final_exam_review

* This was given as the final exam by Prof. Kuzma in 2014*.

Solutions are available on D2L.

Physics 203 T-10 Exam 3 - Final | Name _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ |

**PLEASE PRINT YOUR NAME ABOVE!!!**- Please put away everything except:
- this exam and your Scantron form (use back sides of these pages for drafts you don’t want to be graded).
- one double-sided sheet of notes (formulas, etc.) that you prepared for this exam
- calculator (must have no connection to internet and no text/picture storage…)
- pen/pencil/eraser
- clock/watch (no connection to internet, no memory functions)

- Please make sure you are not sitting next to someone else, unless there is no such seats available.
- No exchange of information between students is allowed during the exam!
- Please do not open this booklet until you are instructed to do so. Please do not unstaple this booklet!
- Out of 4 problems, please cross one out, and attempt the other 3.
**New in 2015**: Failure to cross out a problem will cause the grader to cross one out for you randomly.

- For full credit, show your work!
- If you need to take a quick bathroom break:
- please leave this booklet with one of the proctors on your way out
- pick it up on your way back in.

- If you have a question:
- please read the problem carefully again
- If you think some information is missing, please walk down to a proctor with your question.

- Please hand in your finished exam to one of the proctors on your way out.

- speed of light $c$:
- $c_{\text{vacuum}} = 3 \times 10^8$ m/s

- $G = 6.67 \times 10^{-11}\,\frac{{\text N}\cdot{\text m}^2}{{\text{kg}}^2}$
- $h = 6.626 \times 10^{-34}\;{\text J}\cdot{\text s}$
- $N_A = 6.02 \times 10^{23}\;{\text{mol}}^{-1}$
- electron rest mass:
- $m_0 = 9.11 \times 10^{-31}$ kg

- 1 year = $3.156 \times 10^7$ s
- 1 eV = $1.602 \times 10^{-19}$ J
- $1^\circ=\frac{\pi}{180}\,{\text{rad}}$

Part of the exam | Score | …out of maximum |
---|---|---|

Question 1 | 2.5 | |

Question 2 | 2.5 | |

Question 3 | 2.5 | |

Question 4 | 2.5 | |

Question 5 | 2.5 | |

Question 6 | 2.5 | |

Question 7 | 2.5 | |

Question 8 | 2.5 | |

Question 9 | 2.5 | |

Question 10 | 2.5 | |

Problem 1 | 25 | |

Problem 2 | 25 | |

Problem 3 | 25 | |

Problem 4 | 25 | |

Total: | 100 |

A monochromatic beam of light of wavelength 750 nm is incident normally on a diffraction grating with a slit spacing of 3.0 μm. What is the angle for the 2^{nd} order maximum?

- [….] A) $\;\;0$
- [….] B) $\;\;14.5^\circ\;\;(\,0.253\;{\text{rad}}\,)$
- [….] C) $\;\;30^\circ\;\;(\,{\text{or}}\;\frac{\pi}{6}=0.524\;{\text{rad}}\,)$
- [….] D) $\;\;60^\circ\;\;(\,{\text{or}}\;\frac{\pi}{3}=1.05\;{\text{rad}}\,)$
- [….] E) $\;\;90^\circ\;\;(\,{\text{or}}\;\frac{\pi}{2}=1.57\;{\text{rad}}\,)$

A certain astronomical telescope has a diameter of 1.20 m. According to Rayleigh criterion, what is the minimum angle of resolution for this telescope at a wavelength of 550 nm?

- [….] A) $\;\;4.58 \times 10^{-7}$ rad
- [….] B) $\;\;5.59 \times 10^{-7}$ rad
- [….] C) $\;\;6.71 \times 10^{-7}$ rad
- [….] D) $\;\;2.18 \times 10^{6}$ rad
- [….] E) $\;\;2.66 \times 10^{6}$ rad

Two spaceships approach the Earth from the same direction. One has a speed of $0.33\,c$ and the other a speed of $0.44\,c$, both relative to Earth. What is the magnitude of the speed of the faster spaceship relative to the slower?

