Exam 1 F09 Practice Exam Key

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Feb 20, 2024

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Print Last Name: ____________________ First Name: ____________________ 1 PLEASE NOTE: This practice exam for Fl09 is not the same as the original exam given in 2008, because the topics covered on the exam are slightly different this year in Chem 111A. The practice exam consists of questions covering representative topics, but not necessarily all of the subject matter discussed so far this semester.

Print Last Name: ____________________ First Name: ____________________ 2 Please Print First Name:_________________________ Please Print Last Name:________________________ Signature: ____________________________________ Student ID #: _________________________________ Chemistry 111A FIRST EXAM October 6, 2008 Professors Loomis/Frey 6:30 - 8:00 p.m. 1.5 HOURS will be allowed for the exam. No books, notes, other students, other students’ exams, or any other resource materials are allowed. No form of white-out may be used on the exam. The examination contains (8) questions on (17) numbered pages. Some of the questions contain more than one part. You must clearly and neatly show your work and your reasoning used to arrive at the correct answer in the space provided, in order to receive full credit. Clearly identify your final answers. Partial credit will be given for partially correct reasoning in support of incorrect or correct final answers. No credit will be given unless the answers are legibly written in the spaces provided. If you run out of space, use the scratch page at the end of the exam, you may not use your own scratch paper. Clearly indicate on the page where the problem is written where the continuation of your response is located. Do not write on the backside of the exam pages, they will not be graded. PLEASE WRITE YOUR NAME ON EACH PAGE OF THE EXAM. TURN IN THE ENTIRE EXAM WHEN YOU ARE FINISHED! REGRADE INFORMATION: Read the syllabus section on regrades on the general information sheet. Regrade requests must be placed in the wood cabinet labeled “Regrades,” by the racks on the 2 nd floor of the Lab Sciences Building. Attach the regrade form to the front of the exam stating which questions you would like regraded and the grading errors that have been made. DO NOT MAKE ANY MARKS OR CHANGES ON THE EXAM ITSELF. DO NOT WRITE ANYTHING ON THE EXAM! Regrade requests must be turned in by 4:30 PM on Monday, October 13, 2008. For Grading Only: 1. __________ (6 pts) 5. __________ (14 pts) 2. __________ (8 pts) 6. __________ (17 pts) 3. __________ (17 pts) 7. __________ (10 pts) 4. __________ (18 pts) 8. __________ (10 pts) Total Points: __________ (100 pts)

Print Last Name: ____________________ First Name: ____________________ 3 1. (6 Total pts.) The ancient Greek philosopher Democritus observed that the sand on the seashore when viewed from afar looks to be a single body, a “beach.” Therefore, so also may all matter be made up of tiny little granules of matter – the smallest of which he called atoms, which in Greek means, “uncuttable.” By virtue of the name itself, an atom in Democritus’ theory was the smallest thing in existence. a) (3 pts.) In two sentences or less, describe one way that Democritus’ observation is consistent with Dalton’s atomic theory of matter. Only the first two sentences will be graded. ( Grading: Overall, 2 pts for Dalton’s theory and 1 pt for beach analogy). According to Dalton, atoms are the smallest indivisible particles ( 2 pts ) _characteristic of a specific element, just like the grains of sand are the smallest indivisible pieces of a beach. A beach appears to be a single body from a distance, but upon closer examination, it is composed of multiple pieces (grains of sand) ( 1 pt ) . OR: Molecules and crystals (nonmolecular compounds) consist of smaller entities called atoms, which Dalton considered to be indivisible. ( 2 pts ) Similarly, a beach consists of smaller entities (sand particles) which Democritus considered to be indivisible. ( 1 pt ) b) (3 pts.) In two sentences or less, use Democritus’ observation to give one contradiction to Dalton’s atomic theory of matter. Only the first two sentences will be graded. (Grading: Overall, 1 pt for Dalton’s theory and 2 pts for beach analogy). According to Democritus, grains of sand are the smallest possible particles of the beach and the smallest possible granules of matter are atoms, “uncuttable”. But, we know that granules of sand are divisible into even smaller pieces, such as the elements silicon and oxygen. ( 2 pts ) Similarly, atoms are divisible into smaller pieces (protons, electrons, neutrons, or other subatomic particles), and are not the smallest particles in existence. ( 1 pt ) OR: Grains of sand are not identical (can be different sizes). ( 2 pts ) Similarly, isotopes have different masses and so not all atoms of a given element are identical. ( 1 pt ) OR: Something comparing Dalton’s postulate about atoms combining in fixed ratios ( 1 pt ) contrasted with a varying number of grains of sand used to compose a beach. ( 2 pts )

