3.2, 3.3   What is the effect of a diffraction grating on white light?   This will be discussed in Week 7.

 

3.3, 3.5   What do spectra of different emission sources look like?   This will be discussed in Week 7.

 

3.7   Spectroscopic analyses of stellar composition?   

(a)  Yes, H occurs on star Y.  There are lines in the spectrum of Star Y at ~420, 445, 495, 665 nm and these are all found in the spectrum of H.

(b) Star X appears to contain H, He, Na.  The lines in the spectrum of Star X match the lines in the individual spectra for these elements.

(c)  The composition of Star Y includes H, He, and Hg (again, match the lines).  There are some lines that do not match lines in any of the other spectra (e.g., the two lines near 750nm).

(d) The elements common to both X and Y are H and He.

 

3.8   Relative abundance of elements in Figure 3.6   It looks as if line B is about half as ÔintenseÕ as line A, so the abundance of B is about 50 if element A has an abundance of 100.

 

3.10   Relative abundances of elements in the universe   

(a)  Using Figure 3.7, the log abundance of Ni is 4.6 and that of Co is 3.4.  The relative abundance is just the ratio of the antilogarithms of these numbers or

 

     abundance Ni/abundance Co = 104.6/103.4 =  104.6-3.4 =  101.2 =  ~16

 

(b) H is the most abundant element.  It has the highest log abundance in Figure 3.7.

(c)  Si has a log abundance of 6, so we need to look for elements that have a log abundance of 7 or greater (107/106 = 10).  The elements H, He, C, and O all have log abundances above 7.

(d)  Elements that are between 10-100 times less abundance than Si will have log abundances between 5 and 4 (Na, Al, P, Ca, Mn, Cr, and Ni).

 

3.11   What are the trends in elemental abundance?   

(a)  In Figure 3.7, the log abundance decreases as atomic number increases.

(b)  There are some upward ÔspikesÕ that are counter to the general trend (e.g., Cr to Fe).

 

3.12   What do you already know about atoms?   

Atoms are indivisible in that if they are divided, they lose their characteristic properties.  The subatomic particles are protons, neutrons, and electrons.  Most of the mass of the atom is contained in its nucleus.  Electrons surround the nucleus and occupy most of the space of the atom...

 

3.14   Writing atomic symbols   

(a) Lead is element #82, so it has 82 electrons and 82 protons.  Lead-206 has 206-82 = 124 neutrons.  The symbol is 20682Pb (the mass number should be directly on top of the atomic number).

(b) Cobalt is element #27, so it has 27 electrons and 27 protons.  Cobalt-59 has 59-27 = 32 neutrons.  The symbol is 5927Co.

 

3.16   Identifying isotopes   

Isotopes are the same element, different numbers of neutrons.  So, (b) is the pair of isotopes (both are barium).

 

3.18

The differences are that the second isotope has two more neutrons than the first.  The two isotopes also happen to be ions of different charge.  The second ion has lost 4 electrons, the first has lost only two electrons.

 

3.19   How much of the universe is hydrogen and helium?     

The key to solving this problem is remembering that the abundance scale of Figure 3.7 is logarithmic.  There are ~10x more H than He atoms.  He is 100x more abundant than the next most abundant element (O) and more than 100x more abundance than all the others.  So, H and He account for just under 100% of all the atoms. 

 

Since ~100% of all atoms are H or He, consider just these two elements.  The abundances from Figure 3.7 are 1010.5 for H and 109.5 for He.  The %He is about

 

   100*109.4/(1010.5 + 109.4) = 100*2.5 x 109/(3.2 x 1010+ 2.5 x 109) = ~7%

 

   (a very rough estimate since it is hard to read the graph accurately)

 

 

3.20   Correlating reactions and temperatures in atom formation     

(a)  A nucleus is capturing an electron.  The temperature must be below a few thousand Kelvin (see the text, just before equation (3.4)).

(b) This is a nuclear fusion and requires higher temperatures (to overcome nuclear repulsions).  According to the text, this happens between 109 and 107 Kelvin.

