Unformatted text preview: he ice
phase to the liquid phase is
N ′ 6.7 × 10 25 molecules
=
= 32 molecules / photon
N
2.1 × 10 24 photons 15. SSM REASONING The angle θ through which the Xrays are scattered is related to the
difference between the wavelength λ ′ of the scattered Xrays and the wavelength λ of the
incident Xrays by Equation 29.7 as 1520 PARTICLES AND WAVES λ′ − λ = h
(1 − cos θ )
mc where h is Planck’s constant, m is the mass of the electron, and c is the speed of light in a
vacuum. We can use this relation directly to find the angle, since all the other variables are
known.
SOLUTION Solving Equation 29.7 for the angle θ, we obtain cos θ = 1 − mc
(λ′ − λ )
h ( 9.11×10−31 kg ) ( 3.00 ×108 m/s ) ( 0.2703 ×10−9 m − 0.2685 ×10−9 m ) = 0.26
= 1−
6.63 ×10−34 J ⋅ s θ = cos −1 ( 0.26 ) = 75°
16. REASONING The momentum of the photon is related to its wavelength λ and Planck’s
constant h. The momentum (nonrelativistic) of the ball depends on its mass m and speed v.
We can set the two momenta equal and solve directly for the speed.
SOLUTION The momentum pphoton of the photon and the momentum pball of the ball are
pphoton = h λ (29.6) and pball = mv (7.2) Since pphoton = pball, we have h
1 24
4λ 3
Momentum
of photon = mv
1 24
43
Momentum
of ball or v= h
6.63 ×10−34 J ⋅ s
=
= 4.2 ×10−25 m/s
−9
−3
λ m 720 ×10 m 2.2 × 10 kg ( )( ) 17. REASONING The frequency f of a photon is related to its energy E by f = E/h
(Equation 29.2), where h is Planck’s constant. As discussed in Section 29.4, the energy E is
related to the magnitude p of the photon’s momentum by E = pc, where c is the speed of
light in a vacuum. By combining these two relations, we see that the frequency can be
expressed in terms of p as f = pc/h. SOLUTION
a. Substituting values for p, c, and h, into the relation f = pc/h gives Chapter 29 Problems f= 1521 pc ( 2.3 × 10−29 kg ⋅ m/s ) ( 3.00 ×108 m/s )
=
= 1.0 × 1013 Hz
−34
h
6.63 ×10
J ⋅s b. An inspection of Figure 24.9 shows that this frequency lies in the infrared region of the
electromagnetic spectrum. 18. REASONING Before the scattering, the electron is at rest and has no momentum. Thus, the
total initial momentum consists only of the photon’s momentum, which points along the +x
axis. The total initial momentum has no y component. Since the total momentum is
conserved, the total momentum after the scattering must be the same as it was before and,
therefore, has no y component.
The total momentum after the scattering is the sum of the momentum of the scattered
photon and that of the scattered electron, and it only has an x component. But the scattered
photon is moving along the –y axis, so its momentum has no x component. Therefore, the
momentum of the electron must have an x component.
The total momentum after the scattering is the sum of the momentum of the scattered
photon and that of the scattered electron, and it has no y component. But the scattered
photon is moving along the –y axis, so its momentum points along the –y axis. Therefore,
this contribution to the total final momentum must be cancelled by part of the momentum of
the scattered electron, which must have a component along the +y axis.
SOLUTION Since the total momentum is conserved and since the scattered photon has no
momentum in the x direction, the momentum of the scattered electron must have an x
component that equals the momentum of the incident photon. According to Equation 29.6,
the magnitude p of the momentum of the incident photon is p = h/λ, where h is Planck’s
constant and λ is the wavelength. Therefore, the momentum of the scattered electron has a
component in the +x direction that is px = h λ = 6.63 × 10 −34 J ⋅ s
= 7 .37 × 10 −23 kg ⋅ m / s
−12
9.00 × 10
m The momentum of the scattered electron has a component along the +y direction. This
component cancels the momentum of the scattered photon that points along the –y direction.
To find the momentum of the scattered photon, we first need to determine its wavelength λ′,
which we can do using Equation 29.7: 1522 PARTICLES AND WAVES h
λ′ = λ +
1 − cos 90.0° 43
mc 1 24 =0 = 9.00 × 10−12 m + 6.63 × 10−34 J ⋅ s (9.11 × 10 −31 )( kg 3.00 × 10 m/s
8 ) = 1.14 × 10−11 m Again using Equation 29.6, we find that the momentum of the scattered electron has a
component in the +y direction that is py = h λ = 6.63 × 10 −34 J ⋅ s
= 5.82 × 10 −23 kg ⋅ m / s
−11
1.14 × 10 m 19. REASONING There are no external forces that act on the system, so the conservation of
linear momentum applies. Since the photon is scattered at θ = 180 ° , the collision is
"headon," and all motion occurs along the horizontal direction, which we take as the x axis.
The incident photon is assumed to be moving along the +x axis. For an initially stationary
electron, the conservation of linear momentum states that: p
=
− p′ + pelectron
1
23
1 24
43
1 24
43 Momentum
of incident
photon Momentum
of scattered
photon Momentum
of recoil
electron where the momentum of the scattered photon is negati...
View
Full Document
 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

Click to edit the document details