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### Chapter 18

Nuclear Chemistry

Introduction

- So far we’ve studied chemical reactions where only electrons have changed.
- Chemical properties are determined by electrons!
- Nucleus was not primarily important in these reactions, as it did not undergo any changes.
- Identities remained the same in chemical reactions because protons remained the same.

- This is no longer true in nuclear reactions!

Introduction

- Nucleus is extremely small, dense, and contains a huge amount of energy.
- Millions of times more E than chemical reaction.

- Nucleus = neutrons + protons
- Made of even smaller parts, such as quarks.

- AZX where A = mass # & Z = charge/# of protons.
- Isotopes = atoms of the same element with different #’s of neutrons (protons stay the same).

Types of decay

- Already discussed these in the packet you completed.
- Additional note: decay types can be broken into two categories: those that change mass # and those that don’t.
- Changes mass #: alpha emission.
- Giving off a He atom decreases mass.
- Example: 23892U 42He + 23490Th

- This is a type of spontaneous fission – splitting a heavy nuclide into 2 lighter nuclides.

- Giving off a He atom decreases mass.

Types of Decay

Do not change mass #: particle emitted/used has no mass (mass # = 0).

- Beta emission: 13153I 0-1e + 13154Xe
- Gamma ray emission:
23892U 42He + 23490Th + 00

- Positron emission:
2211Na 01e + 2210Ne

- Electron capture:
20180Hg + 0-1e 20179Au + 00

Penetration Ability

- Radiation emitted has different levels of penetration.
- The more penetrating the emission, the more dangerous.
- The order is as follows:
alpha < beta/positron < gamma ray

- Therefore, alpha particles are least penetrating and gamma rays are by far most penetrating.

AP Practice Question

Which of the following statements is true about beta particles?

- They are electrons with a mass number of 0 and a charge of -1.
- They have a mass number of 0, a charge of -1, and are less penetrating than α particles.
- They are electrons with a charge of +1 and are less penetrating than α particles.
- They have a mass number of 0 and a charge of +1.

AP Practice Question

When 22688Ra decays, it emits 2 α particles, followed by a β particle, followed by an α particle. The resulting nucleus is:

a) 21283Bi c) 21482Pb

b) 22286Rn d) 21483Bi

- Total of 3 α particles, so subtract 12 from mass # and 6 from atomic #: 21482Pb
- β particle means get rid of a neutron and add a proton (and an electron): 21483Bi

AP Practice Question

The formation of 23090Th from 23492U occurs by:

- Electron capture.
- α decay.
- βdecay.
- Positron decay.

AP Practice Question

An atom of 23892U undergoes radioactive decay by α emission. What is the product nuclide?

- 23090Th.
- 23490Th.
- 23092U.
- 23091Pa.

Nuclear Stability

- If a nucleus is unstable it will undergo radioactive decay to become stable.
- Can be tricky to determine if a nuclide is stable and how it will decay, but several generalizations have been.
- Note: nuclide = a specific nucleus of an isotope or atom.

Nuclear Stability

- Of 2000 known nuclides, only 279 are stable.
- Tin has the greatest number of stable isotopes at 10.

- Ratio of neutrons: protons determine stability.

- What is the stable ratio of n:p+ at the lower end of the belt?
- What is the stable ratio of n:p+ at the upper end of the belt?
- Based on this information, what can you conclude about the ratio of n:p+ in stable nuclides?
- This can be found on pg. 842 in your textbook.

Belt/Zone of Stability

- Developed by plotting # of neutrons vs. # of protons of known, stable isotopes.
- Low end of the belt shows a stable ratio of about 1n:1p+.
- As the belt gets higher (more protons), the stable ratio begins to increase to about 1.5n:1p+.
- For isotopes with less than 84 p+, the ratio of n:p+ is a good way to predict stability.

Ways to Predict Stability

- For light nuclides (<20 p+), 1:1 ratio of n:p+ are stable.
- For heavier nuclides (20 to 83 p+) ratio increases (to ~1.5:1).
- Why?
- More neutrons needed to stabilize repulsive force of more protons.

- All nuclides with 84 or more p+ are unstable (because they’re so big).
- Alpha decay occurs- giving off a He atom lessens both mass and atomic #.

- Even #’s of n & p+ are more stable than odd #’s.

Ways to Predict Stability Cont.

- Magic Numbers: certain #’s of n or p+ give especially stable nuclides.
- 2,8,20,28,50,82,126
- A nuclide with this number of n or p+ would be very stable.
- If there is a magic number of n & p+, this is called a double magic number (usually seen in heavier nuclides were extra stability is needed).

- The stability of magic numbers is similar to atoms being stable with certain #s of e-
- 2(He), 8(Ne), 18(Ar), 36(Kr), 54(Xe), 86(Rn)

Examples: Predicting Stability

1) Is the isotope Ne-18 stable?

