# Physics 203 at Portland State 2014

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## Announcements

past announcements have moved to sidebar and can be found here.

### All grades are in, final exam solutions posted!

Please check your final grades, as well as the final exam solutions on D2L. Congratulations to everyone who put the effort into this course and did well! — Nicholas Kuzma 2014/06/13 23:04

### All grades except for the final are in

• I just finished grading papers and projects, and updated my record of the wiki edits.
• Please respond by Monday morning if you have any concerns…. — Nicholas Kuzma 2014/06/13 16:31

• Grades on http://d2l.pdx.edu are current as of June 3. Please check.
• One practice exam is posted here (to be reviewed tomorrow), another – on D2L
• Please complete the survey (see the next announcement below)

### Exam 2 solutions

Exam 2 solutions are posted on D2L now under “Current exam solutions”. Please download! — Nicholas Kuzma 2014/05/16 19:51

### Workshop volunteers needed

Please email your professor if you are interested in becoming a Workshop leader next year. — Nicholas Kuzma 2014/05/10 14:15

### Office hours

From now on, office hour will be held at CH71, same location as the class. The time is still Thursdays 9-10am.

### Project extensions

Project summaries will be due on Wednesday, May 28th. Paper outlines are still due May 20th.

### Summer course: Physics in Biomedicine

• Thinking Of A Career In Medicine?
• Want to find out how much physics is used in the biomedical field?

#### Physics in Biomedicine

##### offered June 23 – July 13 10:30am-12:50pm (MTWRF)

4 credits at Portland State University

• Topics include: MRIs, functional MRI, PET scan, X-ray imaging, CT scans, Radiation Therapy, Microscopy and more!
• Textbook: “Introduction to Physics in Modern Medicine” 2nd edition by Suzanne Amador Kane
• Instructor: Ralf Widenhorn, ralfw@pdx.edu

### For you, wiki editors!

New! At the bottom of the sidebar on the left, there is a new link to the list of ”tasks to do” for our wiki editors (that means you!). Tick it off (by entering your username into the table) and do it!

## Lectures and Notes

Use WolframAlpha for calculations

## Ch.13,14, 25-27 have moved

Previous chapters have been moved to the sidebar:

## Ch.28: Interference and Diffraction (May 20-22)

### Concepts Ch.28

• Constructive and destructive interference
• Phase change during reflection

### Units Ch.28

• Micron ($\mu$m) = $10^{-6}$ m
• (Arc)minute = (1/60)$^\circ$
• (Arc)second = (1/3600)$^\circ$

### Math Ch.28

 $\cos\alpha+\cos\beta=$ $2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)$ $\cos\alpha=\cos(-\alpha)$ $\sqrt{1+\epsilon}\approx$ $1\!+\!\frac{1}{2}\!\epsilon$ $\sqrt{1-\epsilon}\approx$ $1\!-\!\frac{1}{2}\!\epsilon$ For small $|\epsilon|\ll 1$ $\tan[\arcsin(x)] = \frac{x}{\sqrt{1-x^2}}$ $\sin\alpha=-\sin(-\alpha)$ $\tan[\arccos(x)] = \frac{\sqrt{1-x^2}}{x}$

### Lecture notes May 20,22

• page 9 was updated: forgot to divide by 2 (6 cm $\rightarrow$ 3 cm) in the 1st problem, and $m=1\;\rightarrow\;m=0$ in the last problem

### Equation Sheet Ch.28

1. Conditions for bright fringes (constructive interference) in a two-slit experiment:
• $d\sin\theta = m\lambda$
• $d$ is the slit separation
• $m=0,\,\pm 1,\,\pm 2,\,\pm 3,\,...$
• $m\!=\!0$ occurs at $\theta\!=\!0$, this is the central bright fringe
• Positive values of $m$ are above the central bright fringe, negative values are below
• Solving for $\theta$:
• $\theta =$ $\arcsin\left(m\frac{\lambda}{d}\right)$
2. Conditions for dark fringes (destructive interference) in a two-slit experiment:
• $d\sin\theta =$ $\left(m-\frac{1}{2}\right)\lambda$
• $m = 1,\,2,\,3\,...$ (above the central bright fringe)
• $d\sin\theta =$ $\left(m+\frac{1}{2}\right)\lambda$
• $m = -1,\,-2,\,-3\,...$ (below the central bright fringe)
• Solving for $\theta$:
• $\theta =$ $\arcsin\left[\big(m \pm \frac{1}{2}\big)\!\frac{\lambda}{d}\right]$
• “$+$” or “$-$” depending on location: above or below the central fringe
3. Linear distance from the central fringe:
• $y = L \tan\theta$.
• L is the distance to the screen.
• Solving for $\theta$ of a bright fringe:
• $\theta = \arctan\left(m\frac{y}{L}\right)$
• Solving for $\lambda$:
• $\lambda = \frac{d}{m}\sin\theta$
• Solving for $\theta$ of a dark fringe:
• $\theta = \arcsin\left[\big(m\pm\frac{1}{2}\big)\!\frac{\lambda}{d}\right]$.
• “$+$” or “$-$” depending on location: above or below the central fringe
4. Conditions for dark fringes in single-slit diffraction:
• $w\sin\theta = m\lambda$.
• $m = \pm 1,\,\pm 2,\,\pm 3\,...$
• $w$ is the slit width
• Solving for $\lambda: \lambda = \frac{Wsinθ}{m}$ \
• Solving for $θ: θ = sin^{-1}(\frac{m\lambda}{W})$
5. First Dark Fringe for the Diffraction Pattern of a Circular Opening:
• $sinθ = 1.22\frac{\lambda}{D}$
6. Rayleigh's Criterion:
• $θ_{min} = 1.22\frac{\lambda}{D}$
• Note: $\lambda$ is dependent on the diffraction of the material that the light is traveling through. If the diffraction is $n, \lambda$ becomes $\frac{\lambda}{n}$
7. Constructive Interference in a Diffraction Grating:
• $d sin θ = m\lambda$. $m = ±1,±2,±3...$
• Solving for $d: d = \frac{m\lambda}{sinθ}$

Destructive Interference: Waves cancel

$v = c/n$

Wavelength $\lambda$, of light in a medium of index of refraction $n$ greater than 1: $\lambda_n = \lambda$ (in vacuum)$/ n$

