1.) Which of the following are high potential energy arrangements for two electric charges? (3 correct responses!)  (Ans: Remember that it takes work to pull attractive objects apart or to push repulsive objects together!)
        a.)     -                         -          (two negative charges far apart)
        b.)    +                         +          (two positive charges far apart)
        c.)     ++                                 (two positive charges close together) [Took work to push these together]
        d.)     +-                                  (a positive charge near a negative charge)

        e.)     +                        -           (a positive charge far from a negative charge)[Took work to pull these apart]
        f.)     --                                     (two negative charges close together) [Took work to push these together]

2.) Give a brief description of the entire oxygen atom.  You might want to first discuss the nucleus (size and composition), and then the whole atom (size and composition).  Words like electron, proton, neutron, positive, negative, neutral, and sizes like 10^-10 m or 10^-14 m, could conceivably creep into your description. (Ans: On reserve in the Swain West Library for P120, Romer's, Energy: An Introduction to Physics, p 567 has a nice description of atoms and nuclei.)
[Ans: The carbon nucleus has a diameter of about 10^-14 m (or 10 Fermi's) and must have 8 protons.  The light stable nuclei like Oxygen-16 have equal numbers of protons and neutrons. Here O-16 would have 8 neutrons while the nucleus O-18 has 10 neutrons.  A neutral oxygen atom has a diameter of about 10^-10 m and must have 8 orbiting electrons.  If the oxygen atom has more or less than 8 electrons it is called a oxygen ion.]

3.) Almost all of the mass of the oxygen atom you described in problem 2 is in its nucleus? Explain why. (Ans: Masses are discussed in R&K, p 173. The key lies in discovering how much an electron weighs compared to a proton or neutron ) [Ans: m_proton = 1836 m_electron, m_neutron = 1838 m_electron.  So, about 99.9% of the mass of an atom is in its nucleus.]

4.) An atom has a certain number of protons a certain number of neutrons in its nucleus and a certain number of electrons in orbit around the nucleus.  Which of these numbers determines the name of the element?  Try out your answer on the two isotopes of Carbon, stable ¹²₆C₆ and radioactive ¹⁴₆C₈.  Explain how your answer does the job. [Hint: See page 174 in R & K] [Ans: The name is determined only by the number of protons in the nucleus here 6 in each case.]

5.) Consider the radioactive Uranium-235 nucleus, ²³⁵U.  How many protons and how many neutrons does it contain?  [Ans: You need a periodic table or a chemistry friend who knows which element Uranium is.  Hydrogen has 1 proton, helium has 2 protons, ...] [Ans: Uranium is the ninety-second element with atomic number 92, therefore 92 protons in its nucleus.  The number of neutrons must then be 143 to give 235 total "nucleons."]

6.)Neutrons bombard uranium (²³⁹₉₄Pu).  If a neutron and this Plutonium nucleus stick together to form a new nucleus, the new nucleus must be:
        a.) ²⁴⁰₉₄Pu₁₄₆        b.) ²³⁹₉₄Pu₁₄₅        c.)²³⁹₉₅Pu₁₄₄        d.) ²⁴⁰₉₄Am₁₄₆        e.) ²⁴⁰₉₅Am₁₄₇
[Hint: This problem has two parts.  First, you must decide on the number of protons and neutrons in the new nucleus, and then you must use the naming convention of problem 4 to figure out what to call it.] [Ans: Adding a neutron changes 239 to 240.  Since the number of protons is not affected and is still 94, the name must still be Pu (plutonium).]

7.) Consider a nuclear reaction where the total rest mass before the reaction is greater than the total rest mass after.  Do you have to supply energy to make this reaction happen (endothermic) or does the reaction release energy when it takes place (exothermic)?  [Hint: Einstein's DE = Dmc² where DE = E_after - E_before and Dm = m_after - m_before]. [Ans: The reaction is exothermic because initial mass lost must be converted into energy.]

