Chapter 18
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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.

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Chapter 18

Chapter 18

Nuclear Chemistry


Introduction

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!


Introduction1

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).


Chapter 18

18.1

Radioactive Decay &

Nuclear Stability


Types of decay

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.


Types of decay1

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

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

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 question1

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) 21283Bic) 21482Pb

b) 22286Rnd) 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 question2

AP Practice Question

The formation of 23090Th from 23492U occurs by:

  • Electron capture.

  • α decay.

  • βdecay.

  • Positron decay.


Ap practice question3

AP Practice Question

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

  • 23090Th.

  • 23490Th.

  • 23092U.

  • 23091Pa.


Nuclear stability

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 stability1

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.


Chapter 18

  • 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

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

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.


Chapter 18

Pg. 843 in textbook


Ways to predict stability cont

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

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

Section 1 Homework

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


Chapter 18

18.2

The Kinetics of Radioactive Decay


Rate of 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 decay1

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

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 example1

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

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 example1

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).

k


Half life harder example2

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

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

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

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 relationships1

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

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

Section 2 Homework

  • Pg. 870 # 21, 24, 35


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