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spectrum. ultraviolet region of the electromagnetic Chapter 29 Problems 3. 1513 SSM REASONING According to Equation 29.3, the work function W0 is related to the
photon energy hf and the maximum kinetic energy KEmax by W0 = hf – KE max . This
expression can be used to find the work function of the metal.
SOLUTION KE max is 6.1 eV. The photon energy (in eV) is, according to Equation 29.2,
hf = 6.63 × 10 –34 J ⋅ s 3.00 × 10 15 Hz c c
h G × 10 J
hF 1 eV J I = 12.4 eV
H
K
1.60
–19 The work function is, therefore, W0 = hf – KE max = 12.4 eV – 6.1 ev = 6.3 eV 4. REASONING According to Equation 29.3, the relation between the photon energy, the
maximum kinetic energy of an ejected electron, and the work function of a metal surface is hf
K
W0
{ = 1E max + {
43
24 Photon
energy Maximum
kinetic energy
of ejected
electron Work
function Equation 16.1 relates the frequency f of a photon to its wavelength λ via f = c/λ, where c is
the speed of light in a vacuum. The maximum kinetic energy KEmax is related to the mass m
and maximum speed vmax of the ejected electron by KEmax =
these substitutions, Equation 29.3 becomes
hf = KE max + W0 or hc λ 1
2 mv 2 (Equation 6.2). With 2
1
= 2 mvmax + W0 SOLUTION Solving Equation (1) for the wavelength gives λ= hc
1 mv 2
max
2 + W0 ( 6.63 × 10−34 J ⋅ s )( 3.00 × 108 m/s )
λA =
=
2
1 9.11 × 10 −31 kg 7.30 × 105 m/s + 4.80 × 10 −19 J
)(
)
2( 2.75 × 10−7 m (1) 1514 PARTICLES AND WAVES ( 6.63 × 10−34 J ⋅ s ) ( 3.00 × 108 m/s )
=
λB =
2
1 9.11 × 10 −31 kg 5.00 × 105 m/s + 4.80 × 10 −19 J
)(
)
2(
5. 3.35 × 10−7 m SSM REASONING The energy of the photon is related to its frequency by Equation
29.2, E = hf . Equation 16.1, v = f λ , 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 λ= 6. c
c
hc
(6.63 × 10 –34 J ⋅ s)(3.0 × 10 8 m / s)
=
=
=
= 3.1 × 10 −7 m =
f
( E / h)
E
6.4 × 10 –19 J 310 nm REASONING The photons of this wave must carry at least enough energy to equal the
work function. Then the electrons are ejected with zero kinetic energy. Since the energy of a
photon is E = hf according to Equation 29.2, where f is the frequency of the wave, we have
that W0 = hf. Equation 16.1 relates the frequency to the wavelength λ according to f = c/λ,
where c is the speed of light. Thus, it follows that W0 = hc/λ.
SOLUTION Using Equations 29.2 and 16.1, we find that W0 = 6
c.63 × 10
= hc λ −34 h
c J ⋅ s 3.00 × 10 8 m / s 485 × 10 −9 m h 4.10 × 10
= −19 J Since 1 eV = 1.60 × 10–19 J, it follows that c W0 = 4 .10 × 10 −19 J 7. F eV J
G×
h1.601 10 J I =
H
K
−19 2 .56 eV REASONING AND SOLUTION Equation 29.3 gives
KE max = hf − Wo =
= 6
c.63 × 10 hc λ
−34 − Wo h
c J ⋅ s 3.00 × 10 8 m / s 215 × 10 −9 m h− b.68 eV g1.60 × 10 J I = 3.36 × 10
F
3
G 1 eV J
H
K
−19 −19 J Chapter 29 Problems 1515 Converting to electron volts c KE max = 3.36 × 10 −19 J 8. F
h1.601×eV J I =
G 10 J
H
K
−19 2 .10 eV REASONING The intensity S = 680 W/m2 of the photons is equal to the total amount of
energy delivered by the photons per second per square meter of surface area. Therefore, the
intensity of the photons is equal to the energy E of one photon multiplied by the number N
of photons that reach the surface of the earth per second per square meter:
S = NE (1) The energy E of each photon is given by E = hf (Equation 29.2), where h = 6.63×10−34 J·s
c
is Planck’s constant and f is the frequency of the photon. We will use f = λ (Equation 16.1) to determine the frequency f of the photons from their wavelength λ and the
speed c of light in a vacuum.
SOLUTION Solving Equation (1) for N, we obtain N = S/E. Substituting E = hf
(Equation 29.2) into this result yields
N= Substituting f = c S
S
=
E hf (2) (Equation 16.1) into Equation (2), we find that λ ( )( ) 680 W/m 2 730 × 10 −9 m
S
S
Sλ
N=
=
=
=
hf c hc
6.63 × 10 −34 J ⋅ s 3.00 × 108 m/s
h λ ( )( ) = 2.5 × 10 21 photons/ ( s ⋅ m 2 ) 9. SSM REASONING AND SOLUTION The number of photons per second, N, entering the
owl's eye is N = SA / E , where S is the intensity of the beam, A is the area of the owl's
pupil, and E is the energy of a single photon. Assuming that the owl's pupil is circular, ch
2 1
A = π r 2 = π 2 d , where d is the diameter of the owl's pupil. Combining Equations 29.2
and 16.1, we have E = hf = hc / λ . Therefore, 1516 PARTICLES AND WAVES c h
2 1
(5.0 × 10 –13 W / m 2 ) π 2 8.5 × 10 –3 m (510 × 10 –9 m)
SAλ
N=
=
=
hc
(6.63 × 10 –34 J ⋅ s)(3.0 × 10 8 m / s) 73 photons / s 10. REASONING The wavelength λ of the light shining on the surface is related to the
maximum kinetic energy KEmax of the electrons ejected from the surface by
c
hf = KE max + W0 (Equation 29.3), where h is Planck’s constant, f = (Equation 16.1) is
λ
the frequency of the light, and W0 is the work function of the surface. Substituting Equation
16.1 into Equation...
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 Spring '13
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 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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