Answer Key, Problem Set 10

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Chemistry 122
Mines, Spring 2012
Answer Key, Problem Set 10
1. NT1.
(a) What is the zone (valley) of stability?
It is a graphical way to indicate which combinations of protons and neutrons result in a nucleus that
is stable (and, by extension, those that are “reasonably stable”) with respect to spontaneous nuclear
decay. It is typically illustrated as a plot of #neutrons vs #protons (N vs. Z), in which a point appears
on the plot if the combination results in a stable nucleus. It is called a “zone” (or valley) because the
plot does not yield a line or a smooth curve, but rather a “zone” (or valley) containing many points
that cluster together in a pattern that resembles a curve with “width” (like a valley).
(b) Stable light nuclides have about equal numbers of neutrons and protons. What happens to the neutron–to-proton
ratio for stable nuclides as the number of protons increases?
n/p (Tro writes this as N/Z) for stable nuclides gets larger as p increases [up to Z = 83]. It is ~1.0 for
“light” nuclides (Z ≤ ~20), but increases somewhat steadily, but not linearly, to ~1.5 for stable “heavy”
nuclides (Z  80). (For example, when Z is ~50, a stable nuclide has an n/p ratio of about 1.3.
(c) Nuclides that are not already in the zone of stability undergo radioactive processes to [ultimately] get to the zone of
stability. If a nuclide has too many neutrons, which process(es) can the nuclide undergo to become more stable?
-decay (also called -emission) is the preferred mode of decay in this instance, because when a particle is produced, a neutron turns into a proton, thus lowering the n/p ratio. (Note: If a neutron
turns into a proton, the number of neutrons goes down by one unit while the number of protons goes
up by one unit. That means n/p must decrease (numerator decreases and denominator increases
means value of fraction decreases.)
(d) Answer the same question (as in (c)) for a nuclide having too many protons.
This question is actually a bit ambiguous. As such, there are two answers.
For nuclides that really have “too many protons” (i.e., those with Z > 83), the most common result is
-decay. I call this being “beyond” the zone (or valley) of stability, although Tro chooses not to
stress (or even address) this for some reason. Losing an -particle causes both n and p to
decrease, which brings the nuclide “down and left” on the N vs. Z plot, which is in the general
direction of the zone of stability if one is, in fact, “beyond” it. Perhaps Tro does not focus on this
because a fair number of nuclides in this region also undergo -decay. However, I still believe it is
still reasonable to predict -decay here.
For nuclides that have less than 83 protons, but have “too many protons relative to the number of
neutrons in a stable nuclide of that Z value” (i.e., an n/p ratio that is smaller than the nuclides in the
zone of stability, a situation that I call being “below” the zone of stability), positron emission and
electron capture are the most common options. This is because when either of these processes
occurs, a proton turns into a neutron, thus raising the n/p ratio. (It should be pointed out that -decay also
occurs for a small percentage of nuclides below the zone of stability. Note that if n/p > 1, the loss of 2 p’s and 2 n’s does raise
the n/p ratio.)
2. NT2.
Write an equation describing the radioactive decay of each of the following nuclides. (The particle produced is
shown in parentheses, except for electron capture, where an electron is a reactant.)
68
(a) Ga (electron capture);
Answers: (a)
(c)
68
31
Ga 
0
-1
Fr 
208
85
212
87
e 
At 
62
(b) Cu (positron);
68
30
(c)
Cu 
Zn
(b)
62
29

(d)
129
51
4
2
212
Sb 
Fr ();
62
28
Ni 
(d)
Sb ()
0
1
Te 
129
52
129
e
0
1
e
Strategy and Application of Strategy:
In all cases, first write out the reactant nuclide on the left and the product particle on the right
(except for the electron capture case, where it is a second reactant. Then figure out what the
product nuclide must be using the two conservation rules (for mass number and charge).
PS10-1
Answer Key, Problem Set 10
(a)
68
31
Ga 
e 
0
-1
Zn
68
30
In “electron capture”, the electron is a reactant. Sum of lower left subscripts on left is 31 + (-1) =
+30, so it must be so on the right side as well (conservation of charge). Since there is only one
product, 30 must be the number of protons in the product nuclide  Zn. Mass number is 68 + 0
on the left, so it must be 68 on the right as well. In short:
Conservation of Mass Number: 68 + 0 = x  x = 68
31 + (-1) = y  y = +30
Conservation of Charge:
(b)
62
29
Cu 

