Chapter 20

Nuclear Chemistry

Glenn T. Seaborg 1912-1999.*

Transuranium elements.

Pierre and Marie Curie.

1859-1906,* 1867-1934.**

Discovered radium; defined “radioactivity.”

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Basics: Radioactivity

• • • • • •

Nuclear Equations

Nucleons : particles in the nucleus: p + : proton n: neutron.

Mass number A : the number of p + + n.

Atomic number Z : the number of p + .

Symbol: A z X ; e.g. 14 6 C is “carbon-14” Isotopes : have the same number of p + and different numbers of n (and therefore, different mass) In nuclear equations, the total number of nucleons is conserved: 238 92 U



234 90 Th + 4 2 He 2

Radioactivity

There are three

– –

types of radiation which we consider:



Radiation is the loss of 4 2 He from the nucleus,



Radiation is the loss of an electron from the nucleus

–

(electrons represented as either 0 -1 e or 0 -1 β)



Radiation is the loss of high-energy photon from the nucleus.

Example of α emission : Example of β emission :

238 92 U  234 90 Th + 131 53 I  131 54 Xe +  0 1 e 4 2 He

Example of γ emission :

43 Tc  99 43 Tc + 0 0 hν

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Radioactivity -

Separating the types of radiation (-) (++) α β γ _ 4 2 He nucleus electron high energy photons Charge 2+ 1 Mass(g) 6.64x10

-24 9.11x10

-28 0 0 Rel. mass 7,300 1 0 Rel. penetration 1 100 10,000 4

Radioactivity

Complete the following nuclear reactions :

32 16 S + 0 1 n  1 1 p + ____ 7 4

Be

+  0 1

e



_____

235 92 U + 0 1 n  135 54 Xe + 2 0 1 n + ____ 2 1

H

+ 2 1

H

 2 3

He

+

____

98 42 Mo + 2 1 H  0 1 n + ____

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Patterns of Nuclear Stability

Neutron-to-Proton Ratio

Neutron/proton ratio increases as atoms become larger Above 83 Bi, all nuclei are unstable and belt of stability ends.

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Radioactive Series

238

U Series

238 U 92  α  α 226 Ra 88 234 90 Th  α  β 222 Rn 86 234 91 Pa  α  β 234 U  α 92 218 Po 84  α 230 214 Pb 82 90 Th  214 83 Bi  β  β 210 Po 84  α 214 Po 84 206 82 Pb  α 210 82 Pb  β 210 83 Bi

Stable 7

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Nuclear Transmutations

Using Charged Particles - cyclotron

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Rates of Radioactive Decay

• • • •

Calculations Based on Half-Life

Radioactive decay is a first order process: Rate = kN In radioactive decay the constant,

k

, is called the decay constant , and

N

is number of nuclei.

The rate of decay is called activity (disintegrations per unit time).

If N 0 is the initial number of nuclei and N

t

number of nuclei at time t, then is the

ln

N o N t



kt

with half-life t 1/2 =0.693/k 10

Rates of Radioactive Decay

5.0

2.5

1.25

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Rates of Radioactive Decay

• • • • •

Dating

Carbon-14 is a radioactive isotope of carbon and is used to determine the ages of organic compounds.

We assume the ratio of 12 C to 14 C has been constant over time.

For us to detect 14 C, the object must be less than 50,000 years old.

The half-life of 14 C is 5,730 years.

Its abundance is <1% (the most common isotope of carbon is C-12) 12

Rates of Radioactive Decay

14 C is created in upper atmosphere by bombardment of nitrogen with cosmic neutrons:

14 7 N + 0 1 n  14 6 C + 1 1 p

14 C itself decays to stable 14 N by β emission:

14 6 C  14 7 N +  0 1 e

with a half-life of t 1/2 = 5715 yr The amount of 14 C in the environment is constant. 13

How does 14 C dating work?

•

When an organism dies, it no longer takes in carbon compounds but its 14 C continues to decay.

•

5715 years (or one half-life) after the death of the organism (or 1 half-life), the relative amount of 14 C is half that found in living matter.

•

11,430 years after the death of the organism (two half-lives), the relative amount of 14 C is ¼ that found in living matter.

First order rate law:

ln [ 14 C] 0 [ 14 C] t  kt

and t 1/2 =.693/k Since [ 14 C] is proportional to radiation emitted (in counts/min or cpm), the law become:

cpm(initia ln cpm(curren l) t)  kt

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Rates of Radioactive Decay 14 C Dating

21.37 Artifact has 14 C activity of 24.9 counts/m, compared to current count of 32.5 counts/m for a standard. What is the age of the artifact? Given: t 1/2 = 5715 year.

k=.693/t 1/2 = .693/5715 yr = 1.21x10

-4 yr -1 Use first order expression

ln N o N t  kt ln 32.5

 (1.21x10

 4 yr 1 )t 24 .

