Chapter 19 - Nuclear Chemistry Applications

Rates of Radioactive Decay

Half-life - The time it takes for half of the parent nuclides in a radioactive sample to decay to the daughter nuclides

The amount that remains after one half-life is always one-half

of what was present at the start.

The amount that remains after two half-lives is one-quarter of what was present at the start.

A radioactive sample does not decay to zero atoms in two half-lives—You can’t add two half-lives together to get a “whole”

life.

The Concept of Half-life

A plot of the number of Th-232 atoms in a sample initially containing 1 million atoms

as a function of time.

Th-232 has a half-life of 14 billion years.

Half-Life

Each radioactive nuclide has a unique half-life that is not affected by physical conditions or chemical environment.

Radioactive Decay Half-Life

“I-131 decays by beta emission with a half life of 8 days.” ?

131 53

I0

-1e

131 54

Xe➔ +

1) What is meant by

131 53 I1.000 g

131 54 Xe

131 53 I0.500 g

0.500 g131

54 Xe

131 53 I0.250 g

0.750 g131

54 Xe

131 53 I0.125 g

0.875 g

8 days

8 days

8 days

2) The half-life of the beta particle emitter tritium, 3H, is 12 years. How much of a 1.00 g sample of 3H remains after 48 years?

1.00 g ➝0.50 g ➝0.250 g ➝ ➝0.125 g 0.0625 g

Radioactive Decay

n  =  t/t½        (t½  =  half-­‐life)  Nt/No  =  0.5n  

Half-Life

half of the radioactive atoms decay each half-life

First Order Reactions

Rate = k[A] ln[A] = -kt +ln[A]0 (integrated rate law)

graph of ln[A] vs t is straight line (slope = -k and y intercept = ln[A]0)

t½ = ln2/k = (0.693)/k (constant half-life)

•  Rate = k[A]0 = k

 constant rate reactions

•  [A] = -kt + [A]0

•  graph of [A] vs. time is straight line with slope = -k and y-intercept = [A]0

•  t ½ = [A0]/2k

•  when Rate = M/sec, k = M/sec

[A]0

[A]

time

slope = - k ln[A]

ln[A]0

ln[N]t - ln[N]0 = -kt

Radioactive Decay-A First Order Process

[N]t

[N]0ln = -kt [N]t = [N]0 x e-kt

t½ = 0.693 k

k = 0.693

t½

ln[N]t = -kt + ln[N]0 [N] = number of radioactive nuclei [N] = intensity of radioactivity

ln[A]t = -kt + ln[A]0 [A]t = conc A at time t[A]0 = conc A at time 0

3) If you have a 1.35 mg sample of Pu-236, calculate the mass of Pu-236 that will remain after 5.00 years.

t½ = 0.693

k k =

0.693 t½ =

0.693 2.86 y = 0.242 yr-1

ln = Nt N0

-kt

Nt = N0 e-kt = N0 e-(0.242 yr-1)(5.00 yr)

Nt = N0 e-kt = (1.35 mg)e -(0.242 yr-1)(5.00 yr)

Nt = N0 e-kt = 0.402 mg

4) Radioactive radon-222 decays with a loss of one α particle. The half-life is 3.82 days. What percentage of the radon in a sealed vial would remain after 7.0 days?

t½ = 0.693

k k = 0.693 t½ =

0.693 3.82 d = 0.181 d-1

ln = Nt N0

-(0.181 d-1)(7.0 d)

Nt N0

= e-kt = e-(0.181 d-1)(7.0 d)

= e -(1.27)

Nt N0

= 0.28 = 28%

Artifact Dating

Mineral (geological)

Compare amount of U-238 (t½ = 4.5 x 109 yr) to Pb-206

Compare amount of K-40 (t½ = 1.25 x 109 yr) to Ar-40

Archeological (once living materials)

Compare amount of C-14 (t½ = 5730 yr) to C-12

While a substance is living C-14/C-12 ratio is constant (CO2 exchange with the atmosphere continues).

When an organism dies, C-14/C-12 ratio decreases.

Useful to up to about 50,000 yr

5) An artifact contains 12.5% of the original amount of C-14. How old is this sample? (C-14 half-life is 5730 years.)