0
1

e
62
29
Cu 
62
28
Ni 
0
1
e
In “positron emission”, the positron (positive electron) is a product.
Conservation of Mass Number: 62 = x + 0  x = 62
Conservation of Charge: 29 = y + (+1)  y = 28  Ni
(c)
Fr 
212
87
 or
4
2

4
2
4
2


212
87
Fr 
208
85

At 
4
2

He is a product.
Conservation of Mass Number: 212 = x + 4  x = 208
+87 = y + (+2)  y = +85  At
Conservation of Charge:
(d)
129
51
Sb 

0
1

e
129
51
Sb 
Te 
0
1
129
52
e
 is a product, and is an electron
Conservation of Mass Number: 129 = x + 0  x = 129
Conservation of Charge: 51 = y + (-1)  y = 52  Te
3. NT3.
In each of the following nuclear reactions, supply the missing particle.
73
(a) Ga →
(d)
241
73
Ge + ____
Cm + ____ →
0
1
Answers: (a)
e
241
Am
(b)
4
2

(b)
192
Pt →
188
(e)
137
Ba →
137
(c)
0
1
Os + ___
205
Bi →
205
Pb + ___
Ba + ____
(d)
e
(c)
0
-1
(e)  
e
Strategy and Application:
In each case, I used the element symbols to find the elements on the periodic table to get the atomic
numbers (#p’s; subscript). Then I used the conservation rules as noted above (Problem 2).
(a) 73Ga →
73
Ge + ____ 
73
31
Ga 
73
32
Ge  ____ 
0
1
e
Conservation of Mass Number: 73 = 73 + x  x = 0
Conservation of Charge: 31 = 32 + y  y = -1
Check: -emission is associated with a n → p conversion. Mass number stays the same, but
proton number increases by one. That fits what happened here.
(b) 192Pt →
188
Os + ___ 
192
78
Pt 
188
76
Os  ____ 
PS10-2
4
2

Answer Key, Problem Set 10
Conservation of Mass Number: 192 = 188 + x  x = 4
Conservation of Charge: 78 = 76 + y  y = 2
Check: -emission is associated with a loss of 2 p’s and 2 n’s. Mass number decreases by 4,
and proton number decreases by two. That fits what happened here.
(c) 205Bi →
205
Pb + ____ 
205
83
Bi 
205
82
Pb  ____ 
0
1
e
Conservation of Mass Number: 205 = 205 + x  x = 0
Conservation of Charge: 83 = 82 + y  y = +1
Check: positron emission is associated with a p → n conversion. Mass number stays the same,
but proton number decreases by one. That fits what happened here.
(d) 241Cm + ____ →
241
Am 
241
96
Cm  ____ 
241
95
Am 
0
-1
e
Conservation of Mass Number: 241 + x = 241  x = 0
96 + y = 95  y = -1
Conservation of Charge:
Check: Electron capture is associated with a p → n conversion. Mass number stays the same,
but proton number decreases by one. That fits what happened here.
(e) 137Ba → 137Ba + ____ 
4. NT4.