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Print Last Name: ____________________ First Name: ____________________ 4 2. (8 Total pts.) Your goal in this problem is to properly identify 1, 2, 3, and 4, shown in the apparatus, as either parallel electric field plates or magnetic poles so that the desired outcomes of an electron (with an initial velocity of v i ) hitting the detector would be achieved. In so doing, either 1 and 2 would form an electric field and 3 and 4 would form a magnetic field, or vice versa. Desired outcomes: (For both experiments the electric field is on .) – With the magnetic field turned off , the particle should hit the detector at location A. – With the magnetic field turned on , the particle should hit the detector at location B. a) (6 pts.) Please construct the apparatus that will achieve the desired outcomes (listed above) by putting the appropriate number (1, 2, 3, and 4) in the blanks below. Solution: Positive electric field plate ___ 3 ____ ( 1 pt ) Negative electric field plate ___ 4 ____ ( 1 pt ) North magnetic pole ___ 1 ____ ( 1 pt ) South magnetic pole ___ 2 ____ ( 1 pt ) b) (2 pts.) What can you say about the relative magnitude of the two forces when the particle hits location B? Soln: The magnitude of the force from the electric field is larger ( 2 pts , all or nothing) than that from the magnetic field. If the force from the magnetic field were stronger, the particle would hit the target on the other side of the (vertical) center dotted line. e - v i detector Same detector rotated 90° A B 1 2 3 4 Apparatus 1 pt for identifying the electric field plates as 3 and 4. 1 pt for identifying the magnetic poles as 1 and 2.

Print Last Name: ____________________ First Name: ____________________ 5 3. (17 Total pts.) In an alternate universe, on the planet Kellogg, the force of gravity is much greater than on Earth. It is observed that the same electric field strength used by Millikan on Earth is required to stop an oil drop with the same mass and number of electrical charges from falling. (Assume all other constants and equations on Earth are valid on Kellogg.) a) (3 pts.) What result is obtained from the experiments performed on Kellogg? Briefly justify your answer. Soln: The charge of an electron is determined on Kellogg by the Millikan experiment. The magnitude of the charge on an electron, q e will be greater in the alternate universe than in our universe. ( 1 pt ) We know that m oil g = q oil E holds in both universes, and the electric field and mass of an electron also remain the same, but gravity is greater. Therefore, as gravity increases, so must the charge on an electron (and q oil which is nq e ) because they are directly proportional. ( 2 pts ) b) (8 pts.) A scientist, Dr. Koko Krisp, studying the structure of the atom on Kellogg performed a Millikan experiment and obtained the following data. Based on these data, determine the number of electrons in each oil drop and the charge (in whatsits) of the electron in this universe. Measurement (whatists) Number of electrons 40.12 17 21.06 9 7.05 3 9.41 4 16.45 7 Charge of the electron on Kellogg = ____ 2.35 _(or 2.36 ) ____ whatsits ( 2 pts ) 40.12/7.05 = 5.70 x 3 = 17 40.12 - 21.06 = 19.06 and so 40.12/2.36 = 17 21.06/7.05 = 2.99 x 3 = 9 OR 21.06 - 16.45 = 4.61 and so 21.06/2.36 = 9 16.45/7.05 = 2.33 x 3 = 7 16.45 – 9.41 = 7.04 and so 16.45/2.36 = 7 9.41/7.05 = 1.33 x 3 = 4 9.41 – 7.05 = 2.36 and so 9.41/2.36 = 4 7.05/7.05 = 1.00 x 3 = 3 7.05/2.36 = 3 2.36 whatsits 40.12/17 = 2.36 21.06/9 = 2.34 16.45/7 = 2.35 9.41/4 = 2.35 7.05/3 = 2.35 q e= ૛.૜૟ା૛.૜૝ା૛.૜૞ା૛.૜૞ା૛.૜૞ ૛ = 2.35 whatsits 2 pts 2 pts 1 pt 1 pt for recognizing factor of 3 Alt soln: 2 pts 1 pt 1 pt for ID of smallest difference

