Chapter 30:  1  12

13  14  17  25

30  28  35  36
 37  38  39  40
 41  43  47 
31: 1  4  5 
6  Go up
Nuclear Physics and Radioactivity  by Eric Hess,
2003: Year of the Chicken Breast
"Say it with me now, 'newclear'."
1. What is the rest mass of an a particle in MeV/c^{2}?
The book just uses the mass of a He atom in au, which is 4.002602.
4.002602 x 931.5 = 3728 MeV/c^{2}
12. Use Appendix F to calculate the binding energy of ^{2}_{1}H (deuterium)?
Deuterium is made of one neutron, and one normal hydrogen atom. So, add their atomic masses together:
1.008665u + 1.007825u = 2.01649 u.
This number is what the atomic mass should be with all the pieces. However, the table in the back of the book has its mass at 2.014102 u. The binding energy is just the difference.
2.01649  2.014102 = .002388u x 931.5 (to get the energy in MeV) = 2.224422, or 2.224 MeV
(Table of contents)
13. Calculate the binding energy per nucleon for a ^{14}_{7}N nucleus?
^{14}_{7}N is made of 7H, and 7n.
7(1.007825) + 7(1.008665) = 14.11543 u
The book offers a measly 14.003074 u, making a difference of .112356 u between the two.
.112356 x 931.5 = 104.659614 MeV.
But that's for the whole atom. The question asks for the energy per nucleon. There are 14 nucleons in ^{14}_{7}N.
104.659614/14 = 7.477 MeV/nucleon
(Table of contents)
14. Calculate the total binding energy and the binding energy per nucleon for ^{6}_{3}Li. Use Appendix F?
So there are 3H and 3n.
3(1.007825) + 3(1.008665) = 6.04947 u
The book offers 6.015121 u, making a difference of .034349 u between the two.
.034349 x 931.5 = 31.996 MeV total biding energy.
But that's for the whole atom. The question also asks for the energy per nucleon. There are 6 nucleons in ^{6}_{3}Li.
31.996/6 = 5.333 MeV/nucleon.
(Table of contents)
17. How much energy is required to remove (a) a proton, (b) a neutron, from ^{16}_{8}O? Explain the difference in your answers.
a) ^{16}_{8}O minus a proton would be like a transition from ^{16}_{8}O > ^{15}_{7}N + H
15.994915 > 15.000108 + 1.007825 = 16.007933 u
16.007933  15.994915 = .013018 u x 931.5 = 12.1 MeV
b) To remove a neutron, on the other hand is ^{16}_{8}O > ^{15}_{8}O + n
15.994915 > 15.003065 (from the appendix) + 1.008665 = 16.6631725 u
16.6631725  15.994915 = .016815 u x 931.5 = 15.7 MeV
There is a difference because the neutron has more mass and thus takes more energy to remove.
(Table of contents)
25. A ^{232}_{92}U nucleus emits an a particle with KE = 5.32 MeV. What is the final nucleus and what is the approximate atomic mass (in u) of the final atom?
Start by converting the kinetic energy to u:
5.32 MeV/931.5 MeV/u = .005712185 u
The mass of the Uranium is 232.037131 u (from the appendix), and the alpha particle is like a Helium atom with a mass of 4.002602. Subtract the mass lost from the alpha particle and the kinetic energy given to it from the Uranium's mass to find the remaining mass.
232.037131  (4.002602 + .005712185) = 228.0288168 u.
Since it lost 2p and 2n, it leaves an atom of ^{228}_{90}Th, with an approximate mass of 228.02882 u.
(Table of contents)
30. What is the energy of the a particle emitted in the decay ^{210}_{84}Po > ^{206}_{82}Pb + a?
So, look the masses up in the back of the book. Use He as the alpha particle:
209.982848 > 205.974440 + 4.002602 + KE
KE = 209.982848  209.977042 = .005806.
