Physics: Principles and Problems
9th Edition
ISBN: 9780078458132
Table of contents
Textbook solutions
All Solutions
Page 737: Section Review
Exercise 23
Step 1
1 of 2
Scattering of electron beams on a crystal showed a diffraction like picture on the screen which was at that point a result observed only for light or more generally, waves. This made scientist conclude that also particles posses some wavelike properties.
Result
2 of 2
Scattering of electrons on a crystal.
Exercise 24
Step 1
1 of 2
The de Broglie wavelengths of macroscopic objects are very small and a way beyond of the scope of measurability.
Result
2 of 2
Wavelengths of the macroscopic objects are unmeasurable since they are too small.
Exercise 25
Step 1
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In order to solve this problem we are going to use the conservation of energy to find the speed of the electron first and then we are going to use de Broglie relation to find the de Broglie wavelength.
$$
frac{mv^2}{2}=eV
$$
which makes that the speed is given as
$$
v=sqrt{frac{2eV}{m}}=sqrt{frac{2times 1.6times 10^{-19}times 125}{9.11times 10^{-31}}}
$$
Finally,
$$
v=6.63times 10^6, frac{textrm{m}}{textrm{s}}
$$
Now, de Broglie’s wavelength is given by
$$
lambda=frac{h}{mv}=frac{6.63times 10^{-34}}{9.11times 10^{-31}times 6.63times 10^6}
$$
which gives that
$$
boxed{lambda=0.11times 10^{-9}textrm{ m}}
$$
Result
2 of 2
$$
lambda=0.11times 10^{-9}textrm{ m}
$$
lambda=0.11times 10^{-9}textrm{ m}
$$
Exercise 26
Step 1
1 of 2
The wavelength of a photon in a Compton scattering process will increase since the photon will lose part of its energy.
Result
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The wavelength of a photon in Compton effect will increase.
Exercise 27
Step 1
1 of 2
The uncertainty principle tells us that we cannot know the momentum and position at the same time. For that reason if the distance between the slits is appropriate we cannot say with certainty where did an electron go so we get the interference pattern on the screen typical for the waves.
Result
2 of 2
We cannot know the position and the momentum at the same time.
Exercise 28
Step 1
1 of 2
In order to solve this problem we have to understand that for a diffraction the wavelength of the passing waves is connected to the spacing of the slits as
$$
lambda=dsintheta
$$
and since the angle $theta$ is considered to be small for first few minima we can take that $sinthetaapprox0.2$
$$
lambdaapprox250times 0.2=boxed{50textrm{ nm}}
$$
Result
2 of 2
$$
lambdaapprox 50textrm{ nm}
$$
lambdaapprox 50textrm{ nm}
$$
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