Gamma emission does not change the identity of the nuclide.
(a) The only stable isotope of fluorine is F-19. Predict possible modes of decay for F-21, F-18, and F-17. (b) Also
210
195
predict possible modes of decay for Po and (c) Au.
Answers, with most brief explanations:
(a) F-21, too many neutrons  -decay (actual is 
F-18 and F-17, too few neutrons  positron emission or electron capture
(b) Po-210, Z > 83  -decay (actual is 
(actual is P.E. for both)
(c) Au-195, too few neutrons  positron emission or electron capture (actual is EC
Strategy, General Approach (with Explanation):
1) Look to see if the number of protons is greater than 83. If it is, predict alpha decay (even though this won't
always be the actual decay). If p is less than (or equal to) 83, then:
2) First calculate the n/p ratio for the nuclide (although see Note in #2 below for the absolute “quickest”
approach, although you need to make sure you don’t forget what an n/p ratio is!).
3) Then assess the approximate "stable" n/p ratio for the proton number of the nuclide that you're dealing with.
(I.e., find out "where you are" on the "zone of stability" plot.)
Note: As I mentioned in class (and as Tro mentions on p. 873), there is another (“shortcut”) way to get a rough idea of
where the approximate “stable” n/p ratio for a nuclide is. You can look at the average atomic mass of the element on
the periodic table, round it to the nearest integer, and assume that the nuclide with that mass number is likely to be
stable! For example, I told you that near Z = 83, the stable n/p ratio is about 1.5. If you look at Au, and round its
average atomic mass of 196.967 up to 197 and take that as the mass number of a stable nuclide of Au (which has 79
p’s in every nucleus), the number of neutrons in that nuclide would be 197-79 = 118, and thus the n/p ratio would be
1.49, which is very close to 1.5! This works most of the time because the average atomic mass is the weighted
average of the naturally occurring isotopes on Earth, which means those isotopes that are reasonably stable and thus
lie in or near the zone of stability.
3) Compare the actual n/p ratio of each nuclide with the "stable" n/p ratio to see whether or not it has "too many
neutrons" (actual n/p larger than "stable" n/p) or "too few neutrons" (actual n/p smaller than "stable" n/p).
Then you can make your conclusion:
PS10-3
Answer Key, Problem Set 10
a) If there are "too many neutrons" to be stable, the nuclide is expected to turn a neutron into a proton,
which is associated with beta decay.
b) If there are "too few neutrons" to be stable, the nuclide is expected to turn a proton into a neutron, which
is associated with either positron emission or electron capture. NOTE: Alpha decay is also a possibility since
it will also get you somewhat closer to the zone of stability, but it is much rarer, and you need never predict it.
Application of Strategy:
(a) For the F example, clearly Z < 83 since Z = 9. So no -decays are predicted. There is no need to
calculate an n/p ratio since they tell you that the only stable isotope is F-19. That being said, F-21
must have “too many neutrons” to be stable, and so -decay (n → p) is predicted. And similarly,
F-17 and F-18 must have “too few neutrons”, and so positron emission or electron capture are
predicted.
(b) For 210Po, the atomic number is 84, so -decay is predicted.
(c) For 195Au, the atomic number is 79 which is < 83, so no -decay is predicted. Using the “shortcut” method, the assumed stable isotope of Au is Au-197 since the average atomic mass is close
to 197 amu. So Au-195, which has fewer neutrons than in Au-197, is predicted to have “too few
neutrons” and is predicted to decay by the process that turns a p into an n—positron emission or
electron capture. If asked to consider the n/p ratio, you should be able to do so. Here, n/p =
(195 – 79)/79 = 1.47. 79 is very close to the “end of the line” which is at Z = 83 and at which n/p is
close to 1.5. Although you might say this one is “too close to call”, if pushed to guess, you’d have
to say that n/p was too small—it clearly isn’t too large! So again, you’d predict the decay that
would turn a proton into a neutron.
5. NT5.
Complete the following nuclear reactions. [All have been used to synthesize elements.]
1
12
1
(a) ____  42 He  243
(b) 238
97 Bk  0 n
92 U 
6 C  ____  6 0 n
(c)
249
98
Cf  ____ 
240
95
Answers: (a)
Am
260
105
(b)
(d)
Db  4 01 n
244
98
Cf
(c)
15
7
249
98
N