 

3.21   Mass of H atoms in our Milky Way galaxy     

Answers are in back of text.

 

3.22   Nuclear emissions     

 

(a) The nucleus emitting the positron (shown in the web companion example) does not have the same number of gray and blue balls as the one emitting a beta particle. The difference is in the ball that does the emitting. The blue ball is a proton and it changes to a gray neutron when the positron is emitted.   A gray ball changes to blue when a beta particle is emitted. Thus, the graphic is consistent with the observation that nuclei that emit positrons are different from the nuclei that emit electrons.

 

3.24   Balancing nuclear reactions:  particles emitted    

(a) 150Gd -> 146Sm + 42He

Gd is element #64 and Sm is #62, so 2 positive charges have been lost.  The mass number has decreased by 4.  A particle with 2+ charges and a mass of 4 is an alpha particle.

(b) 65Zn -> 65Cu + 01e+

There is no change in mass number, but the atomic number has decreased by 1 going from Zn to Cu (30 to 29).  A positron has been emitted.

 

3.26   Balancing nuclear reactions:  nuclear product

(a) 189F -> 188O + 01e+

(b) 21483Bi -> 21484Po + 0-1e-

 

3.28   Balancing nuclear reactions

(a) 136C +  1on  -> 146C

(b) 11H +  11H  -> 21H  +  01e-

(c) 24998Cf +  188O -> 263106Sg  + 4 10n

(d) 2010Ne +  42He -> 2412Mg

 

3.30    How do emissions from a radioactive sample differ?

Here are some actual results for smoke detector radiation measurements with a Geiger counter: 

 

         no shield, 318 counts per minute

         cardboard, 257 cpm

         Al shield, 251 cpm

         Pb shield, 1 cpm

 

These show that cardboard reduces the number of emissions reaching the detector, but Al doesnÕt reduce them much more.  A Pb shield reduces the measured emissions almost to zero. 

 

3.31    What are the emissions from decay of americium-241?

(a)  Yes, there were some changes in measured emissions.  See 3.30.

(b)  Since the cardboard shield reduced the number of emissions, there must be alpha particle being emitted.  Since Al doesnÕt significantly reduce the emissions further, there are probably no beta emissions.  Since the cpm is still relatively high for Al and drops to ~0 for Pb, we know that gamma radiation is emitted.

 

3.32   Daughter nuclei in radioactive decays

(a) 23190Th -> 23191Pa +  0-1e-

(b) 24195Am -> 23793Np +  42He  + gamma ray

 

 

3.33, 3.34   How can you count a moderately large sample?   You will have to try this one on your own!  Since ~half of the candies should end Ôheads upÕ with each round and these will be removed, the number of candies will be reduced in half with each round.  So, the reductions would follow the pattern 100 -> 50 -> 25 -> 12 or 13 (12.5) -> 6 or 7 (6.25) -> 3 or 4 (3.125) and so on.  If you count whatever number you have at the end and keep doubling it (double once for each round), you can estimate the size of the original sample.

 

 

3.35   What fraction of a sample remains after n half lives?   This will be discussed in Week 7 group work.  Note that equation 3.17 is incorrect.  It should be

This equation describes the curve in Figure 3.14  After two half-lives, the fraction remaining should be (1/2)2 = 0.25, and this is what the graph shows.  To Ôcount backwardsÕ and estimate an original sample size, we need to count the number of half-lives.  If equation 3.17 is rearranged to give N0 = Nná2n, we see that itÕs possible to find N0 before n half-lives passed if we know the value of Nn now.

 

 

3.37   Radioisotope decay and half-life   When fn = 0.01, the P-32 waste is safe to dispose of.  This happens when fn = (1/2)n or when 0.01 = (1/2) n.   Remembering that ln(x)y = y ln x , we have

     ln 0.01 = n[ln (1/2)]

         -4.6 = n[-0.69]

              n = 6.7

So, we need to wait ~7 half-lives or ~7*14.3 days or ~100 days.