8.) Consider the possible forces between a proton and another proton that are very close together.  Three candidate forces, GRAVITIONAL, ELECTRIC, and STRONG.  Between the proton and proton these three forces are respectively:
        a.) attractive, repulsive, zero
        b.) zero, zero, attractive
        c.) attractive, zero, attractive
        d.) attractive, repulsive, attractive
        e.) zero, zero, repulsive
[Hint: It's the mass of two objects that insures that they are gravitationally attractive.  When are two objects attracted or repelled by electric forces?  When are two objects attracted by the strong force that holds the nucleus together?]
[Ans: The gravitational force is always attractive, the electric force is repulsive because the protons have equal positive charge, and the strong force between any two nucleons (protons-protons, protons-neutrons, neutrons-neutrons) is attractive.]

9.) In this fission process what (??) is missing ?        ¹₀n  +  ²³⁵₉₂U  --->   ²³⁶₉₂U  --->  ¹⁴³₅₇La  +  ⁹⁰₃₅Br  +  ??
a.) ³₀H    b.) 3 protons    c.) 2 neutrons    d.) 3 neutrons    e.) none of these.
[Hint: Conservation of charge says that the number of charges on each side of the arrow must balance.  Also conservation of baryon number says that the total number of protons + neutrons must balance.]
[Ans: The answer must be 3 neutrons to  properly balance this equation.]

10.) Calculate the number of joules that can be obtained from the fissioning of one mole (235 grams) of ²³⁵U, assuming 200 MeV average energy release per fission. [Ans: 1.93 x 10⁺¹³ Joules].  Most folks in the US use consume an average of 11,000 watts of power.  How long would the energy from 235 grams of Uranium last you at this rate?
[Ans: 55.4 years]
[Hint: The front cover of R & K gives the energy conversion of 1 eV = 1.60 x 10^-19 Joules, so 1 MeV = 1.60 x 10^-13 Joules.] [Ans: 200 MeV * (1.6 x 10^-13 Joules/MeV) * 6.02 x 10²³ atoms/mole = 1.93 x 10⁺¹³ Joules; t = E/P = (1.93 x 10⁺¹³ Joules)/(11,000 Joules/sec) = 1.75 x 10⁹ sec = 55.4 years.]

11.) Thinking about problem 10, how much natural uranium does it take to provide a mole (235 grams) of ²³⁵U? [Ans: 33.6 kg] [Hint: Only 0.7% of natural uranium is U-235.] [Ans: 0.7% = (mass of U-235)/(mass of U-238), or mass of U-238 = (mass of U-235)/0.007 = (235 grams of U-235)/.007 = 33.6 kg.]

12.) When a neutron hits a Uranium-235 nucleus, ²³⁵₉₂U, the nucleus often absorbs the neutron and the new nucleus fissions, breaking apart into fission fragments and a number of neutrons.  If sufficient ²³⁵₉₂U is present, a chain reaction can take place.  What is meant by a chain reaction? [Ans: R&K, p173]
[Ans: Neutrons from the fission of one U-235 nucleus can create the fission of an other U-235 nucleus, whose neutrons can create an additional fission, whose neutrons can create an additional fission, whose...]

13.) Exponential growth happens when the number of new objects (babies, for example) is proportional to the number of old ones (parents). Exponential decay of a parent sample of radioactive nuclei takes place because the number of decays is proportional to:
        a.) the time allowed for a single decay.
        b.) the mass of the nucleus decaying.
        c.) the number of parent nuclei in the sample. [Mathematically, DN = lNDt]
        d.) the number of daughter nuclei in the sample.
        e.) the number of nucleons in the decaying nucleus.
[Aside: The number of decays per second is also called the activity.  It is the activity that we actually measure, not the number of parent nuclei present.  Both the number of parents and the activity decrease exponentially.]
[Ans: A = DN/Dt = number of decays per second, the measurable quantity.]

14.) As a sample of a radioactive isotope decays, its half-life:
        a.) decreases by 1/2 each half life.
        b.) remains the same.
        c.) doubles each half-life.
        d.) may increase or decrease depending on the isotope.
        e.) would follow a straight line if plotted on semilogarithmic paper.
[Hint: Think about what happens to the doubling time in exponential growth.  Suppose your bank pays 7%/year.  Then, from the rule of 70, your account doubles after 10 years.  Now what is the doubling time of your account?  That is, how long will it take your account to double again?] [Ans: In both exponential growth and exponential decay the constant of proportionality, k or l, is constant.  The rule of 70 tells us that the corresponding doubling time or half life must also be constant.

