Answers to questions and problems for Applying Quantum Tools – Chapter 28 in KJ & F
Questions:
10 a & b: The energy and wavelength of a photon are related by E  hc/ . The power, rate of arrival of photons,
is given by 1 watt = Energy/photon * number of photons/second. By examination of these relationships we see
that since 3(800 nm)  2 (600 nm)  1(400 nm), then E1  E2  E3 and N3  N2  N1.
15. Both particles accelerate through the same potential difference and so have the same kinetic energy. Using
de Broglie’s hypothesis (silly name) and K = ½ mv2 (definition of Kinetic Energy and algebra, v  2K/m. ) you
get:

h
h


mv m 2K/m
h
2Km
This shows that the particle with the smaller mass corresponds to the larger wavelength. Thus, the electron has
the larger de Broglie wavelength
16. The de Broglie wavelength and the speed of the associated particle are related by   h/p  h/(mv). As the
neutron travels upward, it will slow down and, according to the preceding expression, the associated de Broglie
wavelength will increase.
Problems:
16. Given   550 nm. Recall that for photons E  hf .
a) The minimum chemical energy required to generate a photon of wavelength   550 nm is
E  hf 
hc



(663  1034 J  s)(300  108 m/s) 
1 eV
 160  1019 J   23 eV
550 nm
b) The number of ATP molecules needed to produce that much energy is
2.3 eV
 7.5 molecules of ATP
0.30 eV/molecule of ATP
However, the number of ATP molecules must be an integer, so the minimum number is 8.
18. Light of frequency f consists of discrete quanta, each of energy E  hf = hc/. Thus   hc/E.
a) The wavelength of the photon is

hc (4.14 1015 eV  s)(3.0 108 m/s)

 4.14 106 m  4140 nm
E
0.30 eV
to be reported as 4100 nm. This is infrared light.
b) Likewise, for an energy of 3.0 eV, the wavelength is   410 nm and is in the visible region.
c) For an energy of 30 eV, the wavelength is   41 nm and is in the ultraviolet region.
Since E   1, the higher the energy of the photon, the smaller its wavelength.
Next (over)
19. The energy of a photon of frequency
Given red  700 nm and blue  400 nm.
f
and wavelength  is E  hf  hc/ .
Ered
hc/red

400 nm 4

 blue 
  057
Eblue hc/blue
red 700 nm 7
30. The de Broglie wavelength is given by:   h/(mv). and are given
calculation to find m. Also we are given V  2  1013 m3 and   1000 kg/m3.
v  1 mm/s.
You must do a preliminary
m  V  (1000 kg/m3 )(2 1013 m3 )  2 1010 kg

The fraction this is of
150 m
h
663  1034 J  s

 33  1021 m
mv (2  1010 kg)(1 103 m/s)
is
331021 m
 2 1017
150 m
The lesson is clear; even for living things as small as a paramecium moving at biological speeds, the de Broglie
wavelength is a vanishingly small fraction of the size of the object.
44. The Heisenberg uncertainty principle is given by:
xpx  x mvx 
Given
x  1 104 m
h
2
and   1000 kg/m3. A preliminary calculation will give us m. We are also given r
 d/2  25  109 m.
 4
4
m  V    r 3  (1000 kg/m3 )    (25  109 m)3  65  1020 kg
3
 3
Solve for the uncertainty in speed.
vx 
h
663  1034 J  s

 16  1011 m/s
2 mx 2 (65  1020 kg)(1 104 m)
- this means the velocity of the virus is centered on vx = 0
and the range of velocities is +/- 8.0 x 10 -12 m/s or, - 8.0 x 10 -12 m/s ≤ vx ≤ + 8.0 x 10 -12 m/s.
46. Protons are subject to the Heisenberg uncertainty principle. We know the proton is somewhere within the
nucleus, so the uncertainty in our knowledge of its position is at most x = L = 4 fm.
With a finite x, the uncertainty px is given by the uncertainty principle:
px  mvx 
h/2
h
6.631034 J  s
 vx 

 2 107 m/s
x
2 mL 2 (1.67 1027 kg)(4 1015 m)
Because the average velocity is zero, the best we can say is that the proton’s velocity is somewhere in the
range –0.8  107 m/s to 0.8  107 m/s. Thus the smallest range of speeds is 0 to 0.8  107 m/s.
Since the nucleus and hence the uncertainty in the position of the proton is very small we expect a large
uncertainty in the momentum of the protons. Given that the mass of the proton is small, a large uncertainty in
the momentum is the result of a large uncertainty in the velocity.