 

3.38   Half-life determination   On the decay curve, whenever the activity drops by a factor of 2, the same amount of time will be required.  (Very similar to Figure 3.14)

 

3.39   Electron-capture reactions   

(a) 6731Ga +  0-1e-  -> 6730Zn  + gamma radiation

(b) 20181Tl +  0-1e-   -> 20180Hg  + gamma radiation

 

3.40      Energy equivalence of mass

(a) delta E = delta m * c2 = (1 x 10-7kg)(3.00 x 108 m/s)2 = –9 « 109 kgám2/s2 =

          –9 « 109 J         (the units work out since 1 Joule = kg*m2/s2)

 

(b) The process is exothermic since delta E is negative..

 

3.42       Energy release in nuclear fusion

(a) The reaction represented is 4He + 12C -> 16O.

 

change in mass = (26.55291 « 10–27 kg for 16O ) – [(6.64466 « 10–27 kg for 4He) + (19.92101 « 10–27 kg for 12C)]

     = –0.01276 « 10–27 kg

change in energy = (–0.01276 « 10–27 kg)(3.00 « 108 más–1)2 = –1.148 « 10–12 J

 

(b) A light burst occurs when the nuclei meet and fuse.

 

3.43      Nuclear reaction energies

(a)

              311p + 310n -> 321H                   

              21H + 21H -> 32He + 10n              

              32He + 21H -> 42He + 11p            

SUM          211p + 210n -> 42He                                       

 

 

(b) change in mass = 6.64466 « 10–27 kg for 42He –

     [2(1.67262 « 10–27 kg for each 11p) + 2(1.67493 « 10–27 kg for each 10n)]

     = –0.05044 « 10–27 kg

 

     change in energy = (–0.05044 « 10–27 kg)á(3.00 « 108 más–1)2 = –4.54 « 10–12 J

 

(c) The change in energy here is larger in magnitude than in Example 3.41.  When a proton is converted to a neutron (whatÕs happening, in part, in 3.15), 11p + 0-1e-  -> 10n , the change in mass is positive

 

change in mass = mass of neutron – [mass of proton + mass of e-]

     = (1.67493 « 10–27 kg for the 10n) – [(1.67262 « 10–27 kg for the 11p)

         +  (9.10939 « 10–31 kg for the e-)]

 

     = +0.001399061 or +0.00140 kg

 

This means that every time a proton is converted to a neutron (and two protons are converted to neutrons in 3.15), energy must be absorbed (a + mass change means the energy change is +).  This makes the overall energy change in 3.15 less negative than it would otherwise be.  (You can calculate the actual mass and energy changes for 3.15, but this estimate gives you the general idea.)

 

3.45       Mass loss in nuclear fusion

From Table 3.4, the fission of 1 g of U-235 yields an energy change of  –7.0 « 1010 J. For 1 gram of U-235, the change in mass is

change in mass = (–7.0 « 1010 J)/( 3.00 « 108 m/s)2 = –7.8 « 10–7 kg

 

Compared to the mass loss for the fusion of H nuclei to yield helium (7.10 « 10–6 kg, see 3.44 in the text), this mass change is smaller in magnitude and opposite in sign.

 

3.47      Energy released in a hypothetical nuclear reaction

The change in mass is

 

change in mass = (mass Pb-208) – (mass 82 protons + mass 126 neutrons)

    = 345.2788 x 10-27 kg) – [82(1.67262 « 10–27 kg) + 126(1.67493 « 10–27 kg)]

         = –2.9172 « 10–27 kg

 

change in energy = (–2.9172 « 10–27 kg)(3.00 « 108 m/s)2 = –2.63 « 10-10 J

 

This mass change here is larger than the mass change for the synthesis of Al-27 (see 3.46 in the text).

