Q#39.85
Doorway Diffraction. If your wavelength were 1.0 m, you would undergo considerable diffraction in moving through a doorway.
(a) What must your speed be for you to have this wavelength? (Assume that your mass is 60.0 kg.)
(b) At the speed calculated in part (a), how many years would it take you to move 0.80 m (one step)? Will you notice diffraction effects as you walk through doorways?
Answer:
Use $\lambda = \frac{h}{mv}$ to relate your wavelength and speed.
(a) So, $v = \frac{h}{m \lambda}$
$v=\frac{6.626 \times 10^{-34} \ J.s}{(60.0 \ kg)(1.0 m)}= 1.1 \times 10^{-35} \ m/s$
(b) $t=\frac{distance}{velocity}$
$t = \frac{0.80 \ m}{1.1 \times 10^{-35} \ m/s}= 7.3 \times 10^{34} \ s = 2.3 \times 10^{23} \ yr$
Since you walk through doorways much more quickly than this, you will not experience diffraction effects.
A 1-kg object moving at 1 m/s has a de Broglie wavelength λ = 6.6 $\times 10^{-34}$ m which is exceedingly small. An object like you has a very, very small λ at ordinary speeds and does not exhibit wavelike properties.
Q#39.86
Atomic Spectra Uncertainties. A certain atom has an energy level 2.58 eV above the ground level. Once excited to this level, the atom remains in this level for $1.64 \times 10^{-7}$ s (on average) before emitting a photon and returning to the ground level.
(a) What is the energy of the photon (in electron volts)? What is its wavelength (in nanometers)?
(b) What is the smallest possible uncertainty in energy of the photon? Give your answer in electron volts.
(c) Show that $|\Delta E/E|=|\Delta \lambda/ \lambda|$ if $\Delta \lambda/ \lambda$ << 1. Use this to calculate the magnitude of the smallest possible uncertainty in the wavelength of the photon. Give your answer in nanometers.
Answer:
The transition energy E for the atom and the wavelength λ of the emitted photon are related by
$E = \frac{hc}{\lambda}$
Apply the Heisenberg Uncertainty Principle in the form
$\Delta E \Delta t \geq \frac{\hbar}{2}$
Assume the minimum possible value for the uncertainty product, so that $\Delta E \Delta t = \frac{\hbar}{2}$
(a) E = 2.56 eV = $4.13 \times 10^{-19}\ J$, with a wavelength of
$\lambda = \frac{hf}{E}=\frac{(6.626 \times 10^{-34} \ J.s)(2.998 \times 10^8 \ m/s)}{4.13 \times 10^{-19} \ J}$
$\lambda = 4.82 \times 10^{-7} \ m$ = 482 nm
(b) $\Delta E \Delta t = \frac{\hbar}{2}=\frac{1.055 \times 10^{-34} \ J.s}{2(1.64 \times 10^{-7} \ s)}$
$\Delta E = 3.22 \times 10^{-28} \ J$ = $2.01 \times 10^{-9}$ eV
(c) $\Delta E \lambda = hc$, so $\Delta \lambda E + \lambda \Delta E = 0$
and $|\Delta E/E|=|\Delta \lambda/ \lambda|$
So, $\Delta \lambda = \lambda |\Delta E/E|$
$\Delta \lambda = (4.82 \times 10^{-7} \ m) \left(\frac{3.22 \times 10^{-28} \ J}{4.13 \times 10^{-19} \ J}\right)$
$\Delta \lambda = 3.75 \times 10^{-16} \ m = 3.75 \times 10^{-7} \ nm$
The finite lifetime of the excited state gives rise to a small spread in the wavelength of the emitted light.
Q#39.87
You intend to use an electron microscope to study the structure of some crystals. For accurate resolution, you want the electron wavelength to be 1.00 nm.
(a) Are these electrons relativistic? How do you know?
(b) What accelerating potential is needed?
(c) What is the kinetic energy of the electrons you are using? To see if it is great enough to damage the crystals you are studying, compare it to the potential energy of a typical NaCl molecule, which is about 6.0 eV.
(d) If you decided to use electromagnetic waves as your probe, what energy should their photons have to provide the same resolution as the electrons? Would this energy damage the crystal?
Answer:
The electrons behave as waves whose wavelength is equal to the de Broglie wavelength.
The de Broglie wavelength is $\lambda = \frac{h}{mv}$, and the energy of a photon is E = hf = $\frac{hc}{\lambda}$
(a) Use the de Broglie wavelength to find the speed of the electron.
$v=\frac{h}{m \lambda}$
$v = \frac{6.626 \times 10^{-34} \ J.s}{(9.11 \times 10^{-31} \ kg)(1.00 \times 10^{-9} \ m)}$
$v = 7.27 \times 10^5 \ m/s$
which is much less than the speed of light, so it is nonrelativistic.
(b) Energy conservation gives
eV = $\frac{1}{2}mv^2$
V = $\frac{mv^2}{2e}$
V = $\frac{(9.11 \times 10^{-31} \ kg)(7.27 \times 10^5 \ m/s)^2}{2(1.60 \times 10^{-19} \ C)}$
V = 1.51 V
(c) K = eV = 1.51 eV which is about ¼ the potential energy of the NaCl molecule, so the electron would not be too damaging.
(d) E = $\frac{hc}{\lambda}$
E = $\frac{(6.626 \times 10^{-34} \ J.s)(2.998 \times 10^8 \ m/s)}{1.00 \times 10^{-9} \ m}$
E = 1240 eV
which would certainly destroy the molecules under study.
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