Now convert to MeV:
.005806 x 931.5 = 5.41 MeV
(Table of contents)
28. The isotope ^{218}_{84}Po can decay by either a or B^{} emission. What is the energy release in each case? The mass of ^{218}_{84}Po is 218.008965 u.
a decay: The atom loses 2p2n, or equivalently a He atom
^{218}_{84}Po > ^{214}_{82}Pb + a
218.008965 > 213.999798 + 4.002602
The difference in the two sides is:
218.008965  218.0024 = .006565 u
.006565 x 931.5 = 6.12 MeV
B^{} decay: An electron is created out of energy, but to balance the charge, a proton is added to the atom
^{218}_{84}Po > ^{218}_{85}Pb + B^{ }+ v
218.008965 > 218.00868 + E
The difference is:
218.008965  218.00868 = 2.85 x 10^{4}
2.85 x 10^{4} x 931.5 = .27 MeV
(Table of contents)
35. A radioactive material produces 1280 decays per minute at one time, and 6h later produces 320 decays per minute. What is its halflife?
Start with the decay rate formula:
N = N_{o}e^{}^{l}^{t}
The N and N_{o} can be substituted with the 320 and 1280, as long as you use 6h for t:
320 = 1280e^{}^{l}^{6}
320/1280 = e^{}^{l}^{6}
ln(320/1280) = ln(e^{}^{l}^{6})
ln(320/1280) = l6, ln(320/1280)/6, l = .231049
Now use the halflife formula:
T_{1/2} = ln(2)/l = ln(2)/.231049 = 3 hr
(Table of contents)
36. Look up the half life of ^{238}_{92}U in appendix F, and then determine its decay constant.
T_{1/2} = 4.468 x 10^{9} yr
T_{1/2} = ln(2)/l, l = ln(2)/T_{1/2}
l = ln(2)/(4.468x10^{9} x 365 x 24 x 60 x 60) = 4.919 x 10^{18} s^{1}
(Table of contents)
37. The decay constant of a given nucleus is 5.4 x 10^{3} s^{1}. What is its halflife?
T_{1/2} = ln(2)/l = ln(2)/5.4 x 10^{3} = 128.4 sec, or 2.1 minutes.
(Table of contents)
38. What is the activity of a sample of ^{14}_{6}C that contains 4.1 x 10^{20} nuclei?
Use the activity formula:
DN/Dt = lN
l = ln(2)/T_{1/2}, T_{1/2} = 5730 yr (back of the book) x 365 days x 24 hrs x 60 min x 60 sec
l = 3.8359 x 10^{12} s
DN/Dt = 3.8359 x 10^{12} x (4.1 x 10^{20}) = 1.57 x 10^{9 }decays/sec
(Table of contents)
39. What fraction of a sample of ^{68}_{32}Ge, whose halflife is about 9 months, will remain after 3.0 yr?
3 years = 36 months. A halflife is 9 months, so there would be 36/9 = 4 halflifes.
To find the fraction, 1/2^{4} = 1/16 of the sample will remain.
(Table of contents)
40. How many nuclei of ^{238}_{92}U remain in a rock if the activity registers 275 decays per second?
Use the activity formula,
T_{1/2} = 4.468 x 10^{9} yr x 365 x 24 x 3600 = 1.409 x 10^{17} sec
l = ln(2)/T_{1/2} = ln(2)/1.409 x 10^{17} = 4.91932697 x 10^{18} s^{1}
DN/Dt = lN, 275 = 4.91932697 x 10^{18} x N
N = 5.59 x 10^{19} nuclei
(Table of contents)
41. What fraction of a samples left after (a)exactly 4 halflives, (b)exactly 4.5 halflives?
Halflife fractions are just 1/2^{#of halflives}
a) 1/2^{4} = 1/16 = .0625
b) 1/2^{4.5} = .0442
(Table of contents)
43. The iodine isotope ^{131}_{53}I is used in hospitals for diagnosis of thyroid function. If 532 micrograms are ingested by a patient, determine the activity (a) immediately, (b) 1.0 h later when the thyroid is being tested, and (c) 6 months later. Use appendix F.
a) First, look up the halflife, T_{1/2} = 8.04 days x 24 hours x 3600 seconds = 694656 seconds
Find l,
l = ln(2)/T_{1/2} = ln(2)/694656 = 9.97828 x 10^{7} s^{1}
Now, remember back to the Chemistry days of yore...
moles = N/N_{a} = grams you have/molar mass. N is the number of atoms, N_{a} is Avacado's number.