15.) In 1995 the US is electric power production was approximately 450 GWe (A typical large power plant produces about 1 gigawatt of electric power).  Roughly what fraction of that electric power came from geothermal sources in the US in 1995? Where are the major geothermal sources? [Hint: The fraction you will get is very small.  Geothermal energy is discussed in detail starting on p. 158 in R & K.] [Ans: Geothermal power in the US in 1995 was 1750 MWe = 1.75 GWe.  This is a small fraction less than 1%; 1.75/450 = 0.39%]

16.) There is a vast reservoir of thermal energy in the Earth's interior stored as hot rock.   Where did/does this energy come from?
        a.) Energy from the decay of radioactive nuclei like natural uranium.
        b.) Energy from solar neutrinos interacting with the Earth.
        c.) Tidal forces from the Moon-Earth interaction.
        d.) Primordial heat energy remaining from the origin of the solar system.
        e.) Energy from the release of gravitational potential energy as the Earth is gravitationally compressed.
[Hint: We will discuss this in class when we address Geothermal energy.  Also, R&K, p158.] [Ans: In our discussions we will find that each of these contributes to the heating of the Earth's core.]

17.)  A tidal estuary, 1.0 km. wide and 100.0 km. long, has a high tide 10 meters above low tide.  Using the fact that a cubic meter of water has a mass of 1000 kg, find the mass of water available for use in generating power. [Ans:  10¹² kg ]
[Ans: m = rV = density * Volume = (1000 kg/m³)(1000m x 100,000m x 10m) = 10¹² kg.]

18.) Power can be generated both during the 6 1/4 hour period as the tide comes in and again during the 6 1/4 hour period as the tide goes out.  On average, how much power can be generated by the tidal bay of the previous problem?  (Use g = 10 N/kg.) [Ans: 2.22 x 10⁹ Watts.  If you get an answer that disagrees by 2 with this answer, then think carefully about the average height of the water as it goes from high tide to low tide and explain why this matters.]
[Ans: P = Potential Energy/time = mgh/time = (10¹² kg)*(10 N/kg)*(5 meters) / (6.25 hours*3600 sec/hr) = 2.22 x 10⁹ Watts.  N.B. I used 5 meters for the height of the dam, the AVERAGE height of the water, not the maximum height of 10 m or the minimum height of 0 meters.]

19.) To heat your house for one season in Bloomington takes about 10⁷ Kcal.  You could get this heat by cooling off the rock under your yard by, say, 10 ⁰C.  If your yard is 30 m x 30 m, how deep a layer of rock (h) must you cool to extract this much heat?  The specific heat of rock is 0.21 Kcal/kg/⁰C and the density of rock is about 3000 kg/m³.  [Ans: 1.8m.]
[Ans: Q = cmDT.  10⁷ Kcal = (0.21 Kcal/kg/⁰C)*(r*V)*(10 ^oC) = (0.21 Kcal/kg/⁰C)*(3000 kg/m³*30 m*30 m*h)*(10 ^oC).  Solving for h gives h = 1.8 meters.]

20.) In a particular fusion reaction , d  +  t  --->  ⁴₂He  +  n, it is observed that the mass of the final state particles (⁴₂He + n) is less than the mass of the initial state (d + t) by about 3.1 x 10^-29 kg.  How much energy in MeV is realeased in this reaction? 1 MeV = 1.6 x 10^-13 J. [Ans: 2.8 x 10^-12 J or 17.6 MeV , Hint: See problem 7's answer. How does this compare with the energy released in one fission reaction discussed in problem 10?]
[Ans: E = mc² = 3.1 x 10^-29 kg * (3 x 10⁸ m/s)² = 2.8 x 10^-12 Joules = (2.8 x 10^-12 Joules)*(1 MeV / 1.6 x 10^-13 Joules) = 17.6 MeV.  This is only about 17.6 MeV/200 MeV or about 8.8% of a single fission reaction.  On the other hand, remember that it takes only 5 atomic mass units of initial mass to generate 17.6 MeV.  For nuclear fission it takes 236 units to generate 200 MeV.  Per unit mass, fusion releases more energy.]