1. For an air wedge:
1. Constructive interference:
• $.5 + 2d/\lambda = m$
2. Destructive interference:
• Waves cancel (out of phase)
• $.5 + 2d/\lambda = .5 + m$
• Effective path length for ray 1 = .5 * wavelength
• Ray 1 reflects from the air to film interface
• Ray 2 reflects from the film to air interface.
• Effective Ray 2 path length / wavelength in film = 2t / wavelength in film = 2nt / wavelength in vacuum
• t = thickness of film
• Difference in phase changes = (2nt / wavelength in vacuum) - .5
• Destructive Interference: 2nt / wavelength in vacuum = m
• Constructive Interference: (2nt / wavelength in vacuum) - .5 = m
 Refraction index order ($n_1$ and $n_2$ are on either Dark fringes Bright fringes side of the film $n_f$) $\;\;m = 0,\, 1,\, 2,\, 3\,...$ $n_1\!<\!n_f\!>\!n_2\;$ or $\;n_1\!>\!n_f\!<\!n_2$ $t = \frac{\lambda}{2n_f}m$ $t = \frac{\lambda}{2n_f}\left(m+\frac{1}{2}\right)$ $n_1\!<\!n_f\!<\!n_2\;$ or $\;n_1\!>\!n_f\!>\!n_2$ $t = \frac{\lambda}{2n_f}\left(m+\frac{1}{2}\right)$ $t = \frac{\lambda}{2n_f}m$

### Examples Ch.28

#### Problem 6.1

Two sources emitting the light of equal wavelength, phase, and intensity, are located at positions $s_1$ and $s_2$ as shown, a short distance d apart from each other and a distance L away from the vertical screen. Find the locations of bright and dark spots on the screen in terms of the y coordinate along the screen.

Steps:

1. Identify given variables:
• $d$ is the (vertical) distance between the sources $s_1$ and $s_2$
• $L$ is the (horizontal) distance from the sources to the screen
2. Determine relevant equations that relate these variables to the question
• Bright spots:
• $d\sin\theta$ $= m\lambda\;$ for $\;m = 0,\, \pm 1,\, \pm 2,\,$ etc
• Dark spots:
• $d\sin\theta$ $=\big(m-\frac{1}{2}\big)\lambda\;$ for $\;m = 1,\, 2,\, 3,\,$ etc
• $d\sin\theta$ $=\big(m+\frac{1}{2}\big)\lambda\;$ for $\;m = -1,\, -2,\, -3,\,$ etc
• $y = L\tan\theta$
3. Re-arrange the bright spots equation to solve for $\theta$
• $d\sin\theta$ $= m\lambda$
• $\sin\theta$ $=\frac{m\lambda}{d}$
• $\theta$ = $\arcsin\left(\frac{m\lambda}{d}\right)$
4. Re-arrange the darks spots equation to solve for $\theta$
• $d\sin\theta$ $= \left(m\pm\frac{1}{2}\right)\lambda$
• $\sin\theta$ $=\frac{\left(m\,\pm\,\frac{1}{2}\right)\,\lambda}{d}$
• $\theta=$ $\arcsin\left(\frac{(m\,\pm\,\frac{1}{2})\,\lambda}{d}\right)$
5. Insert these expressions for $\theta$ into $\;y = L\tan\theta$
• Bright spots:
• $y =$ $L\tan\big[\arcsin\big(\!\frac{m\lambda}{d}\!\big)\big]=$ $\frac{Lm\lambda}{d\sqrt{1-\left(\frac{m\lambda}{d}\right)^2}}=$ $\frac{m\lambda L}{\sqrt{d^2-(m\lambda)^2}}\;$ for $\;m = 0,\, \pm 1,\, \pm 2,\,$ etc
• Here we used the trig identity
• $\tan[\arcsin(x)] = \frac{x}{\sqrt{1-x^2}}$
• Dark Spots
• $y =$ $L\tan\left[\arcsin\left(\!\frac{(m-\frac{1}{2})\,\lambda}{d}\!\right)\right]=$ $\frac{(m-\frac{1}{2})\,\lambda L}{\sqrt{d^2-\left((m-\frac{1}{2})\,\lambda\right)^2}}\;$ for $\;m = 1,\, 2,\, 3,\,$ etc
• $y =$ $L\tan\left[\arcsin\left(\!\frac{(m+\frac{1}{2})\,\lambda}{d}\!\right)\right]=$ $\frac{(m+\frac{1}{2})\,\lambda L}{\sqrt{d^2-\left((m+\frac{1}{2})\,\lambda\right)^2}}\;$ for $\;m = -1,\, -2,\, -3,\,$ etc
6. You can now use these equations to determine the locations (the y values) of any bright/dark spot of order $m$.
• Note: a given spot exists only if $\big|\sin\theta\,\big|\leq 1$, or, equivalently, the square root in the above expressions takes on a positive argument.

#### Problem 6.2

Two thin slits, 1 mm apart, are illuminated with $\lambda=750\,$nm light. The screen is $L=15\,$m away. Find the locations of the bright fringes on the screen.

Steps:

1. Identify variables and convert to SI units:
• $d =$ $1\,{\text{mm}} \times \frac{1\,{\text m}}{1000\,{\text{mm}}}$ $= 0.001\,$ m
• $\lambda=750\,{\text{nm}} \times \frac{1\,{\text m}}{10^9\,{\text{nm}}}$ $= 7.5 \times 10^{-7}\,{\text m}$
• $L = 15\,$m
2. Determine relevant equations that relate these variables to the available data
• $d\sin\theta$ $= m\lambda\;$ for $\;m = 0,\, \pm 1,\, \pm 2,\,$ etc
• $y = L\tan\theta$
3. In order to combine the two equations, solve $\,d\sin\theta\!=\!m\lambda\,$ for $\,\theta$
• $d\sin\theta$ $= m\lambda$
• $\sin\theta$ $= \frac{m\lambda}{d}$
• $\theta$ $=\arcsin\big(\!\frac{m\lambda}{d}\!\big)$
4. Insert this $\theta$ expression into $y = L\tan\theta$
• $y = L\tan\theta$
• $y = L\tan\big[\arcsin\big(\!\frac{m\lambda}{d}\!\big)\big]$
5. Insert the given values and simplify
• $y = L\tan\big[\arcsin\big(\!\frac{m\lambda}{d}\!\big)\big]$
• $y = (15\,{\text m})\tan\big[\arcsin\big(m\frac{7.5 \times 10^{-7}\,{\text m}}{0.001\,{\text m}}\big)\big]$
• $y =$ $(15\,{\text m})\tan\left[\arcsin\left(m\cdot 7.5 \times 10^{-4}\right)\right]$
6. You can now calculate y for $\;m = 0,\, \pm 1,\, \pm 2,\,$ etc
• For $m = 0$:
• $y =$ $(15\,{\text m})\tan\left[\arcsin\left(0\cdot 7.5 \times 10^{-4}\right)\right]$ $= 0$
• For $m = +1$:
• $y =$ $(15\,{\text m})\tan\left[\arcsin\left(1\cdot 7.5 \times 10^{-4}\right)\right]$ $= 0.01125\,$m $\approx 1.1\,$cm
• For $m = -1$:
• $y =$ $(15\,{\text m})\tan\left[\arcsin\left((-1)\cdot 7.5 \times 10^{-4}\right)\right]$ $= -0.01125\,$m $\approx -1.1\,$cm
• etc

#### Problem 6.3

An air wedge is getting wider along $y$, such that $d(y)=\beta\!\cdot\!y$, with $\beta=10^{-5}\,$(rad). Find the spacing of interference produced by yellow ($\lambda=600\,$nm) light along the $y$ axis.