- 10p+ and 8n 0.8:1
- Ratio is <1:1, so it is unstable!
- Undergoes decay to either increase #n & decrease #p+
2) Is the isotope 126C stable?

- Yes! Ratio = 1:1 (also even # n & p+)
3) Is the isotope sodium-25 stable?

- Ratio of n:p+ = 14:11 1.3:1
- Not stable! How will it become stable?
- Decrease #n beta emission

Section 1 Homework

Pg. 869 #3, 4, 12, 13, 20

18.2

The Kinetics of Radioactive Decay

Rate of Decay

- Rate of decay = - change in #of nuclides
- Negative sign = number decreasing
- Tough to predict when a certain atom will decay, but if a large sample is examined trends can be seen.

- Trends indicate that radioactive decay follows first-order kinetics.
- Thus first-order formulas are used!

change in time

Rate of Decay

- Two formulas are used to solve calculations involving decay and half-life:
- ln[A]t – ln[A]0 = -kt
- t1/2 = ln2
- Usually takes 10 half lives for a radioactive sample to be ‘safe’.
- Half-lives can be seconds or years!

- Formulas above are used when multiples of half lives are not considered.

k

Half Life- ‘Easy’ Example

- Example #1: I-131 is used to treat thyroid cancer. It has a half life of 8 days. How long would it take for a sample to decay to 25% of the initial amount?
- Each half-life cycle decreases the initial amount by half: after 8 days, one half-life, 50% remains. After another 8 days, 25% remains. Thus it would take 16 days.

Half Life- ‘Easy’ Example

- Every half-life cycle will follow the following order of percentages of radioactive isotopes that remain :
100%, 50%, 25%, 12.5%, 6.25%, 3.12%, etc.

- If a question asks about half-life and involves one of these ‘easy’ percentages, you can simply count how many half-life cycles have occurred.
- However, not all are this simple!

Half Life- ‘Harder’ Example

- Example #2: What is the half-life of a radioactive isotope (radioisotope) that takes 15min to decay to 90% of its original activity?
- 90% is not a multiple of half-life cycles, so we need to use the previously mentioned equations to calculate this.

Half Life- ‘Harder’ Example

- Steps:
- Use ln[A]t – ln[A]0 = -kt to solve for the rate constant, k.
- Often times specific amounts/concentrations will not be given! Usually given as percentages of original sample left over. Just assume 100 as the original ([A]0) and use the percent asked about as [A]t.

- Then use t1/2 = ln2 & value of k from above
to solve for the half life (units will vary

depending on what is given in the problem).

- Use ln[A]t – ln[A]0 = -kt to solve for the rate constant, k.

k

Half Life- ‘Harder’ Example

- Example #2: What is the half-life of a radioactive isotope (radioisotope) that takes 15min to decay to 90% of its original activity?
- ln(90) – ln(100) = -k(15min) k = 0.00702/min
- t1/2 = ln2/0.00702min-1 t1/2 = 98.7min

Another Way to Use Half-Life

- If both the half-life of a radioactive isotope is given and the amount of radioactive substance remaining, the amount of time it took for this substance to decay to this point can be calculated.
- First, use t1/2 formula to find k.
- Then, use other formula to solve for t.
- This is how carbon dating (uses radioactive C-14 isotope) is used to determine ages of objects!

Carbon-Dating Example

- If a wooden tool is discovered, and its C-14 activity has decreased to 65% of its original amount, how old is the tool? The half-life of C-14 is 5,730 years.
- 5,730yr = ln2/k k = 1.21 x 10-4/year
- ln65 – ln100 = -(1.21 x 10-4/year)t
t = 3,600 years

Mass-Energy Relationships

- During nuclear reactions and nuclear decay, energy is given off.
- Gamma rays, x-rays, heat, light, and kinetic E

- Why does E always accompany these reactions?
- Small amount of matter is turned into E.
- Law of conservation of matter is not followed during nuclear reactions!

Mass-Energy Relationships

- Einstein’s equation is used to perform calculations involving the mass-energy change in nuclear reactions: E = mc2.
- E = energy released
- m = mass converted into energy (units need to be in kg in order to get J as unit of E)
- c = speed of light = 3.00 x 108 m/s

- Note that the amount of matter turned into E in a nuclear reaction is small, but is amplified by the speed of light to produce a lot of E!

Sample Calculation

- When one mole of uranium-238 decays into thorium-234, 5 x 10-6kg of matter is changed into energy. How much energy is released during this reaction?
E = (5 x 10-6kg)(3.00 x 108m/s)2

E = 5 x 1011 J

Section 2 Homework

- Pg. 870 # 21, 24, 35

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