobium atom is equal to the
number of protons in the nucleus, or .
(c) electrons are in orbit about the niobium nucleus?
Answer: 41 electrons
REASONING
(a) The number of protons in a given nucleus is specified by its atomic number Z.
(b) The number N of neutrons in a given nucleus is equal to the nucleon number A (the number of
protons and neutrons) minus the atomic number Z (the number of protons): (Equation
31.1).
(c) In an electrically neutral niobium atom, the number of electrons in orbit about the nucleus is equal to
the number of protons in the nucleus.
SOLUTION
(a) The number of protons in the uranium nucleus is .
(b) The number N of neutrons in the nucleus is .
(c) The number of electrons that orbit the nucleus in the neutral niobium atom is equal to the
number of protons in the nucleus, or .
REASONING
(a) The number of protons in a given nucleus is specified by its atomic number Z.
(b) The number N of neutrons in a given nucleus is equal to the nucleon number A (the number of protons and
neutrons) minus the atomic number Z (the number of protons): (Equation 31.1).
(c) In an electrically neutral niobium atom, the number of electrons in orbit about the nucleus is equal to the
number of protons in the nucleus.
SOLUTION
(a) The number of protons in the uranium nucleus is .
(b) The number N of neutrons in the nucleus is .
(c) The number of electrons that orbit the nucleus in the neutral niobium atom is equal to the number of
protons in the nucleus, or .
6. The largest stable nucleus has a nucleon number of 209, and the smallest has a nucleon number of 1. If each
nucleus is assumed to be a sphere, what is the ratio (largest/smallest) of the surface areas of these spheres?
*7. The ratio of the radius of an unknown nucleus to the radius of a tritium nucleus is .
Both nuclei contain the same number of neutrons. Identify the unknown nucleus in the form . Use the periodic
table on the inside of the back cover as needed.
Answer:
*8.
An unknown nucleus contains 70 neutrons and has twice the volume of the nickel nucleus. Identify the
unknown nucleus in the form . Use the periodic table on the inside of the back cover as needed.
**9. Refer to Conceptual Example 1 for a discussion of nuclear densities. A neutron star is composed of
neutrons and has a density that is approximately the same as that of a nucleus. What is the radius of a neutron
star whose mass is 0.40 times the mass of the sun?
Answer:
REASONING According to Equation 31.2, the radius of a nucleus in meters is
, where A is the nucleon number. If we treat the neutron star as a uniform sphere, its
density (Equation 11.1) can be written as
Solving for the radius r, we obtain,
This expression can be used to find the radius of a neutron star of mass M and density .
SOLUTION As discussed in Conceptual Example 1, nuclear densities have the same approximate value in all
atoms. If we consider a uniform spherical nucleus, then the density of nuclear matter is approximately given by
The mass of the sun is (see inside of the front cover of the text). Substituting values into the
expression for r determined above, we find
**10.
Suppose that you could pack neutrons inside a tennis ball in
the same way as neutrons and protons are packed together in the nucleus of an atom.
(a) Approximately how many neutrons would fit inside the tennis ball?
(b) A small object is placed 2.0 m from the center of the neutron-packed tennis ball, and the tennis ball
exerts a gravitational force on it. When the object is released, what is the magnitude of the acceleration
that it experiences? Ignore the gravitational force exerted on the object by the earth.
Section 31.3 The Mass Defect of the Nucleus and Nuclear Binding Energy
(Note: The atomic mass for hydrogen is 1.007 825 u; this includes the mass of one electron.)
11. Find the binding energy (in MeV) for lithium .
Answer: 39.25 MeV
REASONING To obtain the binding energy, we will calculate the mass defect and then use the fact that 1 u is
equivalent to 931.5 MeV. The atomic mass given for includes the 3 electrons in the neutral atom. Therefore, when
computing the mass defect, we must account for these electrons. We do so by using the atomic mass of 1.007 825 u
for the hydrogen atom , which also includes the single electron, instead of the atomic mass of a proton.