Steps:

1. A change in phase difference between the two reflected beams of $2\pi$ radians (360$^\circ$) gives the distance to the next fringe along the y axis.
• One of the reflections will accrue an extra $\pi$ of phase change (from air into glass and back into air)
• The bright spots are given by the condition:
• $\frac{\lambda}{2}+2d$ $=m\lambda\;$ where $\;m=0,\,1,\,2,\,3...$
• Consider subsequent fringes $m_1$ and $m_2=m_1+1$:
• $\frac{\lambda}{2}+2d_1$ $=m_1\lambda$
• $\frac{\lambda}{2}+2d_2$ $=(m_1+1)\lambda$
• Subtract these two equations:
• $2(\,d_2-d_1)=$ $(m_1+1)\lambda-m_1\lambda$ $=\lambda$
• $d_2-d_1=$ $\frac{\lambda}{2}$
2. Now convert this difference in $d$ to a difference in $y$:
• $y_2-y_1=$ $\frac{d_2}{\beta}-\frac{d_1}{\beta}=$ $\frac{1}{\beta}(d_2-d_1)=$ $\frac{1}{\beta}\frac{\lambda}{2}=$ $\frac{\lambda}{2\beta}$
3. Plugging in the numbers to find the y distance between the fringes:
• $\Delta y=$ $y_2-y_1=$ $\frac{\lambda}{2\beta}=$ $\frac{600\times 10^{-9}\,{\text m}}{2\,\cdot\,10^{-5}}$ $= 0.03\,$ m, or 3 cm.

#### Problem 6.4

A camera lens ($n_\text{lens}=1.42$) is coated with a thin film ($n_\text{film}=1.55$) to prevent reflections at $\lambda_\text{vac}=600\,$nm. Find the minimum thickness $d$ of the film to achieve this, assuming the (normal) reflections from both surfaces of the film, after emerging into the air, are of equal intensity.

Steps:

1. Note that the reflection from the front of the film is of “low n – high n – low n” type, therefore it incurs an extra $\pi$ of phase:
• air–film–air, $\;n_\text{air}\approx 1$
2. The reflection from the back of the film is of “high n – low n – high n” type, since $\,n_\text{film}>n_\text{lens}\,$. No phase gain there.
• film–lens–film
3. Write the equation for the optical path difference between the two normal reflections for dark-fringe conditions:
• $\Delta{\text{Path}}=$ $2d+\frac{\lambda_\text{film}}{2}$ $=\big(m+\frac{1}{2}\big)\lambda_\text{film}\;$
• The term $\frac{\lambda_\text{film}}{2}$ is to account for the extra half-wavelength ($\pi$ of phase difference) due to reflections being of different type
4. Substitute $m\!=\!1$ for the thinnest film, and express the shortened wavelength in the film in terms of the vacuum value:
• $2d=$ $1\!\cdot\lambda_\text{film}$ $= \frac{\lambda_\text{vac}}{ n_\text{film} }$
5. Solve for $d$:
• $d=$ $\frac{\lambda_\text{vac}}{2\,n_\text{film}}$ $= \frac{600\,{\text{nm}}}{2 \times 1.55 }$ $=194\,{\text{nm}}$

#### Problem 6.5

Use a caliper with $\frac{1\ }{1000\ }$ ${\text inch}$ gap, L$=$15m,

Steps:

• $W$ $=\frac{0.0254\, {\text m}}{1000\ }$ $= 2.54 \times 10^{-5}$
• $\lambda$ = 450 ${\text nm}$ (blue light)
• First dark Fringe:
• $W\sin\theta$ $=\lambda$ (m=1)
• $\sin\theta$ $= \frac{\lambda}{W}$ $=\frac{4.5\times 10^{-7}}{2.54 \times 10^{-5}}$ $=0.0177$ $=1°$
• $y = L\tan\theta$ $=15 \text m \times \tan 1°$ =0.27m
• Note that as W goes down, y goes up and vice versa.

Where is the question?

#### Problem 6.6

Find diffraction limit of the angular resolution (that is, the smallest angle between two stars that can be resolved) for a telescope with a 10-cm diameter objective lens. Assume $\lambda$ $= 600\,{\text{nm}}$

Steps:

1. Convert units to SI:
• $\lambda$ $= 600\,{\text{nm}}$ $=6\times 10^{-7}\,$m
• $D=10\,$cm $=10^{-1}\,$m
2. Use Rayleigh's criterium to calculate the minimally resolvable angle:
• $\theta_\text{min}$ $=1.22 \frac{\lambda}{D}$ $=1.22 \times \frac{6 \times 10^{-7} \text m}{ 10^{-1} {\text m}}$ $= 7.3 \times 10^{-6}$ rad $\approx 1.5\,$arcsec

This will resolve a 1.7-mile object on the Moon.

#### Problem 6.7

A laser of $\lambda=633\,$nm wavelength shines normally through a grating with thin slits every 2$\,\mu$m. Find the angle (from the original beam direction) of the 1st and the 2nd peaks.

Steps:

1. Convert to SI units:
• $633\,{\text{nm}}= 6.33 \times 10^{-7}\,{\text m}$
• $2\,\mu{\text m}=2\times 10^{-6}\,{\text m}$
2. 1st peak:
• $d\sin\theta$ $=\lambda$
• $\sin\theta$ $= \frac{\lambda}{d}$ $= \frac{6.33 \times 10^{-7}\,{\text m}}{2 \times 10^{-6}\,{\text m}}$ $=0.3165$
• $\theta$ $=\arcsin(0.3165)$ $=0.322\;{\text{rad}}=$ $18.5^\circ$
3. 2nd peak:
• $d\sin\theta$ $=2 \lambda$
• $\sin\theta$ $= \frac{2\,\cdot\,6.33 \times 10^{-7}\,{\text m}}{2 \times 10^{-6}\,{\text m}}$ $=0.633$
• $\theta$ $=\arcsin(0.633)$ $=0.685\;{\text{rad}}=$ $39.3^\circ$

1. Q?
• A

## Ch.29: Relativity (May 27-29)

### Constants Ch.29

• Speed of light $c=299792458\frac{\text m}{\text s}$ $\approx 3.00\times 10^8\frac{\text m}{\text s}$
• Gravitational constant $G=6.67384\times 10^{-11}\frac{ {\text N}\cdot{\text m}^2}{ {\text{kg}^2}}$