- [….] A) $\;\;0.096\,c$
- [….] B) $\;\;0.11\,c$
- [….] C) $\;\;0.13\,c$
- [….] D) $\;\;0.67\,c$
- [….] E) $\;\;0.90\,c$

Dr. Kuzma,
For this problem I was a bit confused in class as to which formula we used. I thought since they are approaching earth from the same direction and we want to find a speed relative to the other that we would use the subtraction of velocities? Could you please clarify, thank you!

— *Nichole K.*

- There are two ways of solving it:
- A standard way, where $v_1$ is the 1
^{st}ship's velocity, $v_2$ is the*unknown*2^{nd}ship's velocity relative to the 1^{st}ship, and $v_3$ is the 2^{nd}ship's velocity relative to Earth:- $v_3=\frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}$
- Solve this equation for $v_2$:
- $v_2=\frac{c^2(v_3-v_1)}{c^2-v_1v3}$ $=\frac{v_3-v_1}{1-\frac{v_1v_3}{c^2}}$ $\approx 0.129\,c$

- Using the
*subtraction*of velocities:- Since the ships move in the same direction, their
*relative*speed is the “difference” between their ground-based speeds:- Use the standard formula $v_3=\frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}$
- but plug in
- the positive value $v_1=0.44\,c$ for the faster ship's ground speed
- a
*negative*value for the slower ship's ground speed, $v_2=-0.33\,c$

- $v_3=\frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}=$ $\frac{0.44\,c-0.33\,c}{1-(0.44\cdot 0.33)}$ $\approx 0.129\,c$

— *Prof. Nicholas Kuzma 2015/06/06 15:31*

Light of wavelength 733 nm is incident on a single slit 0.5 mm wide. At what distance from the slit should a screen be placed if the second dark fringe in the diffraction pattern (counting from the central bright spot) is to be 2.0 mm from the center of the screen?

- [….] A) $\;\;0.55$ m
- [….] B) $\;\;0.68$ m
- [….] C) $\;\;0.91$ m
- [….] D) $\;\;1.36$ m
- [….] E) $\;\;2.72$ m

You are moving at a speed $0.33\,c$ relative to George, and George shines a light toward you. At what speed do you see the light passing you by?

- [….] A) $\;\;0.33\,c$
- [….] B) $\;\;0.67\,c$
- [….] C) $\;\;c$
- [….] D) $\;\;1.33\,c$
- [….] E) $\;\;$It depends on whether you are moving away or towards George

The only nuclear power plant in Oregon (closed down in 1992) generated approximately $1.0\times 10^{14}$ J of energy per day. How much matter was converted into energy to produce this yield?

- [….] A) $\;\;1.1$ g
- [….] B) $\;\;13$ g
- [….] C) $\;\;23$ g
- [….] D) $\;\;450$ g
- [….] E) $\;\;1.1$ kg

The wavelength of a light beam is doubled. Which one of the following is correct for the momentum of the photons in that beam of light?

- [….] A) $\;\;$It stays the same
- [….] B) $\;\;$It is quadrupled
- [….] C) $\;\;$It is doubled
- [….] D) $\;\;$It is halved
- [….] E) $\;\;$It is reduced by one-fourth

The wavelength of light in a thin film is 420 nm and the wavelength of the same light in vacuum is 560 nm. What is the index of refraction for the film?

- [….] A) $\;\;0.75$
- [….] B) $\;\;1.00$
- [….] C) $\;\;1.33$
- [….] D) $\;\;1.41$
- [….] E) $\;\;1.50$

An electron is moving with the speed of 3000 m/s. What is its de Broglie wavelength?

- [….] A) $\;\;2.21 \times 10^{-42}$ m
- [….] B) $\;\;7.27 \times 10^{-4}$ m
- [….] C) $\;\;2.2 \times 10^{-37}$ m
- [….] D) $\;\;1.11 \times 10^{-12}$ m
- [….] E) $\;\;2.42 \times 10^{-7}$ m

If the distance between the slits in Young's two-slit experiment is increased, which one of the following statements is true of the interference pattern?

- [….] A) $\;\;$The distance between the maxima decreases
- [….] B) $\;\;$The distance between the maxima stays the same
- [….] C) $\;\;$The distance between the minima increases
- [….] D) $\;\;$The distance between the minima stays the same
- [….] E) $\;\;$Impossible to tell without knowing the wavelength of light in use

A small vertical grating consists of many long, parallel horizontal slits, each spaced a vertical distance $d = 2\,\mu$m apart from the neighboring slits. It is illuminated with a horizontal $\lambda = 510$ nm light beam from behind, and the light is scattered forward at a vertical screen $L = 4$ m away.