Cf
10
5
B 
257
103
Lr  2 ____
(d) 01 n
Work / Reasoning (using conservation rules as in Problem 2):
(a) ____ 
4
2
He 
243
97
Bk 
1
0
n
Answer:
240
95
Am
Conservation of Mass Number: x + 4 = 243 + 1  x = 244 – 4 = 240
(b)
238
92
Conservation of Charge:
y + 2 = 97 + 0  y = 97 – 2 = 95  Am
U 
Answer:
12
6
C  ____  6 01 n
244
98
Cf
Conservation of Mass Number: 238 + 12 = x + 6(1)  x = 250 – 6 = 244
Conservation of Charge:
(c)
249
98
Cf  ____ 
260
105
Db  4 01 n
92 + 6 = y + 6(0)  y = 98  Cf
Answer:
15
7
N
Conservation of Mass Number: 249 + x = 260 + 4(1)  y = 264 – 249 = 15
Conservation of Charge:
(d)
249
98
Cf 
10
5
B 
257
103
Lr  2____
98 + y = 105 + 4(0)  y = 105 – 98 = 7
Answer:
1
0
n
Conservation of Mass Number: 249 + 10 = 257 + 2x  2x = 259 – 257 = 2  x = 2
Conservation of Charge:
98 + 5 = 103 + 2y  2y = 103 -103 = 0  y = 0
PS10-4
Answer Key, Problem Set 10
6. NT6.
-1
Radioactive copper-64 decays with a half-life of 12.8 days. (a) What is the value of k in s ? (b) A sample
64
contains 28.0 mg Cu. How many decay events will be produced in the first second? Assume the atomic mass
64
64
of Cu is 64.0 (amu). (c) A chemist obtains a fresh sample of Cu and measures its radioactivity. She then
determines that to do an experiment, the radioactivity cannot fall below 25% of the initial measured value. How
long does she have to do the experiment?
Answers: (a) 6.27 x 10-7 s-1
(b) 1.65 x 1012 dps
(c) 25.6 days
Work/ Reasoning:
-1
(a) What is the value of k in s ?
k 
ln 0.5 0.693


t 1/2
t 1/2
0.693
 6.266 x 10 -7 s -1  6.27 x 10 - 7 s -1
24 h 60 min 60 s
12.8 d x
x
x
d
h
min
Remember: All nuclear decays are first order, so the above relationship applies.
64
(b) A sample contains 28.0 mg Cu. How many decay events will be produced in the first second? Assume the
64
atomic mass of Cu is 64.0 (amu).
First order  R = kN. or A = kN
NOTE: In this key, A is often used for rate because (radio)activity is
often used as a synonym for “rate of decay”).
Since we know k (part (a)), to get R at any given time, we just need N at that time. In the first
second, N is the initial value—i.e., the number of nuclei in the 28.0-mg sample of 64Cu:
28.0 mg x

0.001 g 1 mol 6.022 x 10 23 atoms Cu 1 nucleus
 2.635 x 10 20
x
x
x
atom
mol atoms
64.0 g
mg

A = 6.266 x 10-7 s-1 2.635 x 10 20
(c)
64
64
Cu nuclei (at t = 0)

Cu nuclei  1.651 x 1014 dps (or cps)
64
A chemist obtains a fresh sample of Cu and measures its radioactivity. She then determines that to do an
experiment, the radioactivity cannot fall below 25% of the initial measured value. How long does she have to
do the experiment?
Explanation, Quick Way (for this problem, because of a special case)
For first order nuclear processes, the half-life is a constant (because it is first order). Thus, the
time it takes for the number of radioactive nuclei to become half of what it was originally is “one
half-life”, and the time it takes for the number to become one half of that, or ¼ the original, is “two
half-lives”. Since decay rate is proportional to number of nuclei, it also takes two half-lives for the
sample’s decay rate (radioactivity) to drop to ¼ (=25%) of its original value. So in this problem,
the chemist has two half-lives’ worth of time to do her experiment. Since the half-life is given as
12.8 days, she has 2 x 12.8 = 25.6 days.
Explanation, More General:
A more general solution, particularly useful when the time is not an integral number of half-lives,
is as follows. Whenever you have a question that relates “time” and “amount (or fraction)
remaining”, you should recognize that the appropriate equation to use is the “integrated rate law”,
and you should UNDERSTAND THE MEANING OF THIS EQUATION, as well as keep track of
units (as always!):
Fraction of
radioactive nuclei
remaining at time t
Nt
 e  kt
No
Remember that there are other equivalent “versions” of
this equation. For example, Tro uses
N
ln t
 No