 

3.48      Binding energy per nucleon for carbon

There are 12 nucleons (6 protons and 6 neutrons) in C-12.  The mass change for the formation of one C-12 nucleus is

 

mass change = mass C-12 – [(mass 6 protons) + (mass 6 neutrons)]

     =  19.92101 « 10–27 kg – [6(1.67262 « 10–27 kg) + 6(1.67493 « 10–27 kg)]

     = –0.16429 « 10–27 kg

 

 

The energy change is = (–0.16429 « 10–27 kg)(3.00 « 108 más–1)2

       = –1.48 « 10-11 J

 

To find the molar binding energy per nucleon, you need to first find the energy change for a mole of C-12:

 

Energy change per mole = (–1.48 « 10-11J/nucleus)(6.022 « 1023 nuclei/mol)

              = -8.91 « 109 kJ/mole

 

The binding energy per nucleon is = -8.91 « 109 kJ/mole)/12 nucleons

= –74.3 « 107 kJ/mole per nucleon

 

This agrees well with the number you can read from Figure 3.18 (–74.3 « 107 kJ/mole per nucleon).

 

3.49       Trends in nuclear stability

(a) Nuclear stability increases (binding energy becomes more negative) from low mass elements to Fe-56.

(b) Stability decreases for elements beyond Fe-56, though itÕs a smaller rate of change.

 

3.50      Energy release in nuclear fusion reactions  From 3.43(b), the nuclear binding energy for He is –4.54 « 10–12 J per nucleus.  The molar binding energy is

 

Molar binding energy = (–4.54 « 10–12 J/nucleus)(6.02 « 1023 nuclei/mole)

           = –2.73 « 1012 J/mole = –273 « 107 kJ/mole

 

The scale on Figure 3.19 is hard to read, but it looks as if each green bar is a little less than 300 kJ/mol in magnitude and negative.  You can verify that the other color bars represent the right energies by doing similar calculations.

 

 

For the conversion of three He-4 nucleiu to one C-12 nucleus, the change in energy is just the energy difference between the blue C-12 bar and the green He-4 bars.  This is hard to read on Figure 3.19, but it looks like it is the difference between ~-900 x 107 kJ/mol( (for C-12) and ~-840 x 107 kJ/mol( (for three He-4 nuclei) or ~-80 « 107 kJ/mol.  The number is negative since the energy of the product (C-12) is lower than the energy of the reactants (three He-4). 

 

The second reaction (He-4 + C-12 = O-16) appears to have a similar change in energy (drop from the green/blue bar to the red bar is about 80 « 107 kJ/mol and is negative.  But it is very hard to read these scales, so there is a lot of uncertainty.

 

 

3.51  Fusion or fission?

(a) Elements above Fe-56 undergo fission, so that is what happens with Pu-240.

(b) Elements below Fe-56 undergo fusion, so that is what happens to Li-6. 

(c)  It is unlikely that Fe-56 will undergo fusion or fission since it is the most stable nucleus.

 

3.52      Collision-induced nuclear fission

Here is just one of the possible reactions:

 

3.53      Chain Reactions  This will be discussed in Week 8 lecture.

 

3.54  How are elemental abundance and nuclear stability related?  We will discuss similar problems during Week 8 lecture.

 

3.55  Cosmic elemental abundance and nuclear reactions  In Figure 3.21, several lower atomic number products like Kr, Rb, and Sr (atomic numbers 36, 37, and 38) form.  There are also some higher atomic number products, Xe, Cs, and Ba (atomic numbers 54, 55, and 56). If U-235 decays to produce these nuclei, then the nuclei would be expected to have accumulated over the years.

 

3.56  The higher abundance of elements with even atomic number 

(a)  He-4 has 2 protons and 2 neutrons, each contributing about 1 mass unit – so the mass of a He-4 nucleus is ~4.  When He-4 nuclei fuse, the masses of the resulting element must be even numbers.  One example of this is equation 3.7.

(b)       126C + 11H -> 137N

This could account for the formation of an odd numbered element, but the problem tells us that N-13 is unstable.  Taking the hint provided in the problem, assu