9

2200 yr = t 15

40 K – 40 Ar Dating

40 K is a strange beast – with a half life of 1.3 x 10 9 yr, it has two simultaneous modes of decay.

88.8% decays by electron-emission to give Ca-40:

40 19 K  40 20 Ca +  0 1 e

11.2% decays by electron-capture (of one of the orbital electrons) to give Ar-40:

40 19 K +  0 1 e  40 18 Ar

40 K constitutes 0.01% of the natural abundance of potassium in the earth’s crust.

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Problem : a mineral is found with a 40 K/ 40 Ar mass ratio of 3/1.

How old is the mineral?

Answer: For the purposes of calculation, let’s assume a femtogram total mass of 40 K and 40 Ar. The respective masses of K-40 and Ar-40 will be 0.75 fg K-40 and 0.25 fg Ar-40. Since the mode of decay giving Ar-40 is 11.2%, the mass of K-40 which decayed is 0.25/0.112 = 2.23 fg. Hence, the original mass of K-40 was 0.75 + 2.23 = 2.98 fg.

ln N o N t  kt k = 0.693/t 1/2 ln 2.98

 (5.33 x 10 -10 0.75

t = 2.58 x 10 9 yr

yr -1 ) t

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Half-lives

Isotope Half-life Type of decay U-238 4.5 x 10 9 U-235 7.0 x 10 8 yr Alpha yr Alpha Th-232 1.4 x 10 10 yr Alpha K-40 1.3 x 10 9 yr Beta-capture or -emission C-14 5715 yr Beta Rn-222 3.825 days Alpha Tc-99 210,000 yr Beta 18

Detection of Radioactivity

• • • • •

Matter is ionized by radiation.

Geiger counter determines the amount of ionization by detecting an electric current.

A thin window is penetrated by the radiation and causes the ionization of Ar gas.

The ionized gas carried a charge and so current is produced.

The current pulse generated when the radiation enters is amplified and counted.

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Energy Changes in Nuclear Reactions

• • • • •

Einstein showed that mass and energy are proportional:

E = mc

2 If a system

loses

mass it

loses

energy (exothermic).

If a system

gains

Since c 2 mass it

gains

energy (endothermic).

is a large number (8.99



10 16 m 2 /s 2 ) small changes in mass cause large changes in energy.

Mass and energy changed in nuclear reactions are much greater than in chemical reactions.

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Energy Changes in Nuclear Reactions

– – –

Consider reaction : 238 92 U



Molar masses: 238.0003 g



234 90 Th + 4 2 He 233.9942 g + 4.0015 g.

–

The change in mass during reaction is 233.9942 g + 4.0015 g - 238.0003 g = -0.0046 g. =



m The process is exothermic because the system has lost mass.

To calculate the energy change per mole of 238 92 U:



E



mc

2  ( 

m

)

c

2   3 .

00  10 8 m/s  2   0 .

0046 g    1 kg 1000 g      4 .

1  10 11 kg m 2   4 .

1  10 11 J s 2

This is a very large number!!

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• •

Nuclear Fission

Splitting of heavy nuclei is exothermic for large mass numbers.

Consider a neutron bombarding a 235 U nucleus:

0 1 n + 235 U 92 142 Ba 56 + 91 Kr 36 + 3 0 1 n 137 52 Te + 97 40 Zr + 2 0 1 n

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• •

Nuclear Fusion

Light nuclei can fuse to form heavier nuclei.

Most reactions in the Sun are fusion.

Fusion processes occurring in sun:

1 1 H + 1 1 H  2 1 H + 0 1 e 2 1 H + 1 1 H  2 3 He 2 3 He + 1 1 H  4 2 He + 0 1 e

Sun currently is about 75% H and 25% He.

Another 15b yrs or so to go before sun “burns” up all its H.

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• • • •

Nuclear Fusion

Fusion of tritium and deuterium requires about 40,000,000K: 2 1 H + 3 1 H



4 2 He + 1 0 n These temperatures can be achieved in a nuclear bomb or a tokamak .

A tokamak is a magnetic bottle: strong magnetic fields contained a high temperature plasma so the plasma does not come into contact with the walls. (No known material can survive the temperatures for fusion.) To date, about 3,000,000 K has been achieved in a tokamak.

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