100 % ➝ 50 % ➝ 25 % ➝ ➝12.5 % 6.25 %

3 x 5730 = 17,200

% C-14 (relative to

living organism)

Number of

Half-Lives

Time

(yrs)

100.0 0 0

50.0 1 5,730

25.00 2 11,460

12.50 3 17,190

6.250 4 22,920

3.125 5 28,650

1.563 6 34,380

Radiometric Datingn = t/t1/2

t = time t1/2 = time for a half-life n = the number of half-lives

Nt/No = 0.5n

No = amount initially present Nt = amount at time t n = the number of half-lives

If we know what fraction of sample is left (Nt/No) and its half-life (t1/2), we can calculate how much time has elapsed.

6) A mammoth tusk containing grooves made by a sharp stone edge (indicating the presence of humans or Neanderthals) was uncovered at an ancient campsite in the Ural Mountains in 2001. The 14C/12C ratio in the tusk was only 1.19% of that in modern elephant tusks. How old is the mammoth tusk?

ln = 1.19 100

-(1.21 x 10-4 yr-1 )(t)

k = 0.693

5730 yr

k = 1.21 x 10-4 yr-1

k = 0.693

t½ ln =

Nt N0

-kt

(-4.43) /-(1.21 x 10-4) = t = 36,600 yr

Nt/No = 0.5n

.0119 = 0.5n

log(0.0119) = nlog(0.5)

-1.92 = (n)(-0.301)

n = 6.38 = # half-lives

yr = (6.38)(5730)

yr = 36,600

7) An ancient skull gives a 4.50 disintegrations / min gC. If a living organism gives 15.3 disintegration / min gC, how old is the skull ?

Kinetics of Radioactive Decay

•  Rate = kN

 N = number of radioactive nuclei

•  t1/2 = 0.693/k

•  the shorter the half-life, the more nuclei decay

every second – we say the sample is hotter

k = 1.21 x 10-4 yr-1

ln rate1 rate2

= -kt t =

4.50 15.3

dis/min gCdis/min gC

ln

-1.21 x 10-4 yr-1

t = 10,000 yr

Measuring Radioactivity

Quantities of Radiation

Parameter' Unit' Descrip/on'

Level'of'radioac/vity' Becquerel'(Bq)*' 1'disintegra/on/s'

Curie'(Ci)'3.7'×'1010'nuclear'disintegra/ons/s'

Ionizing'energy'absorbed'

Gray'(Gy)'1'Gy'='1'J/kg'of'/ssue'mass'

Amount'of'/ssue'damage'

Sievert'(Sv)' 1Sv'='1'Gy'×'RBE**'

*SI'unit'of'radioac/vity;'**Rela/ve'Biological'Effec/veness'

Biological Effects of Radioactivity

Sources of Radiation

Acute&Effects&of&Single&Whole3Body&Doses&of&Ionizing&Radia<on&

Dose&(Sv)&

Toxic&Effect&

0.05–0.25&

No&acute&effect,&possible&carcinogenic&or&mutagenic&damage&to&DNA&

0.25–1.0&Temporary&reduc<on&in&white&blood&cell&

count&

1.0–2.0&Radia<on&sickness:&fa<gue,&vomi<ng,&diarrhea,&impaired&immune&system&

2.0–4.0&Severe&radia<on&sickness:&intes<nal&bleeding,&

bone&marrow&destruc<on&

4.0–10.0&Death,&usually&through&infec<on,&within&

weeks&

>10.0& Death&within&hours&

Biological Effects of Radioactivity

Medical Applications of Radionuclides

Medical Applications of Radionuclides

Therapeutic Agents Imaging Agents

Positron Emission

C-11 B-11

β+

Positron = “antimatter”

Energy of matter/antimatter reaction related to mass defect.

Energy of nuclear reaction released as gamma rays.

What are positrons ?

Positron Emission

β+ antimatter

β- matter

photon 511 kev

photon 511 kev

Detector Det

ecto

r

Positron Emission Tomography

Positron Emission Tomography

PET study revealing differences in brain metabolism in recovering alcoholic (left, 10 days, and right, 30 days,

after withdrawal)

Nuclear Fission

On January 6, 1939, Meitner, Strassmann, and Hahn reported that the neutron bombardment of uranium resulted in nuclear fission—the splitting

of the atom.