**You may specify angles either in degrees or radians, but make sure to indicate your units!**

**A**) [2 pt] At the central point on the screen, nearest to the grating ($\theta = 0$), is there a dark or a bright spot?- …

**B**) [4 pt] Calculate the angle $\theta_1$ of the first bright maximum above the central fringe- …

**C**) [5 pt] What is the distance (in meters) from the 1^{st}bright maximum to the central fringe on the screen?- …

**D**) [3 pt] Calculate the angle $\theta_2$ of the 2^{nd}bright maximum above the central fringe- …

**E**) [3 pt] Calculate the angle $\theta_3$ of the 3^{rd}bright maximum above the central fringe- …

**F**) [5 pt] Is the distance (in meters) between the 3^{rd}and the 2^{nd}bright spots on the screen the same as

the distance calculated in part C)? Either calculate the distance, or give a valid argument.- …

**G**) [4 pt] What is the total number of bright fringes on the screen?

(*Hints:*assume the screen is infinite in size and do not forget to count the possible fringes at and/or below the central fringe location)- …

**Q**: In part**C)**, during the review session and on the answer sheet posted online we are using the value of 0.258 for this equation. I'm not understanding why we wouldn't use 0.255, the answer we go in part**B)**.**A**: 0.255 is the $\sin\theta_1$, which corresponds to $\theta_1=0.258$ rad. For small angles $|\theta|<0.05$ we can use the small-angle approximation $\sin\theta\sim\tan\theta\sim\theta$, but for $\sin^{-1}0.255$ it doesn't work that well any more. —*Prof. Nicholas Kuzma 2015/06/07 15:56*

IK Pegasi (or HR 8210) is a star system in the constellation Pegasus. Just 150 light years away from us, it is the nearest candidate for a Supernova explosion, with potentially deadly consequences for life on Earth. You are preparing an expedition to IK Pegasi to measure critical star parameters there.

**A**) [2 pt] Knowing that sunlight takes 150.0 years to reach IK Pegasi, calculate how far it is (in meters). Assuming the speed of IK Pegasi relative to Earth is negligible for our purposes, is this distance between the Earth and IK Pegasi a proper distance?- …

**B**) [3 pt] If the spaceship you design can travel with the speed of $0.995\,c$, how long (according to Earth clocks) would it take the spaceship to reach IK Pegasi?- …

**C**) [5 pt] In the spaceship frame of reference, both Earth and IK Pegasi seem to be moving relative to the spaceship at $0.995\,c$ in the opposite direction. What is the distance between the Earth and IK Pegasi measured in the spaceship frame?- …

**D**) [5 pt] According to the spaceship clock, how long is the one-way flight from Earth to IK Pegasi?- …

**E**) [5 pt] One of the crew is an identical twin (the other twin will remain on Earth). The ship departs Earth in January 2030, when both twins are 30 years old, and spends 5 Earth years exploring IK Pegasi system after arriving there. At that point, the spaceship goes back to Earth at the same speed. Calculate the biological age of the astronaut twin when (s)he returns on Earth.- …

**F**) [5 pt] Calculate the calendar year in which the ship returns. If human longevity remains roughly the same as today, is the other twin likely to be alive to greet his or her sibling upon arrival?- …

**Q**: ?**A**: …

After several years, the radioactive contamination from the Chernobyl nuclear disaster was dominated by the longer-lived isotopes cesium-137 and strontium-90. The latter (^{90}Sr) has a half-life of 29 years, and spontaneously decays into a highly radioactive ^{90}Y (half-life 64 hours) and an electron. The kinetic energy released in this decay is 0.546 MeV.