  kt


Equation [19.3]
ALSO remember that since the activity (decay rate) is proportional to the amount of nuclide
present (A = kN), the ratio of activities is equal to the ratio of “number of nuclei”. That is:
PS10-5
Answer Key, Problem Set 10
At kNt Nt


 e  kt
Ao kNo No
or in other words,
At
 e  kt
Ao
This means that the fraction of the activity remaining can be used in the integrated rate law just
like “fraction of nuclei remaining” is.
In this problem, the % remaining is 25%, which equals 0.25 in terms of a fraction (or decimal).
Thus:
0.25  e ( 6.27 x 10
2.21 x 10 6 s x
-7
s-1 )t
 ln(0.25) = -(6.27 x 10-7 s-1)t  t =
ln(0.25)
 2.21 x 106 s
6.27 x 10 - 7 s-1
1 hr
1 day
1 min
x
x
 25.6 days (which is the same answer as above [2 x t1/2])
60 s 60 min 24 hr
7. NT7.
3
Fresh rainwater or surface water contains enough tritium ( 1H ) to show 5.5 decay events per minute per 100. g of
water. Tritium has a half-life of 12.3 years. Pretend that the year is 2009 (rather than the current year). You are asked
to check a vintage wine that is claimed to have been produced in 1946. How many decay events per minute should
you expect to observe in 100. g of that wine? (Again, assume it is 2009 right now.)
Answer: 0.16 dpm
Reasoning/Work:
One needs to look at this is as a variant of a “What is t?” kind of problem. That is, typically in this
kind of problem we want to “date” something. So the question really becomes “How long has it
been decaying?” The answer, which is a time, ends up being an “age”. And that time is the “t” in

   0.693 t
At
0.693
t
 kt 

the integrated rate law:
, because k 
for first order processes  . In this problem,
e
e 


Ao
t 1/2


however, we are not asked for time, but effectively are given the time and asked to predict a value
of activity. The same integrated rate law applies. The only “thorny” issues here are that you have
to make some assumptions to solve the problem. One is that you have to assume that 100. g of
wine is essentially 100% water, which is obviously not true, but is probably not too far off (our
bodies are mostly [but not 100%!] water as well!). Also, you must assume that the water ultimately
got into the wine via rainwater or surface water (because it came from grapes, which got their water
from “nature” as the vines grew). The idea is that once the wine is bottled, no new water gets
added, so the tritium in the water just decays as the wine sits. The activity will go away at a rate
dictated by its k (and thus half-life), and thus “now” (i.e., in 2009) the rate will be lower than it was
originally since there aren’t as many radioactive nuclei left (since they’ve been decaying since
1946—about 63 years!). The activity “now” will be related to the original activity (which is assumed
to be equal to the current activity of “fresh” rainwater) by the integrated rate law, so you can plug
into that equation as follows to find the current (2009) activity (predicted):
1/ 2
 0.693 
t
t1 / 2 

Anow
e 
Ao
8. NT8
 0.693 
 0.693 
  2009 1946 y 
 63 y 
 
 
Anow
12.3 y 
12.3 y 

e 
 Anow  5.5 dpm e 
 0.158  0.16 dpm
5.5 dpm
14
12
Assume a constant C/ C ratio of 13.6 counts per minute per gram of living matter. A sample of petrified tree
14
was found to give 1.2 counts per minute per gram. How old was the tree? (t1/2 of C is 5730 years.)
Answer: 20100 y
Work/Reasoning:
Use the integrated rate law again, as in the prior two problems. The ratio of activities is equal to e-kt:
PS10-6
Answer Key, Problem Set 10
 0.693 
t
t1 / 2 

