RADIOACTIVITY

RADIOACTIVITY

Henri Becquerel in 1896 discovered that rays emitted from Uranium compounds had the properties of affecting photographic plates, ionising air and penetrating through substances. The spontaneous emission of such radiations is known as radioactivity and the elements emitting these radiations are called radioactive elements.

Rutherford and Soddy (1903) gave an ingenious interpretation of the radioactive processes and the origin of a and b emissions on the basis of neutron-proton theory.

(i) The atomic nucleii of radioactive elements are unstable.

(ii) The nucleii of radioactive elements are undergoing a process of disintegration forming atoms of new elements called daughter elements which are distinct in physical and chemical properties from parent elements.

(iii) a and b particles are ejected from the nucleus.

(iv) g-rays are secondary products of atomic disintegration.

Radioactivity is found to be of three types a, b and g radiations.

TYPES OF RADIOACTIVE DECAY

${}_{92}^{238}U{\to }_{90}^{234}Th{+}_2^{4}He(\alpha -emission)$

${}_{9o}^{234}Th{\to }_{91}^{234}Pa{+}_{-1}^0(\beta )(\beta -emission)$

${}_{90}^{230}Th\ast {\to }_{90}^{230}Th+\gamma$$\left(\gamma ;-emission\right)$
Illustration 1.;;;Define radioactivity. Who discovered natural radioactivity? Solution:;;;;;;;;;Radioacitvity is a phenomenon of spontaneous emission of invisible radiations by certain substances. Natural radioactivity was discovered by Henri Becquerel.;
CAUSE OF RADIOACTIVITY

Except in the case of ordinary hydrogen all other nuclide contain both neutrons and protons. A look at the stable nuclides show that the ratio N/P (neutron/proton) in them is either equal to 1 or more than 1. The ratio is $\approx$ 1 in all the light stable nuclides up to calcium $\left({}_{20}^{40}Ca\right)$ and thereafter the ratio is greater than 1 for heavy nuclides.
The stable nuclides lie within the shaded area which is called the region or zone of stability. All those nuclides falling outside this zone are invariably radioactive and unstable in nature. Nuclides falling above the stability zone have an excess of neutrons while those lying below have more protons. Both of these cause instability. These nuclides attain stability by making adjustment in the N/P ratio.
Illustration 2.;;;How can we increase the N/Z ratio is ${}_2^{3}He?$Solution:;;;;;;;;The N/Z ratio can be increased by ;;;;;;;;;;;;;;;;;;;;;;(i);;Emission of ${\beta }^{+}$;;;;;;;;;;;;;;;;;;;;;;;;;;;;${}_2^{3}He{\to }_1^{3}H{+}_{+1}^{0}e$;;;;;;;;;;;;;;;;;;;;;(ii);;K – capture;;;;;;;;;;;;;;;;;;;;;;;;;;;${}_2^{3}He{+}_{-1}^{0}e{\to }_1^{3}H+Xray$
THEORY OF RADIOACTIVE DISINTEGRATIONRutherford and Soddy, in 1903, postulated that radioactivity is a nuclear phenomenon and all the radioactive changes are taking place in the nucleus of the atom. They presented an interpretation of the radioactive process and the origin of radiations in the form of a theory known as theory of radioactive disintegration.The disintegration process may proceed in one of the following two ways:(a);a-particle emission: When an $\alpha$-particle $\left[{}_2^{4}He\right]$ is emitted from the nucleus of an atom of the parent element, the nucleus of the new element, called daughter element, possess atomic mass or atomic mass number less by four units and nuclear charge or atomic number less by 2 units, because $\alpha$-particle has mass of 4 units and nuclear charge of two units.;;;;;;;;;;;;;;;;;;;;;;;Parent element $\stackrel{-\alpha }{\to }$ Daughter element ;;;;;Atomic mass;;;;;;;;;W;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;W - 4;;;;;Atomic number Z;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;Z - 2;;;;;e.g.;;;;;;;;;;;;;;;;;$\underset{(Radium)}{{}_{88}^{226}Ra}$;;$\to$;;;;;$\underset{(Randon)}{{}_{86}^{222}Rn{+}_2^{4}He}$;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;$\underset{(Uranium)}{{}_{92}^{238}U}$;;;;;;;$\to$;;;;;$\underset{(Thorium)}{{}_{90}^{234}Th}{+}_2^{4}He$;;;;;;;;;;;;;;;;;;;;;;;;;;;;;${}_{84}^{215}Po$$\to$ ${}_{82}^{211}Pb{+}_2^{4}He$

(b) $\beta$-particle emission: $\beta$ particle is merely an electron which has negligible mass. Whenever a $\beta$-particle is emitted from the nucleus of a radioactive atom, the nucleus of the new element formed, possess the same atomic mass but nuclear charge is increased by one unit over the parent element. $\beta$ - particle emission is due to the result of decay of neutron into proton and electron.