### Math Ch.29

Rotations Relativity
$\cos\left[\arctan\left(\frac{v}{c}\right)\right]$ $=\frac{1}{\sqrt{1+\left(\frac{v}{c}\right)^2}}$ $\cosh\left[{\text{arctanh}}\left(\frac{v}{c}\right)\right]$ $=\frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}}$
$\sin\left[\arctan\left(\frac{v}{c}\right)\right]$ $=\frac{v/c}{\sqrt{1+\left(\frac{v}{c}\right)^2}}$ $\sinh\left[{\text{arctanh}}\left(\frac{v}{c}\right)\right]$ $=\frac{v/c}{\sqrt{1-\left(\frac{v}{c}\right)^2}}$

### Lecture notes May 27,29

• May 27: notes14s.pdf – there is an error on page 9, lines 3 and 4: cos and sin should read cosh and sinh

### Equation Sheet Ch.29

Please update if an equation is not included

1. The laws of physics are the same in all inertial frames of reference
2. The speed of light in a vacuum is the same in all inertial frames of reference
3. The time it takes light to travel a distance 2D is simply $\frac{2D}{c}$. This is proper time, or the time elapsed within the same spatial reference.
4. To the outside observer to a spatial reference system, the time elapsed for them as an object travels 2D at velocity $v$ is
• $\Delta t$ =$\frac{\Delta t_0}{\sqrt{1-\frac{v^2}{c^2}}}$
• $\Delta t_0$ is the Proper time
5. Proper length is the length of an object as measured by an observer within its spatial reference system
6. To the outside observer to a spatial reference system, Contracted length of a object moving at velocity $v$ in the direction of relative motion
• $L=L_0\times\sqrt{1-\frac{v^2}{c^2}}$
• $L_0$ is the Proper length
• $L$ is the Contracted length
• $v=\frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}$
• Here $v$ is the velocity of object 2 relative to some frame
• $v_1$ is the velocity of object 1 relative to the same frame
• $v_2$ is the velocity of object 2 relative to object 1
8. Relativistic subtraction of velocities:
• $v=\frac{v_2-v_1}{1-\frac{v_1v_2}{c^2}}$
• Here $v$ is the relative velocity of object 2 relative object 1
• $v_1$ is the velocity of object 1 relative to some frame
• $v_2$ is the velocity of object 2 relative to the same frame
9. Relativistic momentum
• $p=\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}$
• Here $m_0$ is the rest mass
10. The traditional momentum equation $p = mv$ can be used if we use proper mass $m$:
• $m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$
11. Relativistic energy, $E = mc^2=\frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}$
12. Rest energy, $E_0 = m_0c^2$
• $m_0$ is the rest mass
13. Relativistic Kinetic energy, $K = \frac{m_0c^2}{\sqrt{1 -\frac{v^2}{c^2}}}-m_0c^2$
14. Schwarzschild radius or the radius of a black hole, $R$
• $R = \frac{2GM}{c^2}$
• $G=$ $6.67\times {10^{-11}}\frac{ {\text N}\cdot{\text m}^2}{ {\text{kg}}^2}\;$ is the gravitational constant
• $M$ is the mass of an astronomical body
• $v = c\sqrt{1-(\frac{\Delta t_0}{\Delta t})^2}$

### Examples Ch.29

#### Problem 7.1

How far does the light travel in 1 year? 1 minute? 1 s? 1 ms? 1 $\mu$s? 1 ns?

Steps:

1. Convert the given times to SI units:
• $3.154\times 10^7\,$s, $60\,$s, $1\,$s, $10^{-3}\,$s, $10^{-6}\,$s, $10^{-9}\,$s.
2. Use the simple formula for distance in terms of speed and time:
• $x=c\,t$ $=2.998\times 10^8\,\frac{\text m}{\text s}\cdot t$
3. These distances are SI equivalents of the “light” units for distance:
Distance unit SI equivalent (m) Example
1 light-year $9.5\times 10^{15}$ $\sim\frac{1}{4}$ of the distance to the star nearest to the Sun
1 light-minute $1.8\times 10^{10}$ $\sim\frac{1}{8}$ of the Earth–Sun distance
1 light-second $2.998\times 10^{8}$ $\sim 75\%$ of the Earth–Moon distance
1 light ms $2.998\times 10^{5}$ $186\,$miles: just beyond Seattle (from Portland)
1 light $\mu$s $2.998\times 10^{2}$ $\sim 1000\,$ft from Cramer Hall to Phys. department and back
1 light ns $2.998\times 10^{-1}$ $\sim 1\,$ft about the length of a page of paper

#### Problem 7.2

At a speed $\;v=0.99\,c\;$ one of the two twins travels to Alpha Centauri, the nearest star (system) to our Sun, which is “only” 4.3 light-years away from us. How much older or younger is the space-traveling twin compared to the earth-bound one upon the return of the spaceship?

Steps:

1. In problems like this it is convenient to measure distances in light-years, time in years, and velocities in the units of c (the speed of light). That way all the equations work just fine, and there is no need to convert the units.
2. The time it takes for the ship to fly out to $\alpha$-Centauri and back, measured from Earth, is
• $t=\frac{2d}{v}=$ $\frac{2\,\cdot\,4.3\,c\,(1\,{\text{year}})}{0.990\,c}$ $=4.34\,{\text{years}}\cdot 2=8.68\,$years
• Here we do not take into account the time spent on the distant star, because during that time, both twins age at the same rate
3. In the ship frame, the time of travel $\Delta t_0$ is…
• Use the time dilation equation
• $\Delta t$ =$\frac{\Delta t_0}{\sqrt{1-\frac{v^2}{c^2}}}$
• Substitute the known $\Delta t$ and solve for $\Delta t_0$:
• $\Delta t_0= \Delta t\sqrt{1-\frac{v^2}{c^2}}=$ $8.68\,{\text{years}} \sqrt{1-\frac{(0.990\,c)^2}{c^2}}$ $=8.68\,{\text{years}} \sqrt{1-0.990^2}$ $= 1.22\,$years
4. The difference in the twins' biological age upon return is
• $8.688\,{\text{years}} - 1.228\,{\text{years}} = 7.46\,$years difference in age

#### Problem 7.3

A meter-stick is moving at $\;v=0.7\,c\;$ parallel to its own long dimension. How long does this stick appear from stationary Earth frame?

Steps:

1. Identify given variables
• $L_0 = 1.0\,{\text m}$
• $v=0.7\,c\;$
• Need to find $L$
2. Identify relevant equation(s)
• $L = L_0\sqrt{1-\frac{v^2}{c^2}}$
3. Plug the variables into the equation and solve for $L$. Note: It is not necessary to multiply $v$ by $c$ because the $c$'s will cancel.
• $L = L_0\sqrt{1-\frac{v^2}{c^2}}$
• $L = (1\,{\text m})\,\sqrt{1-\frac{(0.7c)^2}{c^2}}$
• $L = (1\,{\text m})\,\sqrt{1-\frac{0.49c^2}{c^2}}$
• $L = (1\,{\text m})\,\sqrt{1-0.49}$
• $L = (1\,{\text m})\,\sqrt{0.51}$
• $L = 1\,{\text m}\cdot 0.71$
• $L = 0.71\,{\text m}$
4. Note, that if the stick were moving perpendicular to its long dimension, its length would appear unchanged in both frames.