N/6.02 x 10^{23} = 532 x 10^{6}/130.91
N = 2.446444 x 10^{18} nuclei
Now, the activity formula, DN/Dt = lN
9.97828 x 10^{7} x 2.446444 x 10^{18} = 2.44 x 10^{12} decays/second
b)Now use the population growth formula,
N = N_{o}e^{lt}, you can use the decay rate in place of N_{o}
N = 2.44113 x 10^{12} x e^{9.97828 x 10^7 x 3600 s)}
N = 2.43 x 10^{12} decays/second
c)Same as before,
N = N_{o}e^{l}^{t}
N = 2.44113 x 10^{12} x e^{9.97828 x 10^7 x (6 x 30 x 24 x 3600)}
N = 4.48 x 10^{5} decays/second
And they only call it a level (II)
(Table of contents)
47. A sample of ^{233}_{92}U (T_{1/2} = 1.59 x 10^{5} yr) contains 6.50 x 10^{19} nuclei. (a) What is the decay constant? (b) Approximately how many disintegrations will occur per minute?
(a) l = ln(2)/T_{1/2} = ln(2)/(1.59 x 10^{5} yr x 365 days x 24 hours x 3600 seconds) = 1.38236 x 10^{13} s^{1}
(b) Disintegrations just means activity, but they want it per minute, for some reason. Just find decays per second and multiply by 60.
DN/Dt = lN = 1.38236 x 10^{13} x 6.50 x 10^{19} = 8.99 x 10^{6} decays/second
Now multiply by 60,
8.99 x 10^{6} x 60 = 5.39 x 10^{8} decays/minute
(Table of contents)
Chapter 31:
1. Natural aluminium is all ^{27}_{13}Al. If it absorbs a neutron, what does it become? Does it decay by b+ or b? What will be the product nucleus??
Well, absorbing a neutron would increase the number of nucleons, but not the number of protons, so A = A+1, and Z = Z, so you would have Aluminium 28. (^{28}_{13}Al)
If you look in appendix F in the back of the book, you will not find this isotope.(^{28}_{13}Al) But you will see that there is a stable isotope of Si with a mass number of 28. Si has one more proton, and there are no other mass number 28 isotopes listed, and since this is the b decay product of Al (neutron turns to proton plus electron), then I assume that it must decay by b. Of course, I cheated and just looked at the chart of the nuclides on the wall in the physics room, so I am not really guessing.
The nucleus would be that of Si 28 or (^{28}_{14}Si)
4. Does the reaction ^{7}_{3}Li(p,a)^{4}_{2}He require energy or does it release energy?
You just compare the total mass of the left side to the right side. This reaction looks like:
^{7}_{3}Li + p = a + ^{4}_{2}He 7.016003 + 1.007825 4.002602 + 4.002602 You look these masses up in appendix F. I used a hydrogen for the proton. Notice that the number of electrons on the left and right side are balanced out (four on each side)
The total left side mass is 8.023828, and the right side mass is 8.005204, so this reaction seems to have released energy. (8.023828  8.005204 = 0.018624*931.5 = 17.35 MeV released)
(Table of contents)
5. The reaction ^{18}_{8}O(p,n)^{18}_{9}F requires an input of energy equal to 2.453 MeV. What is the mass of ^{18}_{9}F?
Again, you just compare the total mass of the left side to the right side. Since the reaction requires 2.453 MeV of energy, the right side will reflect this by being 2.453/931.5 = 0.002633387 u more massive than the left side.
This reaction looks like:
^{18}_{8}O + p = n + ^{18}_{9}F 2.453/931.5 + 17.999160 + 1.007825 1.008665 + ??.??????? Notice that the number of electrons on the left and right side are balanced out (nine on each side)
The total left side mass is 19.006985, and if the reaction requires or gains 2.453 MeV of energy, it must gain 2.453/931.5 = 0.002633387 u of mass, so the total right side mass must be (19.006985 + 0.002633387 = 19.00961839) So the mass of the F 18 must be this less the mass of the neutron. (19.00961839  1.008665 = 18.000953 u)
6. Calculate the energy released (or energy input required) for the reaction ^{9}_{4}Be(a,n)^{12}_{6}C?
You just compare the total mass of the left side to the right side. This reaction looks like:
^{9}_{4}Be + a = n + ^{12}_{6}C 9.012182 + 4.002602 1.008665 + 12.000000 Look these masses up in the back of the book, and use H 1 for the proton. Notice that the number of electrons on the left and right side are balanced out (six on each side)
The total left side mass is 13.014784, and the right side mass is 13.008665, so this reaction seems to have released energy. (13.014784  13.008665 = 0.006119*931.5 = 5.700 MeV released)