${}_0^{1}n$$\to$ ${}_1^{1}p$ + ${}_{-1}^{0}e$;;;;;The electron produced escapes as a $\beta$-particles leaving proton in the nucleus.;;;;;;;;;;;;;;;;;Parent element $\stackrel{-\beta }{\to }$Daughter element ;;;;;Atomic mass;;;;;;;;;W;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;W;;;;;Atomic number Z;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;Z + 1;;;;;e.g.;;;;;${}_{82}^{214}Pb$$\to$ ${}_{83}^{214}Bi$ + ${}_{-1}^{o}e$

${}_{90}^{232}Th$ $\to$ ${}_{91}^{234}Pa{+}_{-1}^{0}e$

Illustration 3. Arrange the alpha (a), beta ($\beta$) and gama ($\lambda$) radiations in the increasing order of the properties indicated:

(i) penetrating power

(ii) ionizing power

(iii) extent of deflection in a magnetic field.

Solution: (i) Increasing penetrating power order is $\alpha <\beta <\gamma$.

(ii) Increasing ionizing power order is $\gamma <\beta <\alpha .$

(iii) Increasing extent of deflection in a magnetic field is $\gamma <\alpha <\beta$

GROUP DISPLACEMENT LAW

According to this law, "When an $\alpha$ - particle is emitted the daughter element has atomic number

2 units less than that of the parent element. It is consequently displaced two places to the left in the periodic table. When a $\beta$-particle is emitted, the daughter element has an atomic number 1 unit higher than that of the parent element. It is consequently displaced one place to the right in the periodic table.

Counting of number of a and b particles in a radioactive transformation

Parent element Daughter element

${}_{Z_1}{A}^{M_1}$ $\to$ ${}_{Z_2}{B}^{M_2}$

Number of $\alpha$ - particles = $\frac{Change\text{ in mass number}}{4}$= $\frac{(M_1-M_2)}{4}$

Number of b-particles: Let 'x' $\beta$-particles and 'y$\lambda$' $\alpha$ - particles be emitted

Atomic number of parent element – 2y + x = Atomic number of daughter element

Z₁ – 2y + x = Z₂

$\therefore$x = (Z₂ – Z₁ + 2y)

Illustration 4. Which particle emission produces isobars?

Solution: Emission of $\beta$^- particle produces isobars.

${}_{92}^{237}U{\to }_{93}^{237}Np{+}_{-1}^{0}e$

${}_{92}^{237}Uand{}_{93}^{237}Npareisobars$

RATE OF DISINTEGRATION AND HALF-LIFE PERIOD

The radioactive decay of the different radioactive substances differ widely. The rate of disintegration of a given substance depends upon the nature of disintegrating substance and its total amount. The law of radioactive disintegration may be defined as the quantity of radioactive substance which disappears in unit time is directly proportional to the amount of radioactive substances present or yet not decayed.

Rutherford introduced a constant known as half - life period. It is defined as "time during which half the amount of a given sample of the radioactive substance disintegrates".

Half life periods vary from billions of years for some radio isotopes to a fraction of a second.

Half life period is represented as t_1/2. Let the initial amount of a radioactive substance be N_o. After one half life period (t_1/2) it becomes = N_o/2. After two half life periods (2t_1/2) it becomes = N_o/4 and after n half life periods (nt_1/2) it becomes = ${\left(\frac{1}{2}\right)}^n$N₀. Thus, for the total disintegration of a radioactive substance an infinite time will be required.

Amount of radioactive substance left after n half life periods N = ${\left(\frac{1}{2}\right)}^n{N}_o$ and total time T = n x t_1/2. Where n is a whole number.

Illustration 5. A radioisotope has t_1/2 = 5 years. After a given amount of decays for 15 years, what fraction of the original isotope remains?

Solution: t_1/2 = 5 year

t = 15 year $\therefore$ n = $\frac{t}{t_{1/2}}=\frac{15}{5}=3$

Let original amount be No and amount after time t be N

$\therefore$ N = ${\left(\frac{1}{2}\right)}^{n}N\Rightarrow \frac{N}{No}={\left(\frac{1}{2}\right)}^n={\left(\frac{1}{2}\right)}^3=\frac{1}{8}$

Hence (⅛)^th of the original amount left.