Print Last Name: ____________________ First Name: ____________________ 6 c) (6 pts.) Assume the results of the Thomson experiment on planet Kellogg provide a q/m ratio of 5.62  10 27 whatsits/kg for a single electron. (If you were not able to calculate the charge on an electron in part b) of this problem, assume that q e = 3.71 whatsits.) i) (3 pts.) What is the ratio of the total mass of the H atom to the mass of the electron on Kellogg? Solution: m ு ௔௧௢௠ = 1.008 g mol ቀ 1 mol 6.022 x ଵ଴ మయ atoms ቁ ቀ 1 kg 1000 g ቁ = 1.67 x 10 ିଶ଻ kg for H atom ( 1pt ) m e = q e q e m e = 2.35 whatsits 5.62 x 10 27 whatsits/kg = 4.18 x 10 -28 kg ( 1pt ) Therefore, ୫ ಹ ೌ೟೚೘ m e = ଵ.଺଻ ୶ ଵ଴ షమళ ୩୥ 4.18 x 10 -28 kg ൌ ૝. ૙૙ ( 1pt ) Note: if student used 3.71 for charge, then m e = q e q e m e = 3.71 whatsits 5.62 x 10 27 whatsits/kg = 6.60 x 10 -28 kg ( 1pt ) ୫ ಹ ೌ೟೚೘ m e = ଵ.଺଻ ୶ ଵ଴ షమళ ୩୥ 6.60 x 10 -28 kg ൌ ૛. ૞ ( 1pt ) ii) (3 pts.) Compare this ratio on Kellogg to the ratio on Earth. Briefly explain what this difference tells us about the structure of the atom on Kellogg? Soln: Could show 4.00/2000 = 0.002 or say 500 times greater on Earth. (Or, show that m e = 25% (or 1/4) of mass (Kellogg) and m e = 0.05% (or 1/2000) mass (Earth)). Note: if student used 3.71 for the charge in part i, m e = 40% (or 2/5) and 2.5/2000 = 0.00125 or 800 times greater than ratio on Earth) This tells us that the atom’s composition differs substantially from our universe. ( 1 pt ) Significantly more of the mass of the atom is associated with the negative charge on Kellogg than on Earth. ( 1 pt ) While most of the mass of the atom still belongs to the positive charge, a much larger amount is contributed by the negative charge on Kellogg. ( 1 pt )

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Print Last Name: ____________________ First Name: ____________________ 7 4. (18 Total pts.) Answer each of the following questions. Circle either True or False for each of the following: ( 1 pt. each ) a) TRUE or FALSE A neutron with a wavelength of 2.00 nm does not have detectable wave character. b) TRUE or FALSE An atom may have only discrete electronic energy states; therefore, the electron in an atom may have only discrete kinetic energies. c) TRUE or FALSE The momentum of a photon decreases as its frequency increases. d) TRUE or FALSE The wavelength of an object is inversely proportional to its velocity. e) TRUE or FALSE All waves require a medium to propagate. f) TRUE or FALSE In the photoelectric-effect experiment, increasing the photon flux will cause a photocurrent to occur. Briefly answer the following question: g) (4 pts.) In three-four sentences, briefly describe how the Rutherford α -particle scattering experiment disproved the plum-pudding model. Soln: In the Plum-pudding model, the positive charge is equally spread throughout ( 1 pt ) the atom; hence the positive charge is small anywhere in the atom. In the Rutherford scattering experiment, only small scattering angles would be expected ( 1 pt ) for this model. In the actual Rutherford experiment, there were many small scattering angles AND a significant number of large scattering angles (even back-scattering) . ( 1 pt ) These results could only occur if all of the positive charge was concentrated in an area much smaller than the size of the atom. ( 1 pt )

