Chemistry

Nuclear chemistry, sun & lifeSession opener

Session objectives

1. Radioactivity

2. Nuclear reactions

3. Kinetics of radioactive decay

4. Radioactivity equilibrium

5. Fission reaction

6. Fusion reaction

7. Radiocarbon dating

8. Medical applications

Magic numbers

RADIOACTIVITY AND NUCLIDE STABILITY

protons neutrons stable nuclides

even even 157

even odd 52

odd even 50

odd odd 5

“Magic numbers” analgous to the noble gas electronic configurations occur at: 2, 8, 20, 28, 40, 50, 82, 126, 184

The most stable isotopes have “magic numbers” of both protons and neutrons.

n/p too large

beta decay

X

n/p too small

positron decay or electron capture

Y

23.2

Stability of nucleus

It has been observed that for light elements of atomic number upto 20, n/p ratio is 1

When the n/p ratio is too high

This nucleus would be unstable and would tend to lower the ratio by emitting -particles and thus moves towards the zone of stability.

1 1 00 1 1n H e

neutrons proton β - particle (electron)

Stability of nucleus

When the n/p ratio is too low

This is effected either by the emission of an alpha particle or a positron or by capturing an orbital electron

20 20 01011 +1Na Ne + e

positron

82 0 823637 -1Rb + e Kr

235 231 492 90 2U Th + He

particle

m m 4 4z z 2 2A B He

( particle)

Nuclear reactions

Loss of –particle

Loss of –particle

m m 0z z 1 1A B e

Group displacement laws

The emission of an -particle results in the formation of an element which lies two places to the left and the emission of a  -particle results in the formation of an element which lies one place to the right in the periodic table

Illustrative Example

is radioactive and emits a and b particles to form

What is the respective values of a and b?

23892U

20682Pb .

Let x number of particles and y number of particles get emitted.

238 = 4x + 206 or x = 8 92 = 2x – y + 82 or y = 6

Solution

238 4 0 20692 2 1 82U x He y e Pb

212Po decays by alpha emission. Write the balanced nuclear equation for the decay of 212Po.

212Po 4He + AX84 2 Z

212 = 4 + A A = 208

84 = 2 + Z Z = 82

212Po 4He + 208Pb84 2 82

23.1

Solution Cont.

Disintegration series

Name of the Series Initial element Stable element Value of n for initial

element

Value of n for the stable

element Series

4n Thorium series Thorium–232 Lead–208 58 52

4n + 1 Neptunium series Neptunium–237

Bismuth–209 59 52

4n + 2 Uranium series Uranium–238 Lead–206 59 51

4n + 3 Actinium series Uranium–235 Lead–207 58 51

4n series

In the 4n series,all nuclides have mass number that are multiples of 4. 

4n +1 series

In the 4n + 1 series,all nuclides have mass numbers that are one number greater than multiples of 4.

4n+2 series

In the 4n + 2 series, all nuclides have mass number that are two numbers greater than multiples of 4.

4n+3 series

In the 4n + 3 series, all nuclides have mass numbers that are three numbers greater than multiples of 4.

Nuclear Binding Energy

The energy required to break up a nucleus into its component protons and neutrons.

BE = 9 x (p mass) + 10 x (n mass) – 19F mass

E = mc2

BE (amu) = 9 x 1.007825 + 10 x 1.008665 – 18.9984

BE = 0.1587 amu 1 amu = 1.49 x 10-10 J

BE = 2.37 x 10-11J

Nuclear binding energy

Binding energyBinding energy per nucleon=

Number of nucleons

-11-122.37 x 10 J

= =1.25 x 10 J19 nucleons

19 1 19 1 0BE+ F 9 p+10 n

Average Binding Energy as a Function of Atomic Number

Kinetics of radioactive decay

M D

0at t = 0 N

at t = t N

N0=Initial no. of radioactive particles

N=No. of radioactive particles after time t

Kinetics of radioactive decay

The rate of disintegration of M into D is equal to

dNdt

dN dNi.e., N or N (I)

dt dt

N

No

dNor dt

N

on

Nor t (I I)

N

010

N2.303log (iii)

t N

toN N e (iv)

Half life period

1 / 2.693

t

Average or Mean life Period 1 / 2

1 / 2t1

1.44 t0.693

1 curie (C) = 3.7 × 1010 dps

S.I unit is Beequerel (Bq)

(Bq) = 1 disintegration per second.