SOLUTION Noting that the number of neutrons is , we obtain the mass defect as follows:
Since 1 u is equivalent to 931.5 MeV, the binding energy is
12. The binding energy of a nucleus is 225.0 MeV. What is the mass defect of the nucleus in atomic mass units?
13. Determine the mass defect (in atomic mass units) for
(a) helium , which has an atomic mass of 3.016 030 u, and
Answer: 0.008 285 u
(b) the isotope of hydrogen known as tritium , which has an atomic mass of 3.016 050 u.
Answer: 0.009 105 u
(c) On the basis of your answers to parts (a) and (b), state which nucleus requires more energy to disassemble it
into its separate and stationary constituent nucleons. Give your reasoning.
Answer:
More energy must be supplied to than to helium .
14. A 245-kg boulder is dropped into a mine shaft that is deep. During the boulder's fall, the system
consisting of the earth and the boulder loses a certain amount of gravitational potential energy. It would take an equal
amount of energy to “free” the boulder from the shaft by raising it back to the top, so this can be considered the
system's binding energy.
(a) Determine the binding energy (in joules) of the earth–boulder system.
(b) How much mass does the earth–boulder system lose when the boulder falls to the bottom of the shaft?
15. For lead obtain
(a) the mass defect in atomic mass units,
Answer: 1.741 670 u
(b) the binding energy (in MeV), and
Answer: 1622 MeV
(c) the binding energy per nucleon (in MeV/nucleon).
Answer: 7.87 MeV/nucleon
*16. (a) Energy is required to separate a nucleus into its constituent nucleons, as Figure 31.3 indicates; this energy is
the total binding energy of the nucleus. In a similar way one can speak of the energy that binds a single
nucleon to the remainder of the nucleus. For example, separating nitrogen into nitrogen and a
neutron takes energy equal to the binding energy of the neutron, as shown below:
Find the energy (in MeV) that binds the neutron to the nucleus by considering the mass of
and the mass of , as compared to
the mass of .
(b) Similarly, one can speak of the energy that binds a single proton to the nucleus:
Following the procedure outlined in part (a), determine the energy (in MeV) that binds the proton
to the nucleus. The atomic mass of carbon is 13.003 355 u.
(c) Which nucleon is more tightly bound, the neutron or the proton?
*17. Two isotopes of a certain element have binding energies that differ by 5.03 MeV. The isotope with the larger
binding energy contains one more neutron than the other isotope. Find the difference in atomic mass between the two
isotopes.
Answer: 1.003 27 u
REASONING Since we know the difference in binding energies for the two isotopes, we can determine the
corresponding mass defect. Also knowing that the isotope with the larger binding energy contains one more neutron
than the other isotope gives us enough information to calculate the atomic mass difference between the two isotopes.
SOLUTION The mass defect corresponding to a binding energy difference of 5.03 MeV is
Since the isotope with the larger binding energy has one more neutron than the other isotope, the
difference in atomic mass between the two isotopes is
*18. A copper penny has a mass of 3.0 g. Determine the energy (in MeV) that would be required to break all the
copper nuclei into their constituent protons and neutrons. Ignore the energy that binds the electrons to the nucleus and
the energy that binds one atom to another in the structure of the metal. For simplicity, assume that all the copper
nuclei are .
Section 31.4 Radioactivity
19. Write the decay process for each of the following nuclei, being careful to include Z and A and the proper
chemical symbol for each daughter nucleus:
(a)
Answer:
REASONING As discussed in Section 31.4, decay occurs when the nucleus emits a positron,
which has the same mass as an electron but carries a charge of instead of . The general form for
decay is
SOLUTION
(a)
Therefore, the decay process for is .
(b)
Similarly, the decay process for is .
(b)
Answer:
REASONING As discussed in Section 31.4, decay occurs when the nucleus emits a positron,
which has the same mass as an electron but carries a charge of instead of . The general form for
decay is
SOLUTION
(a)
Therefore, the decay process for is .
(b)
Similarly, the decay process for is .
REASONING As discussed in Section 31.4, decay occurs when the nucleus emits a positron, which has
the same mass as an electron but carries a charge of instead of . The general form for decay is
SOLUTION
(a)
Therefore, the decay process for is .