Print Last Name: ____________________ First Name: ____________________ 8 1 pt 1 pt Fill in with the appropriate words to accurately complete the following sentences: h) ( 2 pts ) For non-classical waves, the square of the amplitude is proportional to the probability . i) ( 1 pt ) In the photoelectric effect experiment, when the incident light interacts with the cathode, the nature of the light is best described by a narrow (localized) wavepacket. j) ( 2 pts ) In atomic spectroscopy, the change in the energy of the atom for a transition from a higher lying state to a lower lying state is equal to the negative energy of the emitted photon . Calculate the following: k) (3 pts.) Calculate how many photons are generated in 1.00 hour from a 10.0 Watt, 600.0 nm lightbulb? Solution: J 3.31x10 m 1 nm 10 x nm 600.0 ) 10 Js)(2.998x (6.626x10 λ hc E 19 9 s m 8 34 p      J 36000 sec 1 J 10.0 x hr 1 sec 3600 x hr 1.00 E T   photons 1.09x10 J 3.31x10 J 36000 E E photons # 23 19 p T     1 pt 1 pt 1 pt 1 pt 1 pt

Print Last Name: ____________________ First Name: ____________________ 9 5. (14 Total pts.) The photoelectric effect enables photons to be used to generate electrons and electrical current. a) Image-intensified cameras can amplify light and build up a strong signal even when light levels are low. Essentially, a photon hits the photocathode, ejecting an electron (the photoelectric effect). The electron is accelerated from the photocathode by an electric field, bumping into the walls of the channel as it goes. Each time it strikes the wall, it knocks additional electrons free. The electrons strike a phosphor film, which emits photons in a 1:1 ratio to the electrons. See the figure below. photocathode phosphor film electric plate electric plate detector h  h  ' e i) (1 pt.) Label the direction of the electric field on the schematic. Soln: arrow pointing up, drawn above ii) (5 pts.) The ratio between the number of photons coming out the back end of the device and the number of photons striking the front of the sodium photocathode (  Na = 1.82 eV) is called the gain. This particular apparatus has a gain of 13,700. If the 125-W light source, with a wavelength of 515 nm, is turned on for 11.4 seconds, what is the number of electrons that strike the phosphor film? Please circle your answer. Soln: J 10 x 3.86 m 10 nm 1 nm 515 m/s) 10 x Js)(2.998 10 x (6.626 λ hc E 19 9 8 34 p             For Na, J 10 x 2.92 eV 1 J 10 x 1.6022 eV 1.82 φ 19 19             E p >  so electrons will be ejected and due to the gain, 1 photon provides 13,700 electrons Therefore, electrons 10 x 5.06 photon 1 electrons 13,700 J/photon 10 x 3.86 seconds) W)(11.4 (125 25 19                   E 1 pt 1 pt 1 pt 1 pt 1 pt for having E total /E photon 1 pt for multiplying by the gain

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Print Last Name: ____________________ First Name: ____________________ 10 s m J 2 1 pt 1 pt b) (2 pts.) Using a different light source, a total number of 7.25  10 24 electrons struck the phosphor film in a period of 10.0 seconds. The phosphor film emits at 615 nm. What is the power (in Watts) of the light hitting the detector? Soln: 7.25 x 10 24 electrons ቀ 1 photon 1 electron ቁ =7.25 x 10 24 photons E p = hc  = ൬ 6.626 x 10 -34 Js ൰ (2.998 x 10 8 ms -1 ) 615 x 10 -9 m =3.23 x 10 -19 J/photon (3.23 x 10 -19 J/photon)(7.25 x 10 24 photons) 10.0 seconds = 2.34 x 10 5 W c) (6 pts.) Qualitatively plot the photocurrent produced in a photoelectric cell as a function of wavelength for a constant power over the same area (ie, constant ). Label important features and slopes (if applicable) on your graph. Photocurrent +1 pt for identifying  max  0 Slope = (hc) -1   +1 pt +1 pt +1 pt for having a straight line +1 pt for it pointing to zero +1 pt for having a one- sided plot