#### Problem 7.4

A bus, moving at a speed v, turns on its headlights, emitting light at velocity c relative to the bus' frame. How fast does this light appear to travel relative to Earth?

Steps:

1. Visualize or draw the problem and, if needed, re-word it to simplify what is being asked.
• In this case, it is asking: what is the speed of light relative to Earth when the light is emitted from a moving bus?
• Or, more fundamentally: is the speed of light dependent on the motion of the source?
2. According to Einstein's postulate of special relativity:
• “The speed of light in vacuum is the same in all inertial frames of reference, independent of the motion of the source or the receiver.”
3. Therefore, the answer to this question is: The light appears to travel at velocity $c = 3.00 \times\ 10^8\frac{\text m}{\text s}$ relative to the Earth.
4. Or, alternatively, use the addition of velocities:
• $v=$ $\frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}$
• where $v_1=v_\text{bus}$ is the velocity of the bus
• $v_2=c$ is the speed of emitted light (in bus' frame)
• $v$ is the speed of light (in the Earth frame) that we need to find
• $v=$ $\frac{v_\text{bus}+c}{1+\frac{v_\text{bus}c}{c^2}}=$ $\frac{v_\text{bus}+c}{1+\frac{v_\text{bus}c}{c^2}}\times\frac{c}{c}$ $=\frac{v_\text{bus}+c}{c+v_\text{bus}}\!\cdot\!c$ $=c$

#### Problem 7.5

A spaceship, moving at a speed $\;v=0.5\,c\;$, launches a rocket in the forward direction, also at a speed $\;v=0.5\,c\;$ relative to the spaceship. How fast does the rocket appear to travel relative to our planet? How would this answer change if the rocket were fired in the backward direction, opposite to the spaceship's speed vector?

Steps:

##### Part 1: Rocket fired in forward direction
1. Identify relevant equation(s):
• $v_{23} = \frac{v_{21}+v_{13}}{1+\frac{v_{21}v_{13}}{c^2}}$
2. Identify and label variables:
• Object 1 = spaceship
• Object 2 = rocket
• Object 3 = planet
• $v_{13}= 0.5\,c\;$ is the velocity of the spaceship relative to the planet
• $v_{21}= 0.5\,c\;$ is the velocity of the rocket relative to the spaceship
• $v_{23}\;$ is the velocity of the rocket relative to the planet – the quantity we need to find
3. Plug in the variables and find $v_{23}$.
• Note: It is not necessary to multiply the velocities by c because the c's will cancel and the answer can be expressed in terms of c.
• $v_{23} = \frac{v_{21}+v_{13}}{1+\frac{v_{21}v_{13}}{c^2}}$
.
• $v_{23} = \frac{0.5c+0.5c}{1+\frac{(0.5c)(0.5c)}{c^2}}$
.
• $v_{23} = \frac{1.0c}{1+\frac{0.25c^2}{c^2}}$
.
• $v_{23} = \frac{1.0c}{1+0.25}$
.
• $v_{23} = 0.8\,c$
4. Answer: The rocket appears to be traveling at $\;v=0.8\,c\;$ relative to the planet.
##### Part 2: Rocket fired in backward direction
1. The equation and the set-up are the same, except now the velocity of the rocket is negative: $v_{21} = -0.5\,c$
2. Find $v_{23}$ using this negative value of $v_{21}$:
• $v_{23} = \frac{v_{21}+v_{13}}{1+\frac{v_{21}v_{13}}{c^2}}$
.
• $v_{23} = \frac{(-0.5c)+0.5c}{1+\frac{(-0.5c)(0.5c)}{c^2}}$
.
• $v_{23} = \frac{0\,c}{1+\frac{-0.25c^2}{c^2}}=0$
3. Answer: The rocket appears to be motionless relative to the planet.

#### Problem 7.6

A 50-kg atomic bomb “yields” $4\times 10^{15}\,$J of energy, assuming all of the uranium has fissioned (and not fizzled!). What percentage of the uranium mass is “converted” to energy?

Steps:

1. The data is already in SI units:
• Initial rest mass of the uranium:
• $m_0= 50\,$kg
• Amount of energy converted to kinetic energy, radiation and heat:
• $E_\text{yield}=4\times 10^{15}\,$J
2. Find the initial rest energy:
• $E_\text{rest}=m_0c^2$ $=50\,{\text{kg}}\cdot\left(3\times 10^8\frac{\text m}{\text s}\right)^2$ $=4.5\times 10^{18}\,$J
3. The fraction of the mass converted is the same as the ratio of the yield energy to the rest energy:
• $\frac{m_\text{conv}}{m_\text{rest}}$ $=\frac{E_\text{yield}}{E_\text{rest}}=$ $\frac{4\,\times\,10^{15}\,{\text J}}{4.5\,\times\,10^{18}\,{\text J}}$ $=9\times 10^{-4}$
4. About 0.1% of the initial mass, or about 50 g, is converted to the energy of the blast.
• This mass doesn't disappear, it will contribute to the mass increase of the material that absorbs the energy of the blast

#### Problem 7.7

During the Chernobyl nuclear disaster in the spring of 1986, one of the dominant and most lethal radioactive isotopes released into environment was iodine-131, usually denoted 131I. It has a half-life of 8 days, and spontaneously decays into a stable 131Xe and an electron. The kinetic energy released in this decay is 0.971 MeV. Calculate the speed of the released electron (assuming it, being much much lighter than 131Xe, acquires the vast majority of the kinetic energy). Also find the mass of such an electron, and the recoil velocity of the 131Xe atom.