Anow
e 
Ao
 0.693 

 
 t
1.2 cpm
 e  5730 y   0.08824  e  0.0001209 y t
13.6 cpm
-1
 ln(0.08824) = -(0.0001209 y-1)t  t =
-2.4277
 20080 y  20100 y
- 0.0001209 y -1
9. NT9.
A small atomic bomb releases energy equivalent to the detonation of 20,000 tons of TNT; a ton of TNT releases
9
13
235
4 x 10 J of energy when exploded. Using 2 x 10 J/mol as the energy released by fission of U, (a) approximately
235
what mass of U undergoes fission in this atomic bomb? (b) Does the bomb have more mass or less mass after the
explosion? By how much (in mg)?
(a) 20000 ton TNT x
4 x 10 9 J 1 mol 235 U
235 g
x
x
 940 g
ton TNT 2 x 1013 J mol 235 U
235
U
(b) Answers: Less mass; ~900 mg less
Reasoning: Because of the concept of “mass / energy interconversion” (Einstein), when energy
leaves a system, mass is lost as well! We interpret Einstein’s E = mc2 equation to mean that when
energy leaves a system, it is said to have come from the conversion of a little bit of its mass.
Conversely, when energy is added to a system, mass is added. Any exothermic process must be
associated with a mass loss in the system; any endothermic process results in a main gain.
The magnitude of the mass loss/gain depends on the change in energy according to: E = mc2.
In this case, E = -20000 ton x (4 x 109 J/ton) = -8 x 1013 J (sign is negative because process is
exothermic), so:
m 
E
 8 x 1013 J

c2
2.9979 x 10 8 m/s

- 0.000890 kg x

2
 0.000890 kg
1000 g 1000 mg
x
 890  900 mg  900 mg was lost
kg
g
The negative sign indicates that 900 mg of mass were lost (converted into energy) when the
bomb exploded (reacted). That’s nearly 0.1% of its mass! That is not insignificant!
10. NT10.
(a) What is meant by the terms “binding energy” and “mass defect”?
Binding energy (Eb [my abbreviation]) is the energy required to break apart a nucleus completely into free
protons and neutrons. Mass defect is the mass increase that occurs when a nucleus is completely
broken apart into free protons and neutrons (or the mass loss that accompanies the formation of a
nucleus from free protons and neutrons).
NOTE: Tro’s definition is consistent with mine, although he phrases it differently, as the “difference in mass” between a
nucleus and the “sum of the separated particles”.
(b) Describe clearly in words how you could calculate the binding energy for a nuclide using the precise masses of a
nuclide as well as the mass of a (free) proton and a (free) neutron. Hint: write out the nuclear equation corresponding
to the binding energy.
Binding energy and mass defect are associated with the same process (described in part (a)).
For a nuclide with Z protons, and A – Z neutrons (A is the mass number), the equation that
represents the process associated with both BE and mass defect is:
X  Z11p  (A - Z)01n ;
E  E b ,
d
n
a
A
Z
m  mass defect
Since Eb and m are associated with the same process, they must be related by E  mc2 for a
given nuclide (In fact, this is how Tro defines (nuclear) binding energy). Thus, to calculate binding energy (if
the mass of the nuclide is known), write out the equation corresponding to the break up of the
nucleus of interest into its constituent nucleons. Find the difference in mass between the (sum of
PS10-7
Answer Key, Problem Set 10
the masses of the) free nucleons and the mass of the nucleus. Then calculate binding energy
using (Eb ) E = mc2 (being careful with units!  )
11. NT11.
(a) What is the difference between fusion and fission? (define each)
Fusion refers to the process in which two (or more) nuclei come together to form one (larger)
nucleus. Fission refers to the process in which one nucleus splits apart to form two (or more)
smaller nuclei.
(b) If the binding energy of Ar-40 is 343.8 MeV, what is the binding energy per nucleon?
Eb
binding en ergy
343.8 MeV