U235 92n1

0 Ba142 56 K91

36 n1 03+ + +

Nonradioactive Nuclear ChangesSome nuclei are inherently unstable.

If their nuclei are hit by a neutron, the large nucleus splits into smaller nuclei

This is called fission

Small nuclei can be accelerated to such a degree that they overcome their charge repulsion and smash together.

A larger nucleus is formed.

This is called fusion

Both fission and fusion release large amounts of energy

Nuclear fission:

A nuclear reaction in which the nucleus of an element splits into two lighter nuclei;

The process is usually accompanied by the release of one or more neutrons and energy.

U235 92 n1

0 Ba142 56 K91

36 n1 03+ + +

U235 92 n1

0 Cs138 55 Rb96

37 n1 02+ + +

U235 92 n1

0 Te137 52 Zr97

40 n1 02+ + +

Fissionable Materials

U-235, Pu-239, and Pu-240

Natural uranium is less than 1% U-235

Mostly U-238

Not enough U-235 to sustain a chain reaction

To produce fissionable uranium, natural uranium must be enriched in U-235

to about 3% for “reactor grade”

to about 7% for “weapons grade”

Fission Chain Reaction

A chain reaction occurs when a reactant in a process is also a product of the process.

In the fission process, the neutrons are both.

Only a small number of neutrons are needed to start the chain.

Many neutrons produced in fission are either ejected from the uranium before they hit another U-235 or are absorbed by the surrounding U-238

The minimum amount of fissionable isotope needed to sustain the chain reaction is called the critical mass.

Fission Chain Reaction

Nuclear Power Plants: Controlled Fission

In essence, the Little Boy design consisted of a gun that fired one mass of uranium 235 at another mass of uranium 235, thus creating a supercritical mass. A crucial requirement was that the pieces be brought together in a time shorter than the time between spontaneous fissions. Once the two pieces of uranium are brought together, the initiator introduces a burst of neutrons and the chain reaction begins, continuing until the energy released becomes so great that the bomb simply blows itself apart.

Little Boy

The initial design for the plutonium bomb was also based on using a simple gun design (known as the "Thin Man") like the uranium bomb. The plutonium, however, contained small amounts of plutonium 240, an isotope with a rapid spontaneous fission rate. A gun-type bomb would not be fast enough to work. Before the bomb could be assembled, stray neutrons would have been emitted from the spontaneous fissions, and would start a premature chain reaction, leading to a great reduction in the energy released.

Seth Neddermeyer, a scientist at Los Alamos, developed the idea of using explosive charges to compress a sphere of plutonium very rapidly to a density sufficient to make it go critical and produce a nuclear explosion.

Nuclear Fusion

Nuclear fusion – nuclear reaction in which sub-atomic particles or atomic nuclei collide and fuse together, forming more massive nuclei and releasing energy.

Nuclear Fusion

Fusion is the combining of light nuclei to make heavier nuclei.

The source of the sun’s energy

Requires a high input of energy to initiate the process

Produces 10 times the energy of fission per gram

No radioactive byproducts

The only currently working application is the H-bomb.

Nuclear Fusion

Deuterium-Tritium Fusion Reaction

Artificial Transmutation

Artificial Transmutation

Bombardment of one nucleus with neutrons or another nucleus causing new atoms to form

Requires a “particle accelerator”

Ex: Tc-97 is made by bombarding Mo-96 with deuterium:

Mo96 42H2

1 Tc97 43 n1

0+ +

Formation of Transuranium Nuclides

Pu239 94He4

2 Am240 95 H1

1+ + n1 0+ 2

Pu239 94He4

2 Cm242 96+ + n1

0

Cm244 96He4

2 Bk245 97 H1

1+ + n1 0+ 2

U238 92C12

6 Cf246 98+ + n1

04

Es253 99He4

2 Md256 101+ + n1

0

Cf252 98B10

5 Lr256 103+ + n1

06

Cf249 98H2

1 Es248 99+ + n1

03 25 min

51 hr

163 d

5 d

36 h

76 min

28 sec