Ebook chemistry basis elements part 2

Nuclear Chemistry 9 Nuclear Chemistry Order of Radioactive Disintegration Process Radioactive disintegration is similar to a chemical reaction of the first order, that is, a chemical reaction in which[.]

Nuclear Chemistry

9

Nuclear Chemistry

Order of Radioactive Disintegration Process : Radioactive disintegration is similar to a chemical

reaction of the first order, that is, a chemical reaction in which the rate of the reaction varies as the concentration of one molecular species only

Suppose such a reaction is representation by the equation

A ® B

Let the number of atoms of A originally present be N0 and number of atoms of A, Present at any given

time t, be N Then, the rate of change of A into B is

= …(1)

= Disintegration constant or decay constant

Thus, _ldt = dN/N …(2)

If, dt = 1s

l =

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On Integtrating equation (2)

or In N = -lt + c …(4)

where c is the integration constant

At t = 0,

N = N0

\ c = In N0

Substituting in equation (4) we have

or, In (N/N0) = -lt …(6)

Relation between half-life and decay constant : Suppose, after time t1/2 half of the atoms of the radioactive substance have disintegrated , that is

N = N0/2

Substituting in equation (7) we have

=

or, In 0.5 = -lt1/2

or, 2.303 log 0.5 = -lt

Half-life period of a given radioactive substance is independent of the amount of substance present initially It depends only on disintegration constant of the element

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Half-life nuclide 220 Rn is 54.5 s

l =

We know, 1 mci = 3.7 × 107 disintegration per sec

= lN

N =

Mass of = Number of Rn nuclei

mass of one Rn nucleus

=

or, =

An accident occurs in a laboratory in which a large amount of radioactive material with a known t 1/2

of 20 days becomes embedded in floor Tests show that the level of radiation is 32 times the permissible limit The laboratory can be safely occupied after :

Since the initial level of radiation is 32 times the permissible limit, we have to calculate the time t in which the activity drops to 1/32 of it initial value, i.e.,

=

Disintegration constant,

l =

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Suppose the laboratory becomes safe for use after t days Than

=

=

=

t =

Radioactive Equilibrium

Radioactive equilibrium : If in a disintegration series A, B, C, D etc are some of the intermediates

consecutive atoms i.e.,

a stage may come when the amounts of A, B, C, D etc become constant which is

so because equal If etc., represent the number of atoms of A, B, C, D etc., at equilibrium, then

=

=

Hence = =

or, =

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Thus the amounts present at equilibrium are inversely proportional to their disintegration constant or directly proportional to their half-lives

Radioactive equilibrium may be of two types—

Secular equilibrium : It is limiting case of radioactive equilibrium, in which the half-life of the parent is

many times greater than the half-life of the daughter) usually by a factor of 104 or greater)

After a sufficient long time

NB =

Transient equilibrium : It is similar to secular equilibrium but differs in that the half-lives differ only by

a small factor

After a sufficiently long time

NB =

No of atoms of 226Ra in 1 g of

=

At equilibrium

=

or, =

=

=

=

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Reason Having more Biochemical Importance

is biochemically more important because of its shorter life period Shorter half-life ensures attainment of transient equilibrium faster

barrier height for an a- particle inside the nucleus

Barrier height =

=

A thin sample of gold was irradiated in a thermal neutrons flux of 10 12 neutrons cm 2 sec _1 for 25.6 hrs In the reaction the nuclide 198 Au is produced with a half-life of 64 hours If the thermal neutron absorption cross section is 98 barns, the specific activity of the sample is:

A =

Suppose 1g of the gold was irradiated,

then N =

f =

s =

t1/2 =

Substituting these values, we get

A =

=

Nuclear Chemistry

A sample of 100 mg of a radioactive nuclide decay to 81.85 mg of the same in exactly 7 days The decay constant for this disintegration and the half life of the nuclide Calculated:

Disintegration constant,

l =

t =

l =

Half-life, t1/2 =

A sample of river water was found to contain 8 × 10 _18 tritium atoms, per atoms of ordinary

hydrogen Tritium decomposes radioactively with a half-life of 12.3 Y The ratio of tritium to normal hydrogen atoms 49 Y after the original sample was taken if the sample is stored in a place where additional tritium atoms cannot be formed It would be :

49 years is almost exactly four half-lives (4 × 12.3 = 49.2Y), so we can avoid the log formula The fractions of the tritium atoms remaining after four half-lives would be (1/2)4 = 1/16 The normal

hydrogen atoms are not radioactive, and therefore, their number does not change Thus the final ratio of tritium to normal hydrogen will be

=

Using the approximate equation for the radius of the nucleus, calculate the density of the nucleus of

and density of metallic silver (10.5 g/cm 3 ) Compared :

The approximate equation for the radius is

r = R0 A1/3

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where r is the radius of nucleus of mass number A and Ro is a constant whose value is equal to 1.5 ×