Steps:

1. Convert units to SI:
• Conversion factor:
• $1\,{\text{eV}}=$ $1.6\times 10^{-19}{\text J}$
• $0.971\times 10^{6}\,{\text{eV}} \cdot\frac{1.6\,\times\,10^{-19}\,{\text J}}{1\,{\text{eV}}}$ $=1.55\times 10^{-13}\,{\text J}$
2. Calculate the electron's rest energy
• $E_0=$ $m_0c^2=9.1\times 10^{-31}\,{\text{kg}}\,\cdot\,3.0\times 10^8\frac{\text m}{\text s}$ $=8.19\times 10^{-14}\,{\text J}$
3. Use the relativistic kinetic energy equation to calculate electron's velocity from its kinetic energy $K_e$:
• $K_e=$ $\frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}-m_0c^2$ $=1.55\times 10^{-13}\,{\text J}$
• Here we neglect the kinetic energy of xenon, because it is much heavier than the electron. See the last step below.
4. Now treat this as an equation on $v$ and rearrange the terms to solve for $v$:
• $\frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}=$ $1.55\times 10^{-13}\,{\text J}+m_0c^2$
• $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=$ $\frac{1.55\times 10^{-13}\,{\text J}\,+\,m_0c^2}{m_0c^2}=$ $\frac{1.55\times 10^{-13}\,{\text J}\,+\,8.19\times 10^{-14}\,{\text J}}{8.19\times 10^{-14}\,{\text J}}$ $=2.9$
• $\sqrt{1-\frac{v^2}{c^2}}$ $=\frac{1}{2.9}=0.345$
5. find the velocity $v$
• $1-\frac{v^2}{c^2}=$ $0.345^2$
• $\frac{v^2}{c^2}=$ $1-0.345^2$
• $\frac{v}{c}=\sqrt{1-0.345^2}$
• $v=c\sqrt{1-0.345^2}$
• $v=0.939\,c$
• the speed of this electron is about 94% of the speed of light
6. Calculate the mass of the moving electron
• $m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$ $=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}m_0$ $=2.9\,m_0$
• the mass of a moving electron is 2.9 times higher than the mass of an electron at rest
7. Calculate the speed of xenon-131. The heavy atom moves so slowly that relativistic (Einstein) formulas are not needed (see last step).
• Use the conservation of momentum
• $\mathbf{p}_\text{fin}=$ $\mathbf{p}_e+\mathbf{p}_{^{\,131}{\text{Xe}}}=$ $\mathbf{p}_\text{initial}=0$ $\frac{|}{\big|}$
• $\big|\,\mathbf{p}_e\,\big|=\big|\,\mathbf{p}_{^{\,131}{\text{Xe}}\,}\big|$
• $\big|\,\mathbf{p}_e\,\big|=mv=$ $2.9\,m_0v=$ $2.9\,m_0\times 0.939\,c$ $=7.44\times 10^{-22}\frac{ {\text{kg}}\cdot{\text m}}{\text s}$
• $(m_{^{\,131}{\text{Xe}}})\,(v_{^{\,131}{\text{Xe}}})=$ $7.44\times 10^{-22}\frac{ {\text{kg}}\cdot{\text m}}{\text s}$
• Use Avogadro constant to convert the number of atoms in one mol to mass
• $N_A=$ $6.022 \times 10^{23}\frac{1}{\text{mol}}$
• $m_{^{\,131}{\text{Xe}}}=\frac{M_{^{\,131}{\text{Xe}}}}{N_A}=$ $\frac{0.131\,\frac{\text{kg}}{\text{mol}}}{6.022 \times 10^{23}\,{\text{mol}}^{-1}}$ $=2.17\times 10^{-25}\,$kg
• $v_{^{\,131}{\text{Xe}}}=$ $\frac{7.44\times 10^{-22}\frac{ {\text{kg}}\cdot{\text m}}{\text s}}{2.17\times 10^{-25}\,{\text{kg}}}$ $=3420\,\frac{\text m}{\text s}$
8. Check that the original assumptions make sense:
• relativistic factor for 131Xe:
• $\frac{1}{\sqrt{1-\frac{v^2_{^{\,131}{\text{Xe}}}}{c^2}}}$ $=\frac{1}{\sqrt{1-\frac{3420^2}{(3\,\times\,10^8)^2}}}$ $=1.000000000065$ $=1+6.5\times 10^{-11}$
• this is a very good approximation to 1. We are justified in using non-relativistic equations for the xenon atom
• the kinetic energy of 131Xe:
• $K_{^{\,131}{\text{Xe}}}=$ $\frac{(m_{^{\,131}{\text{Xe}}})\,(v_{^{\,131}{\text{Xe}}})^2}{2}=$ $\frac{(2.17\times 10^{-25}\,{\text{kg}})\,(3420\,\frac{\text m}{\text s})^2}{2}=$ $1.27\times 10^{-18}\,$J
• this is much smaller than $K_e=$ $1.55\times 10^{-13}\,{\text J}$. We are justified in assuming the electron carries most of the decay energy

#### Problem 7.8

Find the Schwarzschild radius of the Earth

Steps:

1. Determine what the question is asking. In this case, the “Schwarzschild radius” is the radius that an astronomical body must have in order to be a black hole. (The radius of the body must be equal to or less than the Schwarzschild radius.)
2. Identify relevant equation(s):
• $R=\frac{2GM}{c^2}$
3. Identify relevant variables:
• R is Schwarzschild radius of the object in question, i.e. what we need to find for the Earth
• $G=$ $6.67\times 10^{-11}\frac{ {\text N}\cdot{\text m}^2}{ {\text{kg}}^2}\;$ is the universal gravitational constant
• $M=$ $5.97\times 10^{24}{\text{ kg}}\;$ is the mass of the object in question, i.e. the mass of the Earth in this case
• $c=$ $3.00\times 10^8\frac{\text m}{\text s}\;$ is the speed of light in vacuum
4. Plug in the variables and find R:
• $R=\frac{2\,\cdot\,6.67\times10^{-11}\,\cdot\,5.97\times10^{24}}{(3.00\times 10^8)^2}\,{\text m}$
• $R=0.00886\,{\text m}$
5. Convert to other units if desired:
• $R=0.00886\,{\text m}\times\frac{1000\,{\text{mm}}}{1\,{\text m}}=8.86\,{\text{mm}}$
6. Answer: The earth would have to condense down to a radius equal to or less than 8.86 mm in order to become a black hole.

#### Problem 7.9

Is flying internationally on a passenger airplane going to delay or accelerate an atomic clock, compared to the identical clock left behind at home? Assume the speed of the airliner $\;v\approx 500\,$mph $\approx 800\frac{\text{km}}{\text{hr}}\;$, the cruising altitude 35,000 ft, and the radius of the Earth $6.4\times 10^6\,$m.