(b)
Similarly, the decay process for is .
20. Write the decay process for carbon , including the chemical symbols as well as the values of Z and A for the
parent and daughter nuclei and the particle.
21.
Osmium is converted into iridium
via decay. What is the energy (in MeV) released in this process?
Answer: 0.313 MeV
REASONING The reaction and the atomic masses are:
When the nucleus is converted into an iridium nucleus, the number of orbital electrons remains the same,
so the resulting iridium atom is missing one orbital electron. However, the given mass includes all 77 electrons of a
neutral iridium atom. In effect, then, the value of 190.960 584 u for already includes the mass of the particle.
Since energy is released during the decay, the combined mass of the iridium daughter nucleus and the
particle is less than the mass of the osmium parent nucleus. The difference in mass is equivalent to the energy
released. To obtain the energy released in MeV, we will use the fact that 1 u is equivalent to 931.5 MeV.
SOLUTION The mass decrease that accompanies the decay of osmium is
. Since 1 u is equivalent to 931.5 MeV, the energy released is
22. Find the energy that is released when a nucleus of lead undergoes decay
to become bismuth .
23.
Find the energy (in MeV) released when decay converts radium into radon
. The atomic mass of an particle is 4.002 603 u.
Answer: 4.87 MeV
24. Lead is a stable daughter nucleus that can result from either an decay or a decay. Write the decay
processes, including the chemical symbols and values for Z and A of the parent nuclei, for
(a) the decay and
(b) the decay.
25. In the form identify the daughter nucleus that results when
(a) plutonium undergoes decay,
Answer:
(b)
sodium undergoes decay, and
Answer:
(c) nitrogen undergoes decay.
Answer:
26. Multiple-Concept Example 7 reviews the concepts needed to solve this problem. When uranium decays, it
emits (among other things) a ray that has a wavelength of . Determine the energy (in MeV) of this
ray.
*27.
Polonium undergoes decay. Assuming that all the released energy is
in the form of kinetic energy of the particle and ignoring the recoil of the daughter
nucleus (lead , 205.974 440 u), find the speed of the particle. Ignore relativistic effects.
Answer:
*28.
Radon produces a daughter nucleus that is radioactive. The daughter, in turn, produces its own radioactive
daughter, and so on. This process continues until lead is reached. What are the total number of particles
and the total number of particles that are generated in this series of radioactive decays?
*29. Review Conceptual Example 5 as background for this problem. The decay of uranium produces thorium
. In Example 4, the energy released in this decay is determined to be 4.3 MeV.
Determine how much of this energy is carried away by the recoiling daughter nucleus and how much by the
particle . Assume that the energy of each particle is kinetic energy, and ignore the
small amount of energy carried away by the ray that is also emitted. In addition, ignore relativistic effects.
Answer: 0.072 MeV (thorium atom) and 4.2 MeV (alpha particle)
**30. An isotope of beryllium emits a ray and recoils with a speed of .
Assuming that the beryllium nucleus is stationary to begin with, find the wavelength of the ray.
**31. Find the energy (in MeV) released when decay converts sodium
into neon . Notice that the
atomic mass for includes the mass of 11 electrons, whereas the atomic mass for includes the mass
of only 10 electrons.
Answer: 1.82 MeV
REASONING Energy is released during the decay. To find the energy released, we determine how
much the mass has decreased because of the decay and then calculate the equivalent energy. The reaction and
masses are shown below:
SOLUTION The decrease in mass is
where the extra electron mass takes into account the fact that the atomic mass for sodium includes the mass of
11 electrons, whereas the atomic mass for neon includes the mass of only 10 electrons.
Since 1 u is equivalent to 931.5 MeV, the released energy is
Section 31.6 Radioactive Decay and Activity
32. In 9.0 days the number of radioactive nuclei decreases to one-eighth the number present initially. What is the half-life
(in days) of the material?