**A**) [1 pt] Convert to SI units the energy of the^{90}Sr initial decay (*Hint*: see page 1 of this exam)- …

**B**) [2 pt] Each mole of^{90}Y weighs 90 g and contains $N_A$ atoms. Calculate the mass of^{90}Y atom in kg.- …

**C**) [1 pt] The electron is released with some unknown velocity $v$. In terms of $v$ and the electron rest mass $m_0$, what is the expression for the kinetic energy of the released electron (formula only)- …

**D**) [6 pt] Neglecting the kinetic energy of the much heavier^{90}Y, equate the kinetic energy (part**C**)

to the energy released in the decay (part**A**) and solve for $v$. What is the velocity of the released electron

(you can express it either in SI units, or as a fraction of the speed of light)- …

**E**) [4 pt] What is the total mass of the released electron compared to its rest mass? (*Hint*: if you are stuck, you may use $E=mc^2$ formula to convert the total energy of the electron to its total mass.)- …

**F**) [3 pt] What is the momentum of the released electron?- …

**G**) [4 pt] Equate the magnitude of the electron momentum found in**F**) to the magnitude of the unknown recoil momentum of^{90}Y. Find the recoil velocity of^{90}Y. (*Hint*: you can use classical (Newtonian) formulas, no need to use relativistic (Einstein) equations for^{90}Y)- …

**H**) [4 pt] Are the assumptions of part**D**) – neglecting the kinetic energy of^{90}Y, and part**G**) – using non-relativistic equations for^{90}Y justified? Support your answers with numerical comparisons or order-of-magnitude error estimates.- …

**Q**: I am a bit confused how to solve part**D)**. I understand that the mass of the electron and $c^2$ are already known values but how are we able to solve for $v$?**A**: in part C) you are supposed to write down the formula for kinetic energy- $E_\text{kin}=mc^2-m_0c^2=$ $\frac{m_0c^2}{\sqrt{1-v^2/c^2}}-m_0c^2$ $=m_0c^2\left(\frac{1}{\sqrt{1-v^2/c^2}}-1\right)$

- in part A) you are supposed to find the total released energy by converting units:
- $E_\text{tot}=8.748\times 10^{-14}$ J

- Equating the total energy to the kinetic energy of the electron, and plugging $m_0$ and $c$, the only remaining unknown is $v$:
- $8.748\times 10^{-14}\,{\text J}=$ $\left(9.11\times 10^{-31}\,{\text{kg}}\right)\left(3\times 10^8\,\frac{\text m}{\text s}\right)^2\left(\frac{1}{\sqrt{1-v^2/\left(3\times 10^8\,\frac{\text m}{\text s}\right)^2}}-1\right)$

- Luckily, $v$ enters this equation only in one place, so it is possible to invert all the functions and find $v$
- The steps are outlined here:
- Divide both sides by $m_0c^2$
- Add 1 to both sides
- Take an inverse ($1/x$) of both sides
- Square both sides
- Subtract both sides from 1

- Etc.

**Q**: I am not sure how to approach part**H)**. It says to support your answers with numerical comparisons or order of magnitude error estimates. I am not certain whether the question is asking us to compare our answer to part**D)**with our answer to part**G)**or if we are to compare something else. I am a little confused. —*Joshua Vandehey 2015/06/06 18:38***A**:I think in a nut shell this question is asking us to compare orders of magnitude of masses and speeds. For the part**D)**we assumed that all energy released during decay went into kinetic energy of an electron. In reality, both particles were “pushed apart” with some velocities inversely proportional to their masses^{1)}. Since mass of an electron is 6 orders of magnitude smaller than mass of proton Y, we can simplify the calculations by assuming that only electron was “pushed away” while proton remain stationary (hence with zero kinetic energy). If I remember correctly, our rule of thumb was 3 orders of magnitude (equal to about 0.1% accuracy): if the difference is greater than that, we can make the simplifying assumption. We can approach the question about part**G)**with similar logic: speed of proton Y was calculated to be about 3313 m/s, which is about 5 orders of magnitude smaller than the speed of light. Since its speed is so small in comparison to*c*, we can use classical formula for momentum. —*Yulia Kotlyar 2015/06/07 14:03*

**Q**: Dr. Kuzma, on the practice exam, Problem 3, part**F)**is confusing me. The answer key shows that the relative momentum equation is used yet the correct answer is found by using $p=mv$ only. Why don't you divide the $mv$ by the $\sqrt{1-(v^2/c^2)}$? Thanks,