Print Last Name: ____________________ First Name: ____________________ 11 + 1 pt + 2 pt for no extra lines + 1 pt + 1 pt + 1 pt + 1 pt + 2 pt for showing calc. once 6. (17 Total pts.) In this problem, estimate the behavior of an electron in a Be 3+ cation using the Bohr Model. a) (6 pts.) Determine the kinetic energy , E k , (in eV) of an electron in an orbit having a radius of 211.6 pm. Hint: at this radius . Soln: Z = 4 and a o = 52.9 pm r n =211.6 pm= ൬ n 2 Z ൰ a o so n 2 = ቀ 211.6 pm(4) 52.9 pm ቁ = 1 and therefore, n = 4 E n = -13.6 eV ൬ Z 2 n 2 ൰ = -13.6 eV ቀ 16 16 ቁ = -13.6 eV E n =V ሺ r ሻ +E k therefore E k = -13.6 eV + 27.22 eV = 13.6 eV b) (11 pts.) Calculate the emission spectrum of the Be 3+ cation with an electron initially in the orbit with r = 211.6 pm. Show all calculations and plot the line spectrum on the axes included below. Make sure to label each transition with n i and n f (n i  n f ). Soln: Since n =4, there will be 3 emission lines (4 → 1, 4 → 2 and 4 → 3). For emission, E photon = - ∆ E atom = - ቈെ2.18 ݔ 10 ିଵ଼ ܬሺ4 ଶ ሻ ቆ 1 ݊ ௙ ଶ െ 1 ݊ ௜ ଶ ቇ቉  = hc E photon = ൬ 6.626 x 10 -34 Js ൰ (2.998 x 10 8 ms -1 ) 3.29 x 10 -17 J ൭ 1 n f 2 - 1 n i 2 ൱  4 → 1 = 6.1 nm  4 → 2 = 30.4 nm  4 → 3 = 117 nm  = hc E photon = ൬ 6.626 x 10 -34 Js ൰ (2.998 x 10 8 ms -1 ) 3.29 x 10 -17 J ൬ 1- 1 4 2 ൰ ൬ 10 9 nm 1 m ൰ ൌ 6.1 nm 0 100 200 300 400 500 600 Wavelength (nm)        eV 22 . 27 pm 6 . 211 ε 4 4 ε 4 V 0 0 2 1        e e r q q r 4 → 1 4 → 2 4 → 3 + 1 pt + 2 pt + 1 pt + 1 pt for use of eqn in some way + 1 pt for each line and label – total of 3 pts If used n = 4 as final instead of initial, max of 5 pts total ( +2 for Δ E = E photon , +1 for calc. effort, +2 for plotting and labeling their lines

Print Last Name: ____________________ First Name: ____________________ 12 + 2 pts + 1 pt + 1 pt + 1 pt + 2 pts for using correct equation 7. (10 Total pts.) As discussed during lecture, particles can be considered as wavepackets comprised of matter waves. For this problem, consider the wavepacket shown below that represents the localization of an electron at a specific time, t 0 , and travelling with a most probable velocity, v  . a) (7 pts.) Use the Heisenberg Indeterminancy (Uncertainty) Principle to estimate the breadth of the wavepacket (in pm) after a time of 1.00 femtosecond (1.00  10 –15 s); ie , at a time of t 0 + 1.00 fs. Solution: From the wavepacket shown above, ∆ x = 137.5 pm - 87.5 pm = 50 pm ∆ x ∆ p ≥ h 4 π and so ∆ p ≥ h 4 π∆ x = 6.626 x 10 -34 Js 4 π (50 x 10 -12 m) = 1.05 x 10 -24 kg m/s ∆ v = 1.05 x 10 -24 kg m/s 9.11 x 10 -31 kg = 1.16 x 10 6 m/s ߂ݒ • ݀ݐ ൌ ሺ1.16 ݔ 10 ଺ ݉/ݏሻሺ1.0 ݔ 10 ିଵହ sሻ ൌ 1160 pm b) (3 pts.) Use the Copenhagen Interpretation to explain the fate of the electron wavepacket and the result of your estimate in part a). Solution: The wavepacket begins to broaden instantly after it is prepared (2 pts) , and 1 fs later it is 23 times its original width. It travels as a wave (1 pt) , as the Copenhagen Interpretation describes. At one instant you can measure very accurately the position of a particle (localized wavepacket), but then it quickly travels (or evolves) as a wave.