Units

Illustrative example

Calculate the activity in terms of dpm of 0.001 g sample of Pu239 (half life = 24300 years).

Solution:

Calculation of decay constant 1 / 2

0.693t

7

1 / 224300 3.15 10

t minutes60

1

7

0.693 60min

24300 3.15 10

dNHence, N

dt

3 238

7

dN 0.693 60 10 6.023 10= 1.37 × 10 dpm.

dt 23924300 3.15 10

3 2310 6.02 10Number N of the nuclei in the given sample

239

Illustrative example

One of the hazards of nuclear explosion is the generation of Sr90 and its subsequent incorporation in bones. This nuclide has a half-life of 28.1 years. Suppose 1 g was absorbed by a new born child, how much Sr90 will remain in his bones after 20 years?

Here a = 10–6 g

a – x = 6.09 × 10–7 g

1

1/ 2

0.693 0.693We have 0.0246 yr .

t 28.1

2.303 a 2.303 aAs t log 20 log

a x 0.0246 a x

a1.64

a x

Solution

Illustrative example

Calculate the weight of Na24 which will give radioactivity of one curie (half-life of Na24 is 15 hr).1C = 3.7 × 1010 dps

N = 2.89 × 1015

6.023 × 1023 of Na24 atoms = 24 g

103.7 10 N .....(i)

5

1/ 2

0.693 0.693Now 1.28 10

t 15 60 60

1515 24 7

23

24 2.89 102.89 10 of Na atoms 1.15 10 g

6.023 10

Solution

Illustrative example

The half-life for the decay of U238 to Th234 is 4.6 × 109 years. How many a particles are produced per second in a sample containing 3 × 1020 atoms of U238 .

Number of particles emit per second

1/ 20.693

We know that t

10 19

0.693or 1.5 10 yr

4.6 10

10 20dNNow N 1.5 10 3 10

dt

104.52 101433

365 24 60 60

10 1dN4.52 10 atoms yr

dt

Solution

Artificial transmutation of elements

14N717O8

4He2 +→ 1H1+

24Al1330P15

4He2 +→ 1n0+

30P

15

30Si

14

+→ 0+1

Shared Nobel Prize 1938

Irene Joliot-Curie.

Rutherford 1919.

Process of transformation of one element into the other by artificial means i.e. by bombarding the nuclei of atom by high speed subatomic particles is called artificial transmutation of elements:

Transuranium Elements

-1

+→+

+→ 0

23892U 10n 23992U

239 Np93

23992U

+→+24998Cf 157N 260105U 4 10n

Element coming after uranium [Z = 92] in the periodic table i.e. with atomic number greater than 92 are called transuranic elements. e.g.

Fission of U-235

Fission of U-235

235 1 23692 0 92U n U

140 93 156 36 0Ba Kr 3 n

144 90 154 38 0Xe Sr 2 n

144 90 155 37 0Co Rb 2 n

The minimum amount of fissionable material required to continue a nuclear chain reaction is called critical mass.

Nuclear Reactors

Breeder Reactors

238U92 →n10+ 1

0-1

239U92

239U92 → 239Np93+

0-1→239Np93 +239Pu94

A breeder reactor is one that produces more fissionable nuclei than it consumes i.e. it increases the concentration of fissionable nuclie.

Fusion

2 2 41 1 2H H He energy.

Difference between fission and fusion

Nuclear fission Nuclear fusion1. It involves breaking up of a heavier nucleus into lighter nucles.

1. It involves fusion of two or more lighter nuclei to form a heavier nucleus.

2. Large number of radioisotopes are formed.

2. It is difficult to control this process.

3. It is a chain process.

3. It is not a chain process.

4. This does not require high temperature

4. This is initiated by very high temperature.

5. It can be controlled and energy is released during the reactions.

5. It cannot be controlled.

Rate law for reactions involving parallel reactions

A

B (80% ) M ain reactionK 1

K 2

A (20% ) Side reaction

1 2 1 2d{A}

Rate K [A] K [A] (K K )[A]dt

Illustrative example

227Ac has a half-life of 22 years with respect to radioactive decay. The decay follows two parallel paths, one leading to 227Th and the other to 223Fr. The percentage yields of these two daughter nuclides are 2 and 98 respectively. What are the decay constants (l) for each of the separate path?