10_13 cm

So r = 1.5 (107)1/3 × 10_13

= 7.12 × 10-13 cm

The volume, mass and density of a single nucleus are calculated from the radius by the following

equations

V =

m =

d =

The nuclear density is 1.18 × 1014 times that of silver metal

The ratio of the nuclear radii of the s - particle and the proton, estimated :

is a-particle

Radius equation = R0 A1/3

= = 41/3 = 1.587

Suppose M represents the mass of and M1 that of Let E be the energy needed to remove the last bound neutron from From the law of conservation of energy

E =

E =

where m n is the mass of neutron

=

=

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A nucleus splits into nuclei, which have mass numbers in the ratio of 2 : 1 Assuming R = 1.5 f, The ratio of radii of nuclei formed :

Following the equation

r = R0 A1/3

R0 is given 1.5 f

For the two nuclei formed

r 1 = R0A11/3; r2 = R0A21/3

Since the mass numbers of nuclei formed are in the ratio of 2 : 1, therefore,

r1 = 1.5 f × (2 × 235/3)1/3 = 8.056 f

r2 = l.5f × (l × 235/3)l/3 = 6.418 f

The average mass of chlorine is 35.5 The mass numbers of two isotopes are 35 and 37 The proportion to which two isotopes are present in ordinary chlorine is :

Let the proportion of CI-35 and Cl-37 be 1 : x

Then the average mass would be

But this is given to be 35.5

So

35 + 37x = 35.5 (1 + x)

35 + 37x = 35.5 + 35.5x

1.5x = 0.5

x = 1/3

Therefore, the ratio is 1 : 1/3 or 3 : 1

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Justified Statement"We need a very powerful crane of lift microscopic nuclear mass."

Suppose volume of nuclear mass of microscopic size to be lifted is 10_4 cm3 Since the mass of a nucleon

is approximately 1 amu, hence the mass (m) of the nucleus of mass number

1.66 × 10_27kg

The nuclear radius is given by

r = R0 A1/3

where, R0 = 1.5 × 10_15m

Nuclear volume, V = 4/3 pr3 = (4/3)p(R0 A1/3)3

= (4/3)p(1.5 × 10_15)3Am3

Hence, the nuclear density,

r =

=

Thus we see the nuclear density is very high and is independent of the mass number of the nucleus

Therefore, all nuclei have approximately the same density

Now

volume of the nuclear mass = 10_4 cm3 = 10_10 m3

(Q 1 m3 = 106 cm3)

Mass = volume × density

= 10_10m3 × 1.17 × l0I7 kg m_3

= 1.17 × l06 kg

Such a high mass can only be lifted only by a powerful crane

Cross Section

The probability of nuclear process is generally expressed in terms of cross-section, which is denoted by s that has the

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dimensions of an area This originates from the simple picture that the probability for the reaction

between a nucleus and impinging particle is proportional to the cross sectional target area presented by the nucleus The cross section for a particular nuclear process is defined by the equation

Ri = lnx s i

This equation applies when there is a well defined beam of particles incident on a target

where, R, = number of processes of the type under consideration occurring in the target per unit time,

I = number of incident particles per unit time,

n = number of target nuclei per cubic centimetre of target,

si = cross section of the specified process, expressed is square centimetres

x = target thickness in cm

The total cross section for collision with a fast particle is never greater than twice the geometrical cross-sectional area of the nucleus and therefore, fast particle cross-sections are rarely much larger than 10_24

cm2 (radii of the heaviest nuclei are about 10_12 cm) Hence a cross-section of 10_24 cm2 is considered as big as barn and 10_24 cm2 has been named the barn (b) and used in expressing cross-sections The

millibarn (mb = 10_3 b), microbarn (mb = 10_6 b) and nanobarn (nb = 10_9 b) are also commonly used

Cross Section and Reaction Rate : For a sample containing N nuclei in a flux of f particles per square

centimetre per second, the rate of reaction of type i, which have a cross section a, is given by

Ri = f Nf i

This is the situation when a sample embedded in a uniform flux of particles incident on it from all

directions and exactly this happens in a nuclear reactor

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How long a 60 mg piece of Co wire has to be placed in a flux of 5 × 10 13 thermal neutrons per square centimetre per second to make 1 mci (1 millicurie = 3.7 × 10 7 dis s _1 ) of 5.22 Y 60 Co, calculated The cross-section for the reaction 59 Co

(x, y) 60 Co is 3.7 b

Following the equation

Ri = f Ns i

R =

= 1.13 × 1012 atom s_1

From equation we calculate that 1 mci of 60Co corresponds to 8.87 × 1015 atoms

Thus it will take

8.87 × 1015/1.13 × 1012 = 7.85 × 103 sec

or, approximately 2.2 hr to produce 1 mci of 60Co

The effective neutron capture radius of a nucleus having a cross section by 1.0 barn, calculated :

l.0 barn = l.0 × l0_24cm2

The area of a circle is given by

A = pr2

Hence r =

= = 5.6 × 10_13 cm

The ratio of number of neutrons to protons decides the so called `belt of nuclear stability' Showing this by drawing a graph and using this, explain the decay process involving :