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Print Last Name: ____________________ First Name: ____________________ 13 8. (10 total pts) A beam of hydrogen atoms traveling at 3.00  10 3 m s − 1 passes through a single slit to give the diffraction pattern partially sketched below. The position of the first minimum in any single-slit diffraction pattern is given by the relationship sin  = C  (where C is a constant dependent on the size of the slit)  Calculate where the first minimum would lie if the experiment were repeated with a beam of helium atoms traveling at the same speed. Solution 1: Need to first determine the wavelength to determine the constant of proportionality. Using the de Broglie relationship: λ H = h/p = h/mv = 6.626  10 − 34 Js /[(1.67  10 − 27 kg)(3.00  10 3 ms − 1 )] = 1.32  10 − 10 m (2 pts.) [mass of H atom = 1.008 g/mol = 1.008g/mol  (1kg/1000g)/(6.022  10 23 mol − 1 ) = 1.67  10 − 27 kg] (2 pts.) C = (sin θ )/ λ H C = sin (19.4  ) / 1.33  10 − 10 m = 2.51  10 9 m − 1 (2 pts.) The momentum of the helium is four times that of the hydrogen therefore the wavelength is ¼ that of hydrogen, m He v = 4m H v λ He = h/4m H v = ¼ λ H λ He = 3.31  10 − 11 m (2 pts.) [or equivalent calculations to find momentum and wavelength] sin θ = (2.51  10 9 m − 1 )(3.31  10 − 11 m) = 0.083 θ = 4.76  (2 pts.) There is an alternative (and neater solution) to this problem: Solution 2: C = sin θ H / λ H = sin θ He / λ He (2 pts.) sin θ H m H v H /h = sin θ He m He v He /h (2 pts.) Therefore: sin θ H m H = sin θ He m He (2 pts.) m He = 4m H sin θ H / sin θ He = 4 (2 pts.) sin θ He = (1/4) sin 19.4 = 0.083 θ He = 4.8  (2 pts.) Intensity Diffraction Angle,  (degrees) 0 19.4 -19.4

Print Last Name: ____________________ First Name: ____________________ 14 +1 pt for concept that E K = +2 pts for this +2 pts for this concept NOTE: This topic was not included on Exam 1 in 2008 but will be covered on Exam 1 this year. 2. (11 Total pts.) a) (6 pts.) An electron confined in a one-dimensional box is in the first-excited state with a speed of 1.29  10 6 m/s. Determine the length of the box in Angstroms (Å). Solution:     J s m kg mv E E k 19 2 6 31 2 2 10 58 . 7 10 29 . 1 10 11 . 9 2 1 2 1                2 31 2 2 34 2 2 2 19 2 10 11 . 9 8 2 10 626 . 6 8 10 58 . 7 L kg Js L m n h E e                 2 19 19 31 2 2 34 2 2 2 2 10 18 . 3 10 58 . 7 10 11 . 9 8 2 10 626 . 6 8 m J kg Js E m n h L e            L = 5.64  10 –10 m = 5.64 Å b) (5 pts.) Determine the wavelength (in nm) of a photon necessary to promote the electron in the box described in part a) to the fifth energy level. If you did not get the length of the box in part a), assume a length of 3.0 Å. Solution:        J m kg Js mL n h E 18 2 10 31 2 2 34 2 2 2 5 10 75 . 4 10 64 . 5 10 11 . 9 8 5 10 626 . 6 8             E = E f – E i = E 5 – E 2 = 4.75  10 –18 J – 7.58  10 –19 J = 3.99  10 –18 J     m J s m Js E hc 8 18 8 34 10 98 . 4 10 99 . 3 10 998 . 2 10 626 . 6            = 49.8 nm ALTERNATE SOLUTION USING L = 3.0 Å        J m kg Js mL n h E 17 2 10 31 2 2 34 2 2 2 5 10 67 . 1 10 0 . 3 10 11 . 9 8 5 10 626 . 6 8                   J m kg Js mL n h E 18 2 10 31 2 2 34 2 2 2 2 10 67 . 2 10 0 . 3 10 11 . 9 8 2 10 626 . 6 8             E = E f – E i = E 5 – E 2 = 1.67  10 –17 J – 2.67  10 –18 J = 1.4  10 –17 J     m 10 41 . 1 J 10 4 . 1 s m 10 998 . 2 Js 10 626 . 6 E hc 8 17 8 34             = 14.1 nm +2 pts +1 pt +1 pt for calculations +1 pt for using correct n (= 2) +1 pt for using correct n (= 5) +1 pt for using correct n (= 5) +1 pt +1 pt +1 pt +1 pt