—*Thomas Ruttger 2015/06/06 15:14***A1**: Not sure what makes you think so. The velocity found earlier was $v=0.875\,c$. Dividing $m_0v$ by the $\sqrt{1-(v^2/c^2)}$, I am getting the correct answer here —*Prof. Nicholas Kuzma 2015/06/07 15:12***A2**: Are you confused because we use the relativistic mass $m=m_0/\sqrt{1-(v^2/c^2)}$? Basically, you need to either use the rest mass $m_0v$ and divide by $\sqrt{1-(v^2/c^2)}$, or just use the relativistic mass times velocity, simply $mv$. Then $m$ already contains the division of $m_0$ by $\sqrt{1-(v^2/c^2)}$ —*Prof. Nicholas Kuzma 2015/06/07 15:18*

**Q**: Dr. Kuzma, can you explain the reasons why each assumption is valid in problem 3 part**H)**of the practice exam. I understand the math but I am not sure under what conditions it is okay to use classical conditions over relativist conditions. Thanks,

—*Thomas Ruttger 2015/06/06 15:22***A**: See another student's question #2 above. Basically, these simplifying assumptions introduce error into the calculations. But in the limits of $m_{(^{90}\text{Y})} \gg m_0$ and $v_{(^{90}\text{Y})}\ll c$, the errors become so small that it's ok to use simplifications at this point.

—*Prof. Nicholas Kuzma 2015/06/07 15:36*

An X-ray (wavelength $\lambda = 2.0000\times 10^{-10}$ m) scatters off of a stationary electron. The scattered X-ray is moving perpendicular to the original X-ray direction ($\theta = 90^\circ = \frac{\pi}{2}$)

**A**) [4 pt] Find the change in X-ray wavelength during scattering ($\Delta\lambda$).- …

**B**) [2 pt] What is the exact wavelength $\lambda'$ of the scattered X-ray? Is it longer or shorter than $\lambda$ ?- …

**C**) [4 pt] What is the energy of the incoming X-ray? What is the energy of the scattered X-ray? Give your answers to 5 significant figures.- …

**D**) [4 pt] Equate the energy difference between the answers in**C**) to the kinetic energy of the scattered electron $\frac{1}{2}m_0v^2$. Solve for the speed of the scattered electron. Is using a non-relativistic formula justified?- …

**E**) [3 pt] Find the momentum of the scattered X-ray from the wavelength you found in**B**).- …

**F**) [1 pt] Equate the magnitude of the momentum in**E**) to the magnitude of the $y$ component of the electron momentum $p_{ey}'$ after scattering. Find $p_{ey}'$.- …

**G**) [4 pt] Find the magnitude of the election's total momentum $\left| p_e'\right|$ from the speed found in**D**)- …

**H**) [3 pt] From the magnitude (part**G**) and the $y$-component (part**F**) of the electron's momentum find the electron scattering angle

(Hint: $\sin\phi=\frac{p'_{ey}}{\left|p'_e\right|}$)

**Q**: I just wanted to clarify something about part**D)**. When it asks “is using a non-relativistic formula justified”, is it simply because using the relativistic formula would yield an answer similar to the non-relativistic formula with an accuracy greater than 3 decimal places? Or is there another explanation as to why it is justified? —*Joshua Vandehey 2015/06/06 19:24***A**: Again, I approached this question as an order of magnitude question: the speed of the scattered electron in only 2 orders of magnitude different from the speed of light. Therefore, using non-relativistic formula would be accurate down to several percent. This is probably not accurate enough for any real-world application, and if I would be answering this question on the exam, I would put “No”, since our rule of thumb is difference by 3 orders of magnitude or greater. You are, correct though: using relativistic formula would produce more accurate results. —*Yulia Kotlyar 2015 14:32***A2**: Yes, using a non-relativistic formula always introduces some error, and the question is how much precision is required. Also remember, that the correction factor $1/\sqrt{1-v^2/c^2}\approx 1+0.5\cdot(v/c)^2$ gives a correction on the order of $(v/c)^2$, so at the speed of a few percent of the speed of light the corrections might be on the order of part per thousand and still ok.

—*Prof. Nicholas Kuzma 2015/06/07 15:41*

final_exam_review.txt · Last modified: 2015/06/07 22:58 by wikimanager

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