33. The half-lives in two different samples, A and B, of radioactive nuclei are related according to
. In a certain period the number of radioactive nuclei in sample A decreases to one-fourth the
number present initially. In this same period the number of radioactive nuclei in sample B decreases to a fraction f of
the number present initially. Find f.
Answer:
REASONING The basis of our solution is the fact that only one-half of the radioactive nuclei present initially
remain after a time equal to one half-life. After a time period that equals two half-lives, the number of nuclei
remaining is of the initial number. After three half-lives, the number remaining is of
the initial number, and so on. Thus, to determine the fraction of nuclei remaining after a given time period, we need
only to know how many half-lives that period contains.
SOLUTION For sample A, the number of nuclei remaining is of the initial number. We can see, then,
that the time period involved is equal to two half lives of radioactive isotope A. But we know that
, so that this time period must be equal to four half-lives of radioactive isotope B. The fraction f of the B nuclei that remain, therefore, is
34. The isotope of phosphorus has a half-life of 14.28 days. What is its decay constant in units of ?
35.
Strontium has a half-life of 29.1 yr. It is chemically similar to calcium, enters the body through the food
chain, and collects in the bones. Consequently, is a particularly serious health hazard. How long (in years) will it
take for 99.9900% of the released in a nuclear reactor accident to disappear?
Answer: 387 yr
36. Two radioactive waste products from nuclear reactors are strontium and cesium
. These two species are present initially in a ratio of . What is
the ratio fifteen years later?
37. Suppose that the activity of a radioactive substance is initially 398 disintegrations/min and two days later it is
285 disintegrations/min. What is the activity four days later still, or six days after the start? Give your answer in
disintegrations/min.
Answer: 146 disintegrations/min
REASONING We can find the decay constant from Equation 31.5, . If we multiply both sides
by the decay constant , we have
where is the initial activity and A is the activity after a time t. Once the decay constant is known, we can use the
same expression to determine the activity after a total of six days.
SOLUTION Solving the expression above for the decay constant , we have
Then the activity four days after the second day is
38.
Iodine is used in diagnostic and therapeutic techniques in the treatment of thyroid disorders. This isotope
has a half-life of 8.04 days. What percentage of an initial sample of remains after 30.0 days?
39. The number of radioactive nuclei present at the start of an experiment is . The number present
twenty days later is . What is the half-life (in days) of the nuclei?
Answer: 8.00 days
REASONING AND SOLUTION According to Equation 31.5, , the decay constant is
The half-life is, from Equation 31.6,
*40. One day, a cell phone company sends a text message to each of its 5800 subscribers, announcing that they have been
automatically enrolled as contestants in a promotional lottery modeled on nuclear decay. On the first day, 10% of the
5800 contestants are notified by text message that they have been randomly eliminated from the lottery. The other
90% of the contestants automatically advance to the next round. On each of the following days, 10% of the remaining
contestants are randomly eliminated, until fewer than 10 contestants remain. Determine
(a) the activity (number of contestants eliminated per day) on the second day of the lottery,
(b) the decay constant (in ) of the lottery, and
(c) the half-life (in d) of the lottery.
*41.
A device used in radiation therapy for cancer contains 0.50 g of cobalt (59.933 819 u). The half-life of
is 5.27 yr. Determine the activity of the radioactive material.
Answer:
*42.
A one-gram sample of radium ( , ) contains
nuclei and undergoes decay to produce radon . The atomic
mass of an particle is 4.002 603 u. The latent heat of fusion for water is . With the energy released
in 3.66 days, how many kilograms of ice could be melted at ?
*43.
The isotope of gold has a half-life of 2.69 days and is used in
cancer therapy. What mass (in grams) of this isotope is required to produce an activity of 315 Ci?
Answer:
*44.
Outside the nucleus, the neutron itself is radioactive and decays into a proton, an electron, and an antineutrino.
The half-life of a neutron outside the nucleus is 10.4 min. On average, over what
distance (in meters) would a beam of 5.00-eV neutrons travel before the number of neutrons decreased to 75.0% of its
initial value?