Steps:

1. Picture the problem:
• The clock will be delayed slightly due to the speed of the airplane itself, but its interaction with the Earth's frame of reference and gravity on the plane will accelerate the clock to a certain extent.
2. Convert the units to SI units:
• $v\approx 800\frac{\text{km}}{\text{hr}}$ $\approx 225\frac{\text{m}}{\text{s}}$
• $h=35000\,{\text{ft}} =$ $10^4\,{\text m}$ is the altitude (height) of the flight relative to Earth's surface
• $r=6.4\times 10^6\,{\text m}$
• $r_0=\frac{2GM_\oplus}{c^2}=$ $0.00887\,{\text m}$ is the Schwarzschild radius of the Earth (mass of the Earth is $M_\oplus=5.97219\times 10^{24}\,$kg)
3. Find the relevant equations:
• Dilation of the clock's time due to speed: $t_0 =$ $t\sqrt{1-\frac{v^2}{c^2}}$
• Dilation of time on Earth from Earth gravity, relative to a distant point: $t =$ $t_\infty\sqrt{1-\frac{r_0}{r}}$
• Dilation of time on the plane from gravity, relative to a distant point: $t_h=$ $t_\infty\sqrt{1-\frac{r_0}{r+h}}$
4. Solve each equation using the converted units:
• Dilation due to speed: $t_0=$ $t\sqrt{1-\frac{v^2}{c^2}} =$ $\left(1- 2.8\times 10^{-13}\right)\,t$
• The relative effect is $\frac{t_0-t}{t}=-0.28$ parts per trillion (the minus indicates slowing down of the traveling clock).
• Dilation on Earth from gravity: $t =$ $t_\infty\sqrt{1-\frac{r_0}{r}} =$ $t_\infty\sqrt{1-\frac{0.00887\,{\text m}}{6.4\times 10^6\,{\text m}}} =$ $\left(1- 6.9297\times 10^{-10}\right)\,t_\infty$
• Dilation on the plane: $t_h =$ $t_\infty\sqrt{1-\frac{r_0}{r+h}} =$ $t_\infty\sqrt{1-\frac{0.00887\,{\text m}}{6.4\times 10^6\,{\text m}\,+\,10^4\,{\text m}}} =$ $\left(1- 6.919\times 10^{-10}\right)\,t_\infty$
5. Find the ratio (relative dilation) between $t_h$ (time at altitude) and $t$ (time on Earth surface):
• $\frac{t_h}{t}$ = $\frac{1- 6.919\times 10^{-10}}{1- 6.9297\times 10^{-10}}\approx$ $1-\left(6.919\times 10^{-10}-6.9297\times 10^{-10}\right) =$ $1+1.0\times 10^{-12}$
• The plus indicates that a stationary clock at altitude goes faster by about 1 part per trillion, due to weaker gravity force up there
6. Interpret the results
• The atomic clock will slow down by 0.28 parts per trillion because of the speed of the airplane, and will accelerate by about 1 part per trillion due to weaker gravity.
• The net effect is acceleration by 0.72 parts per trillion.

## Ch.30: Quantum physics (June 3)

### Constants Ch.30

• Planck constant $h=$ $6.62606957\times 10^{-34}\,{\text J}\!\cdot\!{\text s};$ $\;\;\;\;\;$ $\hbar=\frac{h}{2\pi}=$ $1.054571726\times 10^{-34}\,{\text J}\!\cdot\!{\text s}$

### Units Ch.30

• Micron ($\mu$m) = $10^{-6}$ m
• (Arc)minute = (1/60)$^\circ$
• (Arc)second = (1/3600)$^\circ$

### Equation Sheet Ch.30

An ideal blackbody absorbs all light that's incident on it. Distribution of energy in a blackbody is independent of the material, it depends only upon the temperature.

1. Wien's Displacement Law:
• $f_\text{peak}= (5.88\times 10^{10}\,{\text s}^{-1}\!\cdot\!{\text K}^{-1})\,T$
2. Plank's Quantum Hypothesis:
• Planck constant $h=6.63\times 10^{-34}\,{\text J}\!\cdot\!{\text s}$
• Total energy of radiation at frequency $f$ is
• $E_n=nhf\;$ where $\;\;n=0,\,1,\,2,\,3,\,...$
3. Energy of a photon of frequency $f$
• $E=hf$
• SI units: J
4. Cutoff frequency $f_0$
• $f_0= \frac{W_0}{h}$
• $W_0\;$ is work function
• SI unit: Hz = ${\text s}^{-1}$
• If $\,f>f_0\,$ (frequency of light is above the cutoff frequency), then the electron can be ejected with finite kinetic energy
• If $\,f<f_0\,$, then no electrons are ejected
5. Maximum kinetic energy (K) a photoelectron can have:
• $K_\text{max}= E-W_0$ $=hf-W_0$
6. Rest mass of a photon
• $m_0= 0$
7. Momentum of a photon
• $p= \frac{hf}{c}$ $= \frac{h}{\lambda}$
8. Compton scattering (electron is initially at rest):
1. To conserve energy:
• Energy of incident photon = energy of scattered photon + final kinetic energy of electron
• $hf= hf' +K$
2. Compton Shift Formula
• $\Delta\lambda = \lambda' – \lambda =$ $\frac{h}{m_e c}\left(1-\cos\theta\right)$
• SI unit: m
• When $\theta = 180^\circ$, the change of the photon's wavelength is maximum
• When $\theta = 0^\circ$, no light actually scatters off of electrons, the change in wavelength is zero.
9. de Broglie wavelength (of any particle or object)
• $\lambda= \frac{h}{p}$
• SI unit: m
10. Constructive interference when scattering from a crystal
• $2d \sin\theta= m\lambda$
• $m=1,\,2,\,3,\,...$
11. The Heisenberg uncertainty principle:
1. Momentum and position
• $\Delta p_y\Delta y \ge \frac{h}{2\pi}$
2. Energy and time
• $\Delta E\Delta t \ge \frac{h}{2\pi}$

### Examples Ch.30

#### Problem 8.1

Find the peak emission frequencies of two black-body light sources, one at 4700 K (“cool white”), and another at 2700 K (“warm white”).

Steps:

1. Use Wien's Displacement Law to find
• $f_\text{peak}=$ $(5.88\times 10^{10}\,{\text s}^{-1}\!\cdot\!{\text K}^{-1})\,T$
2. Substitute the given T values to find peak frequencies:
• Peak emission frequency for “Cool White”:
• $f_\text{peak}=$ $(5.88\times 10^{10}\,{\text s}^{-1}\!\cdot\!{\text K}^{-1})\cdot 4700\,{\text K}$ $= 2.76 \times 10^{14}\,$Hz
• Peak emission frequency for “Warm White”:
• $f_\text{peak}=$ $(5.88\times 10^{10}\,{\text s}^{-1}\!\cdot\!{\text K}^{-1})\cdot 2700\,{\text K}$ $= 1.59 \times 10^{14}\,$Hz
• Note, that in these equations s$^{-1}$ becomes Hz, without the need to divide by $2\pi$.

#### Problem 8.2

The work function of potassium is $W_0=2.29\,$eV. What (if any) electron emission is observed when the blue ($\lambda=450\,$nm) or, alternatively, the red ($\lambda=750\,$nm) light strikes the metal surface?