Nuclear Stability : n/p ratio : The stability of nucleus depends on the number of neutrons and protons

present Light

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nuclei upto atomic number 20 form most stable nuclei when the number of protons p is equal to the number of neutrons n i e., n/p = 1 For elements with higher atomic number n/p ratio increases

progressively to about 1.6 In such cases the number of neutrons is greater than that of protons With increase in the number of protons the force of repulsion between them increases and this tends to lower the stability of nucleus

By plotting a graph between number of neutrons and protons for the nuclei of various elements it has been found that most stable nuclei (non-radioactive nuclei) lie within the shaded area which is called the zone or belt of stability because it contains the stable nuclei Nuclei that fall above or below this belt are unstable Nuclei that fall above the stability belt have more neutrons while those lying below have more protons Such unstable nuclei would attain stability by undergoing change that would produce a nucleus

with n/p ratio within the stability zone

Examples :

(1) Nuclei lying above the belt of stability are richer in neutrons and hence they disintegrate is such a

manner that one of their neutrons is converted into a protons i.e.,

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such nuclei emit p-particle For example,

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Radioactive Stable

or, 6p + 8n ® 7p + 7n +

n/p = 8/6 = 1.33 ® n/p = 1/1= 1.00

(2) The nuclei lying below the zone of stability are deficient in neutrons and hence disintegrate in such a way that one of their protons is converted into a neutrons The conversion of a proton into a neutron can

be done by any of the following two ways

(a) Emission of a positron

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(b) Electron-capture process

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or, 12p + 11n ® 11p + 12n +

n/p = 11/12 = 0.91 ® n/p = 12/11 = 1.09

(3) Other nuclei having higher number of protons or neutrons disintegrate by (a),

decay

More unstable of each of the following pairs and in each case, that type of process could the unstable nucleus undergo:

(a) 16 C , 16 N (b) 18 F, 18 Ne

(a) has the ratio of neutron to protons farther above the belt of stability It would emit a particle

to set n/p ratio back to the stable range

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(b) has a lower n/p ratio than It could emit a positron or capture a K electron to get that ratio into the range of stability

Each of the following nucleide classified as `probably stable', `beta emitter' or `positron emitter'

The nucleide near the belt of stability are probably stable, those above the belt are beta emitters, those below, positron emitters Thus the stable hucleides are 208Pb and 120Sn; the beta emitters are 40Ca, 3OA1 and 94Kr, the positrons emitters are l95Hg, 8B and l50Ho

The binding energy per nucleus of 16 O is 7.97 MeV and that of I7 O is 7.75 MeV The energy needed to remove a neutron from 17 O, calculated :

Energy needed to remove a neutron from 17O is given by

E = Energy equivalent of

(mass of i6O + mass of neutron _ mass of l7O)

= (8m p + 8m n _ 7.97 × 16)

+ m n _ (8m p + 9m n _ 7.75 × 17)

where m p = mass of proton, m n = mass of neutron

= (7.75 × 17 _ 7.97 × 16) MeV = 4.23 MeV

The approximate mass of uranium that must undergo fission to produce the same energy as 10 5 kg of coal, calculated Assumed that heat of combustion of coal = 8000 cal g - 1 and one fission of uranium releases 200 MeV

Mass of coal = 105 kg = 108 g

Heat of combustion of coal = 8000 cal g_1

Total heat produced on combustion of 108g coal

= 8000 × l08 = 8 × 1011 cal

= 8 × 4.184 × 1011J = 35.5 × 1011 J

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Thus, we have to calculate the mass of uranium that will undergo fission to produce 33.5 × 1011J of energy

Energy released per fission = 200 MeV

= 200 × l06 × l.6 × l0_19J = 3.2 × 10_11 J

Number of U-235 atoms needed to produce this amount of energy =

\ Mass of 1023 atoms of U-235 =

`K-electron Capture'

When an electron from the nearest orbital i.e., K-shell orbital is absorbed by the nucleus to convert a

proton into a neutron without emitting any particle, the process is known as orbital electron capture Since it is most usually a electron which is captured by the nucleus, the process is also known as K-electron capture Usually an K-electron from higher energy level L, M, N etc drop back to fill the vacancy

in K-shell

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Spallation Reactions

In these reactions, a high speed projectile having high energy (400 MeV) chips a fragment off from a heavy nucleus and a large number of light particles are emitted The daughter nucleus has atomic number 10-20 units less and mass number about 30 units less than the parent nucleus For example,

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Nuclear Chemistry

Nuclear Reactions completed by writing equations

(iii) 3 H (a, n) (iv)

(v) (vi) (D, T)O 18

(vii) (viii) 30 P ( , p) 33 S