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PHYS-271 HOMEWORK #1
Due on Wednesday Feb 1st at 9h25 in class. Late homework will not be graded.
1. On the Planck Constant and Planck Scales
Planck realized the great importance of his constant h beyond a simple fitting parameter
in the quantum theory of light. In fact, he proposed that using the fundamental constant h,
c (the speed of light) and G (Newton’s gravitational constant) one can construct a natural set
of units, or scales, related to length, time and mass. Discuss the scales that you found here in
relation to other scales that you know.
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1
1
2
a) Show that the expressions ( hG
) 2 , ( hG
) 2 , and ( hc
G ) have dimensions of length, time
c3
c5
and mass and compute their numerical values. These quantities are known as Planck length,
Planck time, and Planck mass.
b) Assume now that h is much larger, and given by h = 1 J · s. Re-calculate the Planck
length, time and mass. What do you tentatively conclude?
2. Light Quantization and the Photoelectric Effect
a) Calculate the energy of a photon whose frequency is a) 5 x 1014 Hz, b) 10 GHz and c)
30 MHz. Express your answer in electron volts.
b) Determine the corresponding wavelengths for the photons given in a).
c) An FM radio transmitter has a power output of 100 kW and operates at 94 MHz. How
many photons per second does the transmitter emit?
d) The average power generated by the Sun is 3.74 x 1026 W. Assuming the average wavelength is 500 nm, find the number of photons emitted in 1 s.
e) The photocurrent of a photocell is cut off by a retarding potential of 2.92 V for radiation
at 250 nm. Find the work function for the material.
f) Light of wavelength 500 nm is incident on a metallic surface. If the stopping potential
for the photoelectric effect is 0.45V, find i) the maximum energy of the emitted electrons, ii) the
work function and iii) the cutoff wavelength.
g) A light source of wavelength λ illuminates a metal and ejects photoelectrons with a
maximum kinetic energy of 1.00 eV. A second light source with half the wavelength of the first
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ejects photoelectrons with a maximum kinetic energy of 4.00 eV. What is the work function for
the metal?
3. The de Broglie Waves of Matter
a) Calculate the de Broglie wavelength for a proton moving with a speed of 106 m/s.
b) Calculate the de Broglie wavelength for an electron with kinetic energy i) 50 eV and
ii) 50 keV.
c) Calculate the de Broglie wavelength of a 74 kg person running at a speed of 5 m/s.
Will this person “interfere” with a door 1 meter wide?
d) Calculate what would h need to be for this person to “interfere” with the 1m wide
door.
e) Show that the de Broglie wavelength of an
√ electron accelerated from rest through a
small potential difference V is given by λ = 1.226/ V , where λ is in nanometers and V is in
volts.
f) An electron has a de Broglie wavelength equal to the diameter of the hydrogen atom.
What is the kinetic energy of the electron? Compare it with the the ground state energy of the
hydrogen atom, =13.6 eV.
Phys 271:Assignment 1-Solutions
1. (a)
[c] = LT −1
[h] = L2 M T −1
[G] = L3 M −1 T −2
`p =
hG
c3
1/2
−→ [`p ] = [h]1/2 [G]1/2 [c]−3/2
= (L2 M T −1 )1/2 (L3 M −1 T −2 )1/2 (LT −1 )−3/2
= L(1+3/2−3/2) M (1/2−1/2) T (−1/2−1+3/2)
=L
Similarly, one can verify [tp ] = T and [mp ] = M .
Alternatively, let `p = cx hy Gz . Then
[`p ] = L = [c]x [h]y [G]z
= (LT −1 )x (L2 M T −1 )y (L3 M −1 T −2 )z
= L(x+2y+3z) T (−x−y−2z) M (y−z) .
=⇒
x + 2y + 3z = 1
x + y + 2z = 0
y−z =0
∴
x = −3/2,
y = 1/2,
z = 1/2.
So
`p =
hG
c3
1/2
.
/2
Numerical values are:
`p = 4.0501 × 10−35 m
tp =
1.3510 × 10−43 s
mp =
5.4573 × 10−8 kg
/1.5
(b) Assume that h = 1 J s . Then
`p = 1.5734 × 10−18 m
tp =
5.2483 × 10−27 s
mp =
2.1201 × 109 kg .
/1.5
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2. (a)
i.
E1 = hν1 = 2.0678 eV
ii.
E2 = hν2 = 4.1357 × 10−5 eV
iii.
E3 = hν3 = 1.2407 × 10−7 eV
/1.5
(b)
i.
λ1 =
c
= 5.9958 × 10−7 m
ν1
ii.
λ2 = 2.9979 × 10−2 m
iii.
λ3 = 9.9931 m
/1
(c)
n=
Etotal
Pt
=
= 1.6055 × 1030
Ephoton
hν
/1
(d)
n=
Pt
Ephoton
=
P tλ
= 9.4138 × 1044
hc
/1
(e)
Ephoton =
hc
= 4.9594 eV
λ
ϕ = Ephoton − Kelectron = 2.0394 eV
/2
(f)
Ephoton =
hc
= 2.4797 eV
λ
2
i.
Kmax = 0.45 eV
ii.
ϕ = Ephoton − Kmax = 2.0297 eV
iii.
λcutof f =
hc
= 6.1085 × 10−7 m
ϕ
/1.5
(g)
(1)
hc
λ
hc
(1)
=
= 2Ephoton
λ/2
Ephoton =
(2)
Ephoton
(1)
(1)
Kmax
= Ephoton − ϕ
(2)
(1)
(2)
Kmax
= Ephoton − ϕ = 2Ephoton − ϕ
(2)
(1)
=⇒ Kmax
− Kmax
=ϕ
ϕ = 2.00 eV
/2
3. (a)
h
= 3.9615 × 10−13 m .
mv
λ=
/1
(b)
p2
2m
√
=⇒ p = 2mK
K=
i. For K = 50 eV,
λ= √
h
= 1.7346 × 10−10 m .
2mK
ii. For K = 50 keV,
λ = 5.4854 × 10−12 m
3
/3
(c)
λ=
h
= 1.7908 × 10−36 m
mv
λ << 1 m ∴ the person will not interfere with the door.
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(d) In order for the person to ‘interfere’ with the door the de Broglie wavelength must be approximately of the order of the width of the door, in analogy with light passing through a single slit.
So
h
∼1m
mv
=⇒
h ∼ (1 m )mv = 370 J s
∴ h ∼ 100 J s .
/1
(e)
K = eV
−→
K = |V | in eV
So
−→
p2
K=V =
2m
√
p = 2mV
So
λ=
h
h
hc
1
√
=√
=√
2
p
2mV
2mc V
(1240 eV nm )
1
√
=p
2(5.11 × 105 eV ) V
1
≈ 1.266 √ nm
V
/2
(f) Take the radius of the hydrogen atom to be the Bohr radius.
h
= 2a0
p
h
p=
2a0
λ=
Then
K=
p2
h2
=
= 134 eV
2m
8ma20
/2
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