Problem Set 1
PHY 291 H1S
February 5, 2008
Problem 1-2: Atoms abounding. Given the knowledge that atomic diameters are of the order of 1 or 2 Angstroms, estimate the order of magnitude of
the number of atoms in (a) a pin’s head; (b) a human being; (c) the whole earth.
(a) Let’s say a pin’s head is made of steel, which is mostly iron. Let’s say
is 1 mm3 in volume. The density of steel is ≈ 8 g/cm3 , and one mole of iron is
56 g. Then,
number of atoms = 1 mm3 ×8
3
1 cm
mole of iron 6 × 1023 atoms
g
×
×
×
3
cm
10 mm
56 g
mole
≈ 6 × 1019 atoms
(b) TA Lindsay weighs about 68 kg. She’s probably made mostly of water,
which is 18 g/mol.
number of atoms = 68 kg ×
1000 g mole of H2 0 6 × 1023 atoms
×
×
1 kg
18g
mole
≈ 2 × 1027 atoms
(c) The mass of the earth is ≈ 5 × 102 4 kg. The earth’s crust is almost half
oxygen, which is 16 g / mol.
number of atoms = 5 × 1024 kg ×
1000 g mole of O2
6 × 1023 atoms
×
×
1 kg
16g
mole
≈ 2 × 1050 atoms
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Problem 1-8. Photons and radio waves. A radio station broadcasts at a
frequency of 1 MHz with a total radiated power of 5000 watts.
(a) What is the wavelength of this radiation?
(b) What is the energy (in electron-volts) of the inidifidual quanta that compose the radiation? HOw many photons are emitted per second? per cycle of
oscillation?
(c) A certain radio receiver must have 2 microwatts of radiation power incident
on its antenna in order to provide intelligible reception. How many 1-MHz photons does this require per second? per cycle of oscillation?
(d) Do your answers to parts (b) and (c) indicated that the granularity of electromagnetic radiation can be neglected in these circumstances?
(a) The speed of light is 3.0 × 108 m/s.
c = λf → λ =
3.0 × 108 m/s
c
=
= 300 m
f
1 × 106 s−1
(b) The energy of a single quanta (photon):
E = hf = 6.6 × 10−34 J · s (1 × 106 s−1 ) ×
1 eV
= 4.1 × 10−9 eV
1.6 × 10−19 J
Since the total energy per second is the power, the number of photons per second
number of photons/s = 5000 J/s×
1 photon
≈ 8×1030 photons/s
6.6 × 10−34 J · s (1 × 106 s−1 )
In one oscillation,
number of photons/osc = 8×1030 photons/s×
1s
= 8×1024 photons/osc
1 × 106 oscillations
(c) To generate 2 µW of power,
number of photons/s = 2×10−6 J/s×
6.6 ×
1 photon
≈ 3×1021 photons/s
J · s (1 × 106 s−1 )
10−34
In one oscillation,
number of photons/osc = 3×1021 photons/s×
1s
= 3×1015 photons/osc
1 × 106 oscillations
(d) The answers to parts (b) and (c), being very large numbers, indicate that the
granularity of electromagnetic radiation can be neglected in these circumstances.
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Problem 1-10. The photoelectric effect in lanthanum. The threshold wavelength of light required to eject electrons from the surface of the metal lanthanum
(Z = 57) is 3760 Angstroms.
(a) What is the work function of the metal in electron-volts?
(b) What is the maximum kinetic energy of photoelectrons emitted by this metal
when it is illuminated with ultraviolet light of wavelength 2000 Angstroms?
(a) We assume that at the threshold wavelength, ejected electrons have no
kinetic energy (Kmax = eV = 0). Rearranging formula (1-14) from French and
Taylor, we get
φ=
6.6 × 10−34 J · s(3.0 × 108 m/s)
hc 1
· −V =
= 3.3 eV
e λ
1.6 × 10−19 (3760 × 10−10 m)
(b) Using our answer from (a) and formula (1-14) and the fact that Kmax = eV ,
Kmax =
6.6 × 10−34 J · s(3.0 × 108 m/s)
hc
− eV =
− 1.6 × 10−19 J/eV (3.3 eV)
λ
2000 × 10−10 m
= 4.6 × 10−19 J (= 2.9 eV)
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Problem 1-11. Determination of Planck’s constant. The clean surface of
sodium metal (in a vacuum) is illuminated with monochromatic light of various
wavelengths and the retarding potentials required to stop the most energetic
photoelectrons are observed as follows: (see text) Plot these data in such a way
as to show that they lie (approximately) along a straight line as predicted by
the photoelectric equation, and obtain a numerical value for Planck’s constant.
We can use formula (1-14) from French and Taylor to see how we may plot
these data to extract Planck’s constant. Plotting 1/λ vs. V and fitting the data
to a straight line will give a slope of hc/e, from which we can extract h.
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Retarding Potential (V)
2.5
y = 1.24e−06*x − 2.27
2
1.5
1
0.5
2
2.5
3
3.5
Inverse Wavelength (m−1)
4
6
x 10
Given a slope of m = 1.24 × 10−6 V · m,
h=
me
1.24 × 10−6 V · m (1.6 × 10−19 J/V)
=
= 6.6 × 10−34 J · s
c
3.0 × 108 m/s
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Problem 1-13. Rutherford scattering and the size of the nucleus. The point
nucleus model Rutherford used to describe the alpha-particle scattering results
of Geiger and Marsden fails to describe correctly the results for collisions in
which the minimum separation is smaller than the nuclear radius. These close
encounters occur for high-energy alpha particles and for collisions which are
nearly head-on and result in scattering through large angles. Significant departures from the Rutherford scattering formula occur in such collisions for alpha
particles of 32-MeV energy scattered by gold.
(a) As a crude indication of the size of the nucleus, calculate the distance of
closest approach of a 32-MeV alpha particle in a head-on collision with a gold
nucleus (Z = 79). Neglect the recoil of the gold nucleus. Give a qualitative
argument why we may ignore in this calculation the electrons surrounding the
gold nucleus - or should we?
(b) What is the ratio of the radius of the gold atoms to the radius of the gold
nucleus deduced in (a)? Compare this to the ratio of the average distance of
the earth from the sun (1.5 × 1013 cm) to the radius of the sun (7.0 × 1010 cm).
(c) In a sketch of the atom in which the nucleus is represented by a circle 1 mm
in radius, how large a circle is needed to represent a typical Bohr electron orbit
in correct relative scale?
(a) Approximating the gold nuclues as a point particle of charge q1 79e, the
Coulomb potential energy E is given by
E=
1 q1 q2
,
4π0 r
where r is the separation of the α-particle of charge q2 from gold nucleus.
Converting all kinetic energy E = 32 MeV to potential energy, the minimum
distance d of closest approach for α-particle is
d=
1 q1 q2
= 7.1 × 10−15 m.
4π0 E
(b) Take ratom = 102 pm, the ratio of the radius of the gold atom to its nucleus
is
ratom
102 × 10−9
=
= 1.4 × 107 .
rnuc.
7.12 × 10−15
The ratio of average distance of the earth-to-sun to the sun’s radius is ≈ 200,
which is 105 smaller then the result for the gold atom.
(c) The ratio of cartoon nucleus’s radius to the actual size is equal to the cartoon
electron orbit R to the correct value, i.e.,
10−3
R
= −10 ,
−14
10
10
and R ∼ 10 m.
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