 8.595 MeV/nucleon
nucleon number of nucleons 40 nucleons
NOTE: The number of nucleons
equals the mass number (A),
since A  n + p (by definition)
(c) What is the difference between “binding energy” and “binding energy per nucleon”? Which one helps you predict
whether or not a nuclide will undergo fission or fusion to lower its energy?
Binding energy is defined in NT10; the binding energy per nucleon is the binding energy for a
particular nuclide divided by the number of nucleons in the nuclide (see part (b) above).
It is the binding energy per nucleon which indicates how thermodynamically stable a nuclide is
(compared to other nuclides). If the binding energy per nucleon for a nuclide is smaller than the
maximum in Figure 19.12, then the nuclide is not as stable as it could be by either breaking up (if
the nuclide has a mass number larger than ~56) or fusing together with another (if the nuclide has a
mass number smaller than ~56).
(d) Do larger nuclides always have larger binding energies? Do larger nuclides always have larger binding energies
per nucleon?
Answers: Yes, No
Explanation / Reasoning:
Larger nuclides always have larger binding energies than smaller nuclides (because with each
“extra” nucleon added / bound, energy must be lowered [because nucleons attract one another!]), but they
do NOT always have the largest binding energy per nucleon. After a certain point, it becomes less
stable (per nucleon) to continue to get bigger. This is what is shown in Figure 19.12.
There may seem to be a discrepancy here, but there isn’t. Comparing the binding energy of one
nucleus of U-238 to one nucleus of Fe-56 is like comparing “apples to oranges” since the number of
nucleons is not the same. So even though U-238 has a larger binding energy than Fe-56, it is not
“more stable” because if the nucleons were rearranged into four Fe-56 nuclei (plus some extra
nucleons), energy would be lowered. How can you know this without actually calculating the E for
this hypothetical reaction? You just look at the binding energy per nucleon and that tells you this
must be true! If the same number of nucleons is involved, it is clear that the nucleus having the larger
binding energy per nucleon will be the “lower-energy” (overall) arrangement of nucleons. (because
E/nucleon x # nucleons  Eoverall). But even if the number of nucleons is not the same, the Eb per
nucleon will reflect intrinsic “stability” of a nucleus. LOOK AT THE POWERPOINT SLIDES IN WHICH
I ASK “WHAT IS THE MOST STABLE WAY TO ARRANGE 30 (OR 240) NUCLEONS” TO HELP
MAKE THESE IDEAS CLEAR TO YOU.
(e) Of the nuclides in the table shown on the first page of this PS10 sheet, which one(s) are more thermodynamically
stable than Ar-40? [see part (b)] (Note: You may need to calculate some values omitted from that table.)
Answer: Ni-60 is the only one, since it is the only one that
has a Eb/nucleon greater than 8.595
MeV/nucleon.
Note: You had to calculate the Eb/nucleon values for three of the nuclides
in the table by dividing Eb by the mass number ( # of nucleons)—I
put those in bold underlined italics in the table to the right.
PS10-8
Nuclide
He-5
Be-10
N-15
Mg-30
Ni-60
Cd-120
Cm-240
Mass #
5
10
15
30
60
120
240
Eb (MeV)
24.41
64.98
115.49
241.6
526.8
1015
1810
Eb/nucleon
4.88
6.50
7.70
8.05
8.78
8.46
7.54
Answer Key, Problem Set 10
12. NT3.
3
1
2
3
2
(a) Calculate the binding energy per nucleon for 1 H and 1 H . The atomic masses (in amu) are 1 H , 2.01410, and