Steps:

• If the frequency of the light is greater than $f_0$ (cut-off frequency), then the electron can leave the metal with finite (and positive) kinetic energy.
1. Convert the variables to SI units:
• $W_0=2.29\,{\text{eV}}\cdot 1.60\times 10^{-19}\frac{\text J}{\text{eV}}=$ $3.67\times 10^{-19}\,$J
• $\lambda_\text{blue}=4.50\times 10^{-7}\,$m
• $\lambda_\text{red}=7.50\times 10^{-7}\,$m
2. Calculate cut-off frequency for both red and blue light using the same work function value
• The cut-off frequency is the same for any wavelength, it only depends on the type of metal
• Equation: $\;f_0= \frac{W_0}{h}$
• Both red and blue Light:
• $f_0=\frac{3.67\times 10^{-19}\,{\text J}}{6.63\times 10^{-34}\,{\text J}\cdot{\text s}}$ $= 5.53\times 10^{14}\,$Hz
3. Convert the given wavelengths for red and blue light to frequency.
• Equation: $\;f= \frac{c}{\lambda}$
• Blue light: $\;f_\text{blue}=\frac{3.00\,\times\,10^8\,\frac{\text m}{\text s}}{4.50\,\times\,10^{-7}\,{\text m}}$ $= 6.66\times 10^{14}\,$Hz
• Red light: $\;f_\text{red}=\frac{3.00\,\times\,10^8\,\frac{\text m}{\text s}}{7.50\,\times\,10^{-7}\,{\text m}}$ $= 4.0\times 10^{14}\,$Hz
4. If calculated frequency is greater than cut-off frequency, electron emission will be observed
• The blue photon will be able to eject an electron because it exceeds the cut off frequency $(f_\text{blue}>f_0)$
• The red photon will not be able to eject any electrons, because its frequency is below the cutoff $(f_\text{red}<f_0)$
• The maximum kinetic energy of the electrons ejected by the blue light is
• $K_\text{max}=hf_\text{blue}\!-hf_0=$ $h(f_\text{blue}\!-f_0)=$ $6.63\times 10^{-34}\,{\text J}\cdot{\text s}\cdot(6.66\times 10^{14}-5.53\times 10^{14})\,{\text{Hz}}=$ $7.5\times 10^{-20}\,$J
• The same kinetic energy in eV units is about 0.5 eV

#### Problem 8.3

Find the maximum change in wavelength for blue light ($\lambda_b=450\,$nm) and for X-rays ($\lambda_X=0.1\,$nm) during Compton scattering off of electrons that are initially at rest. What percentage of the incident wavelength is this change in each case?

Steps:

1. The maximum change in the photon's wavelength occurs when it scatters in the reverse direction $(\theta=180^\circ)$, in which case the change is twice the Compton wavelength $\frac{h}{m_ec}$:
• $\Delta\lambda=\frac{h}{m_ec}\big(1-\cos 180^\circ\big)$ $= \frac{h}{m_ec}\big(1-(-1)\big)$ $= \frac{h}{m_ec}\cdot 2$ $=\frac{2h}{m_ec}$
• This change in $\lambda$ doesn't depend on the $\lambda$ itself, so it is the same for both wavelengths:
• $\Delta\lambda=$ $\frac{2h}{m_ec}=$ $4.86\times 10^{-12}\,$m
2. Percent change for blue light
• $\frac{\Delta\lambda}{\lambda_b}=$ $\frac{4.86\times 10^{-12}\,{\text m}}{4.5\times 10^{-7}\,{\text m}}=$ $1.08\times 10^{-5}$ $\approx 0.001\%$
3. Percent change for X-rays
• $\frac{\Delta\lambda}{\lambda_X}=$ $\frac{4.86\times 10^{-12}\,{\text m}}{1\times 10^{-10}\,{\text m}}=$ $4.86\times 10^{-2}$ $\approx 4.9\%$

#### Problem 8.4

In a classical model, an electron is orbiting a proton (the nucleus) in a hydrogen atom, with velocity $v=\sqrt{\frac{k_cq_e^2}{m_er}}$ $\approx 2.2\times 10^6\frac{\text m}{\text s}$, where $r=5.3\times 10^{-11}\,{\text m}$ is the radius of its orbit. Find the uncertainty of the electron's energy and momentum, and compare them to the electron's kinetic energy and momentum.

Steps:

1. Find the momentum from the known velocity and mass (since $v<0.01c$, no need to use Einstein's relativity)
• $\big|\,p\,\big|=m_ev=$ $9.11\times 10^{-31}\,{\text{kg}}\cdot 2.2\times 10^6\frac{\text m}{\text s}=$ $2.00\times 10^{-24}\frac{ {\text{kg}}\cdot{\text m}}{\text s}$
2. Find the kinetic energy of the electron
• $K=\frac{m_ev^2}{2}=$ $0.5\cdot 9.11\times 10^{-31}\,{\text{kg}}\cdot \left(2.2\times 10^6\frac{\text m}{\text s}\right)^2=$ $2.2\times 10^{-18}\,{\text J}$
3. Use the uncertainty principle to find $\Delta p_y$, assuming the uncertainty $\Delta y$ is the radius of the orbit
• $\Delta p_y=\frac{h}{2\pi\Delta y}$ $=\frac{6.63\,\times\,10^{-34}\,{\text J}\cdot {\text s}}{2\pi\,\cdot\,5.3\times 10^{-11}\,{\text m}}=$ $2.00\,\times\,10^{-24}\,\frac{ {\text{kg}}\cdot{\text m}}{\text s}$
4. Use the uncertainty principle to find $\Delta E$, assuming the uncertainty $\Delta t$ is the time it takes to travel 1 radian
• $\Delta t=\frac{r}{v}$ $=\frac{5.3\times 10^{-11}\,{\text m}}{2.2\times 10^6\frac{\text m}{\text s}}$ $=2.4\times 10^{-17}\,$s
• $\Delta E=\frac{h}{2\pi\Delta 4}$ $=\frac{6.63\,\times\,10^{-34}\,{\text J}\cdot {\text s}}{2\pi\,\cdot\,2.4\times 10^{-17}\,{\text s}}=$ $4.4\,\times\,10^{-18}\,{\text J}$
5. In this classical model, the uncertainty in momentum is equal to the momentum itself, and the uncertainty in energy is twice the value of kinetic energy itself

### Homework Questions Ch.30

#### Problem 58

An X-ray scattering from a free electron is observed to change its wavelength by 3.59 pm. Find the direction of propagation of the scattered electron, given that the incident X-ray has a wavelength of 0.530 nm and propagates in the positive x direction.

1. I need help on this homework question
1. From the Compton scattering formula, find the angle $\theta$ of the scattered photon.
2. Find the y component of the photon's momentum after scattering ($p_y'=p'\sin\theta$)
3. Using the momentum conservation along the y axis, find the electron's y-momentum $p_{ey}'$
4. Find the electron's kinetic energy $K$ from the energy conservation. Is this a relativistic case (compare to $m_ec^2$)?
5. Find the electron's total momentum $p_e'$ (after scattering) from its energy $K$
6. Find the electron's scattering angle from its y-component of momentum and the total momentum $\left(\sin\phi=\frac{p_{ey}'}{p_e'}\right)$