Print Last Name: ____________________ First Name: ____________________ 15 Blank Scratch Page

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Print Last Name: ____________________ First Name: ____________________ 16           L x n L x n  sin 2 Formulas, Constants, Conversion Factors, etc. vector quantities are given in bold F = m a v = d x /dt (if v is constant) a =  v /  t (if a is constant) F = m g F = q E F = q v x B V( r ) = r q q 0 4 2 1  E K = m v 2 /2 E = h  speed =  c =  E K = h  –  E n = (–2.18  10 –18 J) Z 2 /n 2 = (–13.6 eV) Z 2 /n 2 p = h /  p = m v  = h /( m v) (  p)(  x)  h/(4  ) d sin  = m  D sin  = m  D sin  = (m+ ½)  L = n  /2 2 2 2 8 E mL h n n  1 Å = 10 –10 m e n m e h a Z a n r 2 2 0 0 0 2    

Print Last Name: ____________________ First Name: ____________________ 17 Formulas, Constants, Conversion Factors, etc. Table B-2 Derived Units in Sl: Quantity Unit Symbol Definition Energy joule J kg m 2 s –2 Force newton N kg m s –2 = J m –1 Power watt W kg m 2 s –3 = J s –1 Pressure pascal Pa kg m –1 s –2 = N m –2 Electric charge coulomb C A s Electric potential difference volt V kg m 2 s –3 A –1 = J C –1 Table B-3 Prefixes in Sl: Fraction Prefix Symbol Factor Prefix Symbol 10 –1 deci- d 10 deca- da 10 –2 centi- c 10 2 hecto- h 10 –3 milli- m 10 3 kilo- k 10 –6 micro-  10 6 mega- M 10 –9 nano- n 10 9 giga- G 10 –12 pico- p 10 12 tera- T 10 –15 femto- f 10 –18 atto- a

Print Last Name: ____________________ First Name: ____________________ 18 Formulas, Constants, Conversion Factors, etc. Physical Constants: Standard acceleration of terrestrial gravity g = 9.80665 m s –2 (exactly) Avogadro’s number N 0 = 6.022137  10 23 Bohr radius a 0 = 0.52917725 Å = 5.2917725  10 –11 m Boltzmann’s constant k B = 1.38066  10 –23 J K –1 Electron charge e = 1.6021773  10 –19 C Faraday constant F = 96,485.31 C mol –1 Masses of fundamental particles: Electron Proton Neutron m e = 9.109390  10 –31 kg = 0.00054857990 u m p = 1.67263  10 –27 kg = 1.00727647 u m n = 1.674929  10 –27 kg = 1.00866490 u Ratio of proton mass to electron mass m p / m e = 1836.15270 Permittivity of vacuum  0 = 8.8541878  10 –12 C 2 J –1 m –1 Planck’s constant h = 6.626076  10 –34 J s Speed of light in a vacuum c = 2.99792458  10 8 m s –1 (exactly) Universal gas constant R = 8.31451 J mol –1 K –1 = 0.0820578 L atm mol –1 K –1 Conversion Factors: Standard atmosphere 1 atm = 1.01325  10 5 Pa = 1.01325  10 5 kg m –1 s –2 (exactly) Atomic mass unit 1 u = 1.660540  10 –27 kg 1 u = 1.492419  10 –10 J = 931.4943 MeV (energy equivalent from E = mc 2 ) Calorie 1 cal = 4.184 J (exactly) Electron volt 1 eV = 1.6021773  10 –19 J = 96.48531 kJ mol –1 Foot 1 ft = 12 in = 0.3048 m (exactly) Gallon (U.S.) 1 gallon = 4 quarts = 3.78541 L (exactly) Liter-atmosphere 1 L atm = 101.325 J (exactly) Metric ton 1 metric ton = 1000 kg (exactly) Pound 1 lb = 16 oz = 0.45359237 kg (exactly)

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