H , 3.01605. melectron  0.000549 amu; mp 1.00728 amu; mn  1.00866 amu
Answers:
2
1
H , 1.78 x 10-13 J OR 1.11 MeV
3
1
(per nucleon)
H , 4.53 x 10-13 J OR 2.83 MeV (per nucleon)
Strategy:
1) Write out the balanced equation corresponding to breaking up the nuclide into free nucleons.
2) Find m  mnucleons - mnuclide .
NOTE: When matom (atomic mass) is given, I find mnuclide by subtracting the mass of the electrons
from matom: mnucleus [w/o e’s]  matom [w/ e’s] - melectrons. This is how I did it in class. I find
this more straightforward than what Tro does.
Tro uses the mass of an H atom in place of the mass of a proton, and then uses the
mass of the atom (without subtracting away the mass of the electron(s)):
m  mH atoms + neutrons - matom
3) Convert m (which will be in units of amu [per nucleus]) into kg [per nucleus] using:
1 amu = 1.6605 x 10-27 kg (to 5 SF)
4) Substitute into E = mc2 (here, it ends up being Eb  m.d. x c2). Recognize that units will be J
(if c is used in units of m/s and m.d. in kg), and that this is still per nucleus.
5) For binding energy per nucleon, recognize that you must divide Eb by the number of nucleons.
You can do that at this point if you want the result in J/nucleon. If you want the result in
MeV/nucleon, you can convert J to MeV first, and then divide by the number of nucleons.
(Actually, it doesn’t matter the order in which you do the conversion from J to MeV. I just prefer
to do my “division by the # nucleons” last.)
Use: 1 MeV = 1.6022 x 10-13 J
Execution of Strategy for 21H :
2
1
H 
p 
1
1
1
0
n;
E  Eb ( 12H ), and m  mass defect(of 12H )
 m.d.  m  m (1p + 1n) - mH-2 nucleus  (1.00728 + 1.00866) – (2.01410 – 0.000549)  0.002389 amu
Convert to kg: 0.002389 amu x
1.6605 x 10 -27 kg
 3.9669..x 10 - 30 kg (per nucleus)
amu
Eb  m.d. x c2  (3.966.. x 10-30 kg)(2.9979 x 108 m/s)2  3.565.. x 10-13 J (per nucleus)
Eb per nucleon =
3.565 x 10 -13 J
 1.7825 x 10 -13  1.78 x 10-13 J/nucleon
2 nucleons
OR (converting Eb to MeV first):
Eb = 3.565.. x 10 -13 J x
Eb per nucleon =
MeV
 2.225.. MeV (per nucleus)
1.6022 x 10 -13 J
2.225.. MeV
 1.112..  1.11 MeV/nucleo n
2 nucleons
PS10-9
(Note: this matches the
value in Fig. 19.12)
Answer Key, Problem Set 10
Execution of Strategy for 31H :
H 
3
1
p  2 01n
1
1
 m.d.  m  m (1p + 2n) - mH-3 nucleus  (1.00728 + 2(1.00866)) – (3.01605 – 0.000549)  0.009099 amu
Convert to kg: 0.009099 amu x
1.6605 x 10 -27 kg
 1.5108..x 10 - 29 kg (per nucleus)
amu
Eb  m.d. x c2  (1.5108.. x 10-29 kg)(2.9979 x 108 m/s)2  1.357.. x 10-12 J (per nucleus)
Eb per nucleon =
1.357 x 10 -12 J
 4.526.. x 10 -13  4.526 x 10 - 13 J/nucleon
3 nucleons
OR (converting Eb to MeV first):
Eb = 1.357.. x 10 -12 J x
Eb per nucleon =
MeV
 8.475.. MeV (per nucleus)
1.6022 x 10 -13 J
8.475.. MeV
 2.825..  2.83 MeV/nucleon
3 nucleons
2
3
(b) Which nuclide is more thermodynamically stable, H or H? Explain briefly.
Answer: 3H is more stable.
Explanation: A greater Eb per nucleon means “more thermodynamically stable”. It represents
an average amount of “energy lowering” per nucleon in that particular nucleus, and lower energy
(per nucleon) implies more stable.
3
3
NOTE: In this case, the Eb itself also happens to be greater for H but that is not why H is more stable. One must always
compare Eb per nucleon to assess relative thermodynamic stability of two nuclei.
PS10-10
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