227 Ac

1

2

227Th (2%)

223Fr (98%)

On solving,0.693total 0.0315 per year

22

1 2 0.0315

2

1

98while 48

2

41 6.3 10 per year

22 3.08 10 per year

Solution

Radioactive equilibrium

A B C D

If the rate of decay of B is the same as its rate of formation from A, then the amount of B will remain constant.

A BdN dNdt dt

A B

B A

NN

Age of minerals and rocks

The end product in the natural disintegration series is an isotope of lead. Each disintegration step has a definite decay control. By finding out the amounts of percent radioactive element and the isotope of lead (e.g. 92U238 and 82Pb206) in a sample of rock and knowing the decay constant of the series, the age of rock can be calculated.

Illustrative ExampleAn uranium containing ore pitch blende contains 0.055 g of 82Pb206 for 1 g of 92U238. Calculate the age of the ore. (Half-life = 4.6 × 109 years)

Here N = 1 g

206 238238Now 0.055g of Pb 0.055 gm of U 0.0635 g

206

oN 1 0.0635 1.0635g

99

0.693We have 0.15 10

4.6 10

oN2.303t log ,

N

89

2.303 1.0635t log 4.10 10 yrs.

10.15 10

Solution

Radio carbon dating

The principle is based upon the formation of C14 by neutron capture in the upper atmosphere.

14 1 14 1N + n C + H7 0 6 1

This carbon-14 is radioactive with a half life period of 5700 years. This is changed into nitrogen by the emission of particles.

14 14 06 7 -1 1

2

C N + e t = 5700 yrs

Radio carbon dating

The age of the object can be estimated.

This method cannot be applied to estimate the age of an object which is about 20,000 to 50,000 yrs old.

14

14

2.303 initial activity (amount of C in a living object)tiem(t) log

Final activity (amount of C in a dead object)

Illustrative Example

A piece of charcoal from the ruins of a settlement in Japan was found to have 14C/12C ratio that was 0.617 times that formed in living organism. How old is this piece of charcoal. Given the half life of 14C is 5,770 years. [log 1.62 = 0.210]

We know

t = 4027 years

o

1/ 2 t

N0.693 2.303log

t t N

0.693 2.303 1log

5770 t 0.617

Solution

Application: Tracers

Radioisotopes are used as tracers to find the reaction mechanism.

For example

6C H C5

O

O H2C H O H3

*6C H C5

O

O C H 3*H O++

23.7

• 1 out of every 3 hospital patients will undergo a nuclear medicine procedure

• 24Na, t½ = 14.8 hr, emitter, blood-flow tracer

• 131I, t½ = 14.8 hr, emitter, thyroid gland activity

• 123I, t½ = 13.3 hr, -ray emitter, brain imaging

Brain images with 123I-labeled compound

Radioisotopes in Medicine

Illustrative Example

The half life of W-238, which decays to Pb-206 is 4.5 x 109 years. A rock containing equal numbers of atoms of two isotopes would be how much old?

238 206W Pb

91

o

t

0.693t / 2 4.5 10 yrs.

kN2.303

k logt N

238o

238t

N Initial no. of atoms of W

N No. of atoms of W after certain time

t Age of the rock

t9

t

9 9

2N0.693 2.303log

t N4.5 10

2.303t 4.5 10 log2 4.5 10 yrs.

0.693

Solution

Illustrative Example

A small amount of solution containing Na24 radioisotope with activity 2 × 103 dps was injected into blood of a patient. After 5 hr, a sample of blood drawn out from the patient showed an activity of 15 dpm ml–1 half-life of Na24 is 16 hr. Find the volume of blood in the patient.

Solution

Let the volume of blood be V ml.

Activity of Na24 in V ml blood = 2 ×103 dps = 2 × 60 × 103 dpm

Activity of Na24 in V ml blood after 5 hr = 15 × V dpm

1

1/ 2

0.693 0.6930.0462 hr

t 16

Now

0N0.693t log

N

30.693 120 10or 5 log

0.0462 15 V

3V 3.713 10

Thank you