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Unformatted text preview: ve since is moves along the −x
direction (the scattering angle is 180°). Using the relation p = h/λ (Equation 29.6), where h
is Planck’s constant and λ is the wavelength of the photon, we can write the expression for
the momentum of the electron as
pelectron = p + p′ = h λ + h
1 1 = h + λ′ λ λ′ SOLUTION Substituting numerical values into the equation above, we have
1
1 pelectron = (6.626 × 10 –34 J ⋅ s) +
= 4.755 × 10 –24 kg ⋅ m/s
−9
−9 0.2750 × 10 m 0.2825 × 10 m Chapter 29 Problems 1523 20. REASONING The wavelength λ of the incident Xrays is related to the wavelength λ′ of
h
the scattered Xrays by λ ′ − λ =
(1 − cosθ ) (Equation 29.7), where h = 6.626×10−34 J·s is
mc
8
Planck’s constant, c = 2.998×10 m/s is the speed of light in a vacuum, m = 9.109×10−31 kg
is the mass of an electron, and θ = 122.0° is the angle at which the Xrays are scattered. The
wavelength λ′ of the scattered photon is found from the magnitude p′ of its momentum
h
(Equation 29.6).
via p′ =
λ′
SOLUTION Solving λ ′ − λ = h
(1 − cosθ ) (Equation 29.7) for λ, we obtain
mc λ = λ′ − h
(1 − cos θ )
mc (1) h
h
(Equation 29.6) for λ′ yields λ ′ = , which, on substitution into
λ′
p′
Equation (1), gives Solving p′ = λ = λ′ − h
h
h
(1 − cos θ ) = − (1 − cos θ )
mc
p′ mc (
( )(
)( )
) 6.626 × 10 −34 J ⋅ s 1 − cos122.0o
6.626 × 10−34 J ⋅ s
=
−
1.856 × 10 −24 kg ⋅ m/s 9.109 × 10−31 kg 2.998 × 108 m/s
= 3.533 × 10 −10 m = 0.3533 nm 21. SSM WWW REASONING The change in wavelength that occurs during Compton
scattering is given by Equation 29.7: λ′ – λ = ( λ ′ – λ )max h
(1 – cos θ )
mc or ( λ ′ λ )max = h
2h
(1 – cos 180° ) =
mc
mc is the maximum change in the wavelength, and to calculate it we need a value for the mass m of a nitrogen molecule. This value can be obtained from the mass per mole
M of nitrogen ( N 2 ) and Avogadro's number N A , according to m = M / N A (see
Section 14.1). SOLUTION Using a value of M = 0.0280 kg/mol , we obtain the following result for the
maximum change in the wavelength: 1524 PARTICLES AND WAVES ( λ ′ – λ )max ( ) 2 6.63 × 10 –34 J ⋅ s
2h
2h
=
=
=
mc
M 0.0280 kg/mol 3.00 × 108 m/s N c 23
–1 A 6.02 × 10 mol ( ) = 9.50 × 10 –17 m 22. REASONING Energy is conserved during the collision. This means that the energy E of
the incident photon must equal the kinetic energy KE of the recoil electron plus the energy
E ′ the scattered photon:
E = KE + E ′
(1)
The energy E of a photon is related to its frequency f by E = hf (Equation 29.2), where h is
Planck’s constant. The frequency, in turn, is related to the wavelength λ by f = c/λ
(Equation 16.1), where c is the speed of light in a vacuum. Substituting f = c/λ into E = hf
gives
(2)
E = hc/λ
The wavelength λ ′ of the scattered photon depends on the wavelength λ of the incident
photon according to Equation 29.7, so that we have λ′ = λ + h
(1 − cos θ )
mc Since the photon is scattered straight backward, θ = 180°, and λ′ = λ + h
(1 − cos180° ) = λ + 2h
mc
mc (3) SOLUTION The kinetic energy of the recoil electron is given by KE = 1 mv 2
2
(Equation 6.2), where m is its mass and v is its speed. Substituting this expression into
Equation (1), we have E = 1 mv 2 + E ′ . Solving for the speed v of the electron gives
2 v= 2 ( E − E′)
m (4) From Equation (2), we also know that E = hc/λ and E ′ = hc / λ ′ . Substituting this expression
into Equation (4), we find that the speed of the electron can be written as Chapter 29 Problems v= Since λ ′ = λ + v= = 1525 2h c 1 1 −
m λ λ′ 2h
[Equation (3)], the speed of the electron is
mc 2h c 1
1
λ −
2h m
λ+ mc 2 ( 6.63 × 10−34 J ⋅ s )( 3.00 × 108 m/s )
9.11 × 10−31 kg × 1
1 − −9
−34
2 ( 6.63 × 10
J ⋅ s) 0.45000 × 10 m 0.45000 × 10−9 m + −31
8 ( 9.11× 10 kg ) ( 3.00 × 10 m/s ) = 3.22 × 106 m/s 23. REASONING AND SOLUTION
a. We have
λ = λ' − (h/mc)(1 − cos 163°) = 0.1819 nm b. For the incident photon
E = hf = hc/λ = 1.092 × 10 –15 J c. For the scattered photon
E ' = hf ' = hc/λ' = 1.064 × 10 –15 J d. The kinetic energy of the recoil electron is, therefore,
KE = E − E ' = 2.8 × 10 –17 J 24. REASONING
a. Consider one square meter of the sail’s surface. Each of the N photons that strike this
square meter in a onesecond interval (∆t = 1.0 s) has an initial momentum p that is
h
determined by its wavelength λ, according to p =
(Equation 29.6), where λ 1526 PARTICLES AND WAVES h = 6.63×10−34 J·s is Planck’s constant. Each photon is fully reflected, so the final
momentum of a photon is equal to –p. The magnitude of the change ∆p in a photon’s
momentum, then, is equal to ∆p = p − (−p) = 2p, and the magnitude ∆P of the net
momentum change undergone by all N photons is given by ∆P = N∆p = 2Np. In order to
cause this momentum change, the sail exerts an impulse of magnitude F∆t = ∆P
(Equation 7.4) on the photons, where F is the magnitude of the force exerted...
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 Spring '13
 CHASTAIN
 Physics, The Lottery

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