Phys 197 Homework Solution 38B
Q12.
In a photoelectric-effect experiment, which of the following will increase the maximum
kinetic energy of the photoelectrons? (a) Use light of greater intensity; (b) use light of higher frequency; (c) use light of longer wavelength; (d) use a metal surface with a larger work function. In
each case justify your answer.
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(a) No. There are more photoelectrons, but the energy distribution, most specifically including
the max energy, is unchanged.
(b) Yes. Higher frequency means higher photon energy. Ephoton = hf .
(c) No. Longer λ means lower frequency.
(d) No. A larger work function means more photon energy goes into freeing the electron, so
less is left over for its KE.
Q15.
Why must engineers and scientists shield against x-ray production in high-voltage
equipment?
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High voltages will lead to arcs which lead to bremsstrahlung x-rays.
P16.
(a) What is the minimum potential difference between the filament and the target of an
x-ray tube if the tube is to produce x rays with a wavelength of 0.150 nm? (b) What is the shortest
wavelength produced in an x-ray tube operated at 30.0 kV?
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(a) The minimum voltage is such that an electron acquires the energy of a photon of the desired
wavelength.
Ephoton = hc/λ = (1240 eV · nm)/(0.150 nm) = 8270 eV.
The potential to give an electron this energy is 8270 V.
(b) This tube produces electrons of 30 keV, so the cutoff photon energy will be 30 keV.
λ = (1240 eV · nm)/(30 000 V) = 0.0413 nm.
P23.
X rays with an initial wavelength of 0.900 × 10−10 m undergo Compton scattering.
For what scattering angle is the wavelength of the scattered x rays greater by 1.0% than that of the
incident x rays?
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Recognize a problem for the Compton formula. We will need: λ = 90.0 pm and λ′ = 90.9 pm.
λ′ − λ = (h/mc)(1 − cos θ). With h/mc = 2.426 pm,
(1 − cos θ) = (90.9 − 90.0)/2.426 = 0.3710.
cos θ = 0.6290 ⇒ θ = 51.0◦ .
P25.
An electron and a positron are moving toward each other and each has speed 0.500 c in
the lab frame. (a) What is the kinetic energy of each particle? (b) The e+ and e− meet head-on and
annihilate. What is the energy of each photon that is produced? (c) What is the wavelength of each
photon? How does the wavelength compare to the photon wavelength when the initial kinetic energy
of the and is negligibly small (see Example 38.6)?
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(a) At 0.5pc, we will use relativistic. K = mc2 (γ − 1), with
γ = 1/ (1 − v 2 /c2 ) = 1.1547 (keep extra sig dig).
Since mc2 = 511 keV,
K = (0.1547)(511 keV) = 79.05 keV.
Incidentally, KNewton = (1/2)m(c/2)2 = mc2 /8 = 63.9 keV. Off by 20%.
(b) Since total energy is conserved, each photon gets the relativistic energy of one electron.
Ephoton = mc2 γ = (511 keV)(1.155) = 590.0 keV.
(c) λ = 1240 eV · nm/Ephoton = (1240 eV · nm)/(590 000 eV) = 2.10 pm.
In annihilation at rest, Ephoton = 511 keV, and the wavelength is proportionately longer.
λann. at rest = (2.10 pm)(590/511) = 2.43 pm.
P35.
An incident x-ray photon of wavelength 0.0900 nm is scattered in the backward direction
from a free electron that is initially at rest. (a) What is the magnitude of the momentum of the
scattered photon? (b) What is the kinetic energy of the electron after the photon is scattered?
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(a) Strategy: Calculate initial and final momenta of the photon, then use that to get pelectron .
Use pphoton = h/λ to get momenta.
λ′ = λ + (h/mc)(1 − cos θ) = (90 pm) + (2.426 pm)(1 − (−1)) = 94.85 pm.
p = h/λ = (6.626 × 10−34 Js)/(90 × 10−12 m) = +7.36 × 10−24 kg m/s.
p′ = (−)6.99 × 10−24 kg m/s. Since p′ is in the opposite direction of p,
∆p = (−)14.35 × 10−24 kg m/s, which is equal to (−)∆pelectron . Since the electron started
from rest,
pelectron = 1.435 × 10−23 kg m/s.
(b) The energy lost by the photon appears as K of the electron.
Ephoton = pc = (7.36 × 10−24 kg m/s)(3 × 108 m/s) = 2.21 × 10−15 J.
E ′ photon = 2.10 × 10−15 J. No negative signs with energy!
∆Ephoton = 1.1 × 10−16 J ⇒ Kelectron = 1.1 × 10−16 J. Or 690 eV.
P39.
Nuclear fusion reactions at the center of the sun produce gamma-ray photons with
energies of about 1 MeV (106 eV). By contrast, what we see emanating from the sun’s surface are
visible-light photons with wavelengths of about 500 nm. A simple model that explains this difference
in wavelength is that a photon undergoes Compton scattering many times – in fact, about 1026 times,
as suggested by models of the solar interior – as it travels from the center of the sun to its surface.
(a) Estimate the increase in wavelength of a photon in an average Compton-scattering event. (b) Find
the angle in degrees through which the photon is scattered in the scattering event described in part
(a). (Hint: A useful approxi- mation is cos φ ≈ 1 − φ2 /2 which is valid for φ << 1. Note that φ is
in radians in this expression.) (c) It is estimated that a photon takes about 106 years to travel from
the core to the surface of the sun. Find the average distance that light can travel within the interior
of the sun without being scattered. (This distance is roughly equivalent to how far you could see if
you were inside the sun and could survive the extreme temperatures there. As your answer shows,
the interior of the sun is very opaque.)
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(a) Calculate the change in wavelength and divide by 1026 .
λ0 = 1240 eV · nm/(106 eV) = 0.00124 nm.
∆λ = (500 − .00124) = 5 × 10−24 nm. Or 5 × 10−33 m.
(b) ∆λ = (1.2 × 10−12 m)(1 − cos θ) ⇒
1 − cos θ = (5 × 10−33 m)/(1.2 × 10−12 m) = 4 × 10−21 .
Making the suggested approximation,
2
1 − cos
θ2 /2 ⇒
√ θ = 1 − (1 − θ /2) =−11
θ = 2 · 4 × 10−21 = 9 × 10
rad = 5 × 10−9 ◦ .
(c) Calculate the total distance D the photon travels in 106 yr, then divide by number of
collisions, given as 1026 .
D = (3 × 108 m/s)(3.15 × 107 s/yr)(106 yr) = 9.5 × 1021 m. Divide by 1026 to get:
d = 9.5 × 10−5 m = 10−4 m = 0.1 mm.