Chapter 30: | 1 | 12 | 13 | 14 | 17 | 25 | 30 | 28 | 35 | 36 | 37 | 38 | 39 | 40 | 41 | 43 | 47
31:
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Nuclear Physics and Radioactivity - by Eric Hess, 2003: Year of the Chicken Breast
"Say it with me now, 'new-clear'."

1. What is the rest mass of an a particle in MeV/c2?

The book just uses the mass of a He atom in au, which is 4.002602.

4.002602 x 931.5 = 3728 MeV/c2

12. Use Appendix F to calculate the binding energy of 21H (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

13. Calculate the binding energy per nucleon for a 147N nucleus?

147N 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 147N.

104.659614/14 = 7.477 MeV/nucleon

14. Calculate the total binding energy and the binding energy per nucleon for 63Li. 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 63Li.

31.996/6 = 5.333 MeV/nucleon.

17. How much energy is required to remove (a) a proton, (b) a neutron, from 168O? Explain the difference in your answers.

a) 168O minus a proton would be like a transition from 168O --> 157N + 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 168O --> 158O + 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.

25. A 23292U 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 22890Th, with an approximate mass of 228.02882 u.

30. What is the energy of the a particle emitted in the decay 21084Po --> 20682Pb + 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

28. The isotope 21884Po can decay by either a or B- emission. What is the energy release in each case? The mass of 21884Po is 218.008965 u.

a decay: The atom loses 2p2n, or equivalently a He atom

21884Po --> 21482Pb + 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

21884Po --> 21885Pb + 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

35. A radioactive material produces 1280 decays per minute at one time, and 6h later produces 320 decays per minute. What is its half-life?

N = Noe-lt

The N and No can be substituted with the 320 and 1280, as long as you use 6h for t:

320 = 1280e-l6

320/1280 = e-l6

ln(320/1280) = ln(e-l6)

ln(320/1280) = -l6, ln(320/1280)/-6, l = .231049

Now use the half-life formula:

T1/2 = ln(2)/l = ln(2)/.231049 = 3 hr

36. Look up the half life of 23892U in appendix F, and then determine its decay constant.

T1/2 = 4.468 x 109 yr

T1/2 = ln(2)/l, l = ln(2)/T1/2

l = ln(2)/(4.468x109 x 365 x 24 x 60 x 60) = 4.919 x 10-18 s-1

37. The decay constant of a given nucleus is 5.4 x 10-3 s-1. What is its half-life?

T1/2 = ln(2)/l = ln(2)/5.4 x 10-3 = 128.4 sec, or 2.1 minutes.

38. What is the activity of a sample of 146C that contains 4.1 x 1020 nuclei?

Use the activity formula:

DN/Dt = lN

l = ln(2)/T1/2, T1/2 = 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 1020) = 1.57 x 109 decays/sec

39. What fraction of a sample of 6832Ge, whose half-life is about 9 months, will remain after 3.0 yr?

3 years = 36 months. A half-life is 9 months, so there would be 36/9 = 4 half-lifes.

To find the fraction, 1/24 = 1/16 of the sample will remain.

40. How many nuclei of 23892U remain in a rock if the activity registers 275 decays per second?

Use the activity formula,

T1/2 = 4.468 x 109 yr x 365 x 24 x 3600 = 1.409 x 1017 sec

l = ln(2)/T1/2 = ln(2)/1.409 x 1017 = 4.91932697 x 10-18 s-1

DN/Dt = lN, 275 = 4.91932697 x 10-18 x N

N = 5.59 x 1019 nuclei

41. What fraction of a samples left after (a)exactly 4 half-lives, (b)exactly 4.5 half-lives?

Half-life fractions are just 1/2#of half-lives

a) 1/24 = 1/16 = .0625

b) 1/24.5 = .0442

43. The iodine isotope 13153I 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 half-life, T1/2 = 8.04 days x 24 hours x 3600 seconds = 694656 seconds

Find l,

l = ln(2)/T1/2 = ln(2)/694656 = 9.97828 x 10-7 s-1

Now, remember back to the Chemistry days of yore...

moles = N/Na = grams you have/molar mass. N is the number of atoms, Na is Avacado's number.

N/6.02 x 1023 = 532 x 10-6/130.91

N = 2.446444 x 1018 nuclei

Now, the activity formula, DN/Dt = lN

9.97828 x 10-7 x 2.446444 x 1018 = 2.44 x 1012 decays/second

b)Now use the population growth formula,

N = Noe-lt, you can use the decay rate in place of No

N = 2.44113 x 1012 x e-9.97828 x 10^-7 x 3600 s)

N = 2.43 x 1012 decays/second

c)Same as before,

N = Noe-lt

N = 2.44113 x 1012 x e-9.97828 x 10^-7 x (6 x 30 x 24 x 3600)

N = 4.48 x 105 decays/second

And they only call it a level (II)

47. A sample of 23392U (T1/2 = 1.59 x 105 yr) contains 6.50 x 1019 nuclei. (a) What is the decay constant? (b) Approximately how many disintegrations will occur per minute?

(a) l = ln(2)/T1/2 = ln(2)/(1.59 x 105 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 1019 = 8.99 x 106 decays/second

Now multiply by 60,

8.99 x 106 x 60 = 5.39 x 108 decays/minute

Chapter 31:

1. Natural aluminium is all 2713Al.  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.  (2813Al)

If you look in appendix F in the back of the book, you will not find this isotope.(2813Al)  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 (2814Si)

4. Does the reaction 73Li(p,a)42He 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:

 73Li              + p               = a +                 42He 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)

5. The reaction 188O(p,n)189F requires an input of energy equal to 2.453 MeV.  What is the mass of 189F?

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:

 188O              + p               = n    +               189F 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 94Be(a,n)126C?

You just compare the total mass of the left side to the right side.  This reaction looks like:

 94Be              + a               = n +                 126C 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)