*45. Two radioactive nuclei A and B are present in equal numbers to begin with. Three days later, there are three
times as many A nuclei as there are B nuclei. The half-life of species B is 1.50 days. Find the half-life of species A.
Answer: 7.23 days
REASONING According to Equation 31.5, the number of nuclei remaining after a time t is .
Using this expression, we find the ratio as follows:
where we have used the fact that initially the numbers of the two types of nuclei are equal . Taking
the natural logarithm of both sides of the equation above shows that
SOLUTION Since when days, it follows that
But we need to find the half-life of species B, so we use Equation 31.6, which indicates that . With
this expression for , the result for becomes
Since , the result above can be solved to show that .
Section 31.7 Radioactive Dating
46. A sample has a activity of 0.0061 Bq per gram of carbon.
(a) Find the age of the sample, assuming that the activity per gram of carbon in a living organism has been
constant at a value of 0.23 Bq.
(b) Evidence suggests that the value of 0.23 Bq might have been as much as 40% larger. Repeat part (a), taking
into account this 40% increase.
47. Review Multiple-Concept Example 11 for help in approaching this problem. An archaeological specimen
containing 9.2 g of carbon has an activity of 1.6 Bq. How old (in years) is the specimen?
Answer:
REASONING According to Equation 31.5, the number of nuclei remaining after a time t is .
If we multiply both sides of this equation by the decay constant , we have . Recognizing that is
the activity A, we have , where is the activity at time . can be determined from the fact that
we know the mass of the specimen, and that the activity of one gram of carbon in a living organism is 0.23 Bq. The
decay constant can be determined from the value of 5730 yr for the half-life of using Equation 31.6. With known
values for and , the given activity of 1.6 Bq can be used to determine the age t of the specimen.
SOLUTION For , the decay constant is
The activity at time is . Since and , the age
of the specimen can be determined from
Taking the natural logarithm of both sides leads to
Therefore, the age of the specimen is
48. The half-life for the decay of uranium is . Determine the age (in years) of a rock specimen
that contains 60.0% of its original number of atoms.
49. Review Conceptual Example 12 before starting to solve this problem. The number of unstable nuclei remaining after a
time is N, and the number present initially is . Find the ratio for
(a) ,
Answer: 0.999
(b) ( ; use , since otherwise the answer is out of the range of your
calculator), and
Answer:
(c)
. Verify that your answers are consistent with the reasoning in Conceptual
Example 12.
Answer: 0.755
50. Multiple-Concept Example 11 reviews most of the concepts that are needed to solve this problem. Material found
with a mummy in the arid highlands of southern Peru has a activity per gram of carbon that is 78.5% of the
activity present initially. How long ago (in years) did this individual die?
*51. When any radioactive dating method is used, experimental error in the measurement of the sample's activity
leads to error in the estimated age. In an application of the radiocarbon dating technique to certain fossils, an activity
of 0.100 Bq per gram of carbon is measured to within an accuracy of . Find the age of the fossils and the
maximum error (in years) in the value obtained. Assume that there is no error in the 5730-year half-life of nor in the value of 0.23 Bq per gram of carbon in a living organism.
Answer:
REASONING According to Equation 31.5, . If we multiply both sides by the decay constant ,
we have
where is the initial activity and A is the activity after a time t. The decay constant is related to the half-life through
Equation 31.6: . We can find the age of the fossils by solving for the time t. The maximum error
can be found by evaluating the limits of the accuracy as given in the problem statement.
SOLUTION The age of the fossils is
The maximum error can be found as follows. When there is an error of ,
, and we have
Similarly, when there is an error of , , and we have
The maximum error in the age of the fossils is .
**52. (a) A sample is being dated by the radiocarbon technique. If the sample were uncontaminated, its activity would
be 0.011 Bq per gram of carbon. Find the true age (in years) of the sample.
(b) Suppose the sample is contaminated, so that only 98.0% of its carbon is ancient carbon. The remaining 2.0%
is fresh carbon, in the sense that the it contains has not had any time to decay. Assuming that the lab
technician is unaware of the contamination, what apparent age (in years) would be determined for the