Disintegration constant: A chemical reaction whose rate varies directly to the concentration of one molecular species only, is termed as first order reaction. Radioactive disintegration is similar to such a chemical reaction as one radioactive species changes into other. This can be represented as.

A $\to$ B

Suppose the number of atoms of a radioactive substance present at the start of observation, i.e., when t = 0, is N_o and after time t the number of atoms remaining unchanged is N. At this instant of very small number of atoms dN disintegrate in a small time dt; the rate of change of A into B is given$-\frac{dN}{dt}$. The negative sign indicates the number of atoms decreases as time increases.

Since rate of disintegration or change is proportional to the number of atoms present at that time, the relation becomes.

$-\frac{dN}{dt}$= .N …(i)

'$\lambda$' is called the disintegration constant or decay constant.

Evidently, $-\frac{dN}{N}$= $\lambda$.dt …(ii)

If dt = 1 second, $\lambda$ =$-\frac{dN}{N}$ … (iii)

Thus, $\lambda$ may be defined as the fraction of the total number of atoms which disintegrates per second at any time.

Integrating equation (ii), $-\int \frac{dN}{N}=\lambda \int dt$

or, – log N = $\lambda$t + C ……… (iv)

where C is the integration constant.

When t = 0, N = N₀

Putting values in equation (iv)

– log N₀ = C

Putting the value of C in equation (iv)

– log N = $\lambda$t – log N₀

or, log N_o­ – log N = $\lambda$t

or, log $\frac{N_o}{N}$ = $\lambda$t

or, 2.303 log₁₀ $\frac{N_o}{N}$ = $\lambda$t

or, $\lambda$ =$\frac{2.303}{t}\log \frac{N_o}{N}$ … (v)

This equation is called kinetic equation and is obeyed by first order reactions.

Relationship between half life period and radioactive disintegration constant

When t = t_1/2 N = $\frac{N_o}{2}$

Putting the values in equation (v)

$\lambda$ = $\frac{2.303}{t_{1/2}}{\log }_{10}\frac{N_o}{N_o/2}$ = $\frac{2.303}{t_{1/2}}{\log }_{10}2$

So, $\lambda$ = $\frac{0.693}{t_{1/2}}$ [$\therefore$ log₁₀2 = 0.3010]

or, t_1/2 = $\frac{0.693}{\lambda }$

Thus, half life period of a given radioactive substance does not depend on the initial amount of a radioactive substance but depends only on the disintegration constant of the radioactive element.

Illustration 6. What mass of C – 14 isotope will have an activity equal to one curie? Given that the half life period of C – 14 is 5730 years.

Solution: Rate of disinitegration $\left(-\frac{dN}{dt}\right)$

= 1 curie = $3.7\times {10}^{10}dps$

$K=\frac{0.693}{5730\times 365\times 24\times 3600}{S}^{-1}$

$Now-\frac{dN}{dt}=KN$

$\therefore 3.7\times {10}^{10}=\frac{0.693}{5730\times 365\times 24\times 3600}\times N$

or $N=9.65\times {10}^{21}atoms$

$\therefore$ Amount of C – 14 in grams

= $\frac{14}{6.023\times {10}^{23}}\times 9.65\times {10}^{21}=0.2244gms$

Average life Period (T)

Since total decay period of any element is infinity, it is meaningless to use the term total decay period for radioelement. Thus the term average life is used which is determined by the following relation.

Average life (T) = $\frac{sum\text{ of lives of the nuclides}}{total\text{ number of nuclides}}$

Relation between average life and half - life

Average life(T) of an element is the inverse of its decay constant, i.e.,

T = $\frac{1}{\lambda }$

Substituting the value of $\lambda$ = $\frac{0.693}{t_{1/2}}$

Average life (T) = 1.44 x Half life (t_1/2)

Illustration 7. Prove that time required for 99.9% decay of radioactive species is almost ten times of its half life period.

Solution: t = $\frac{2.303}{\lambda }$log$\frac{No}{N}$

No = 100, N = 100 - 99.9 = 0.1

t_99.9% = $\frac{2.303}{\lambda }\log \frac{100}{01}$ $\Rightarrow$ $t_{99.9\%}=\frac{2.303}{\lambda }\times 3$ …(1)

t_1/2 = $\frac{0.693}{\lambda }$ …(2)

$\frac{t_{99.9\%}}{t_{1/2}}=\frac{2.303}{\lambda }\times \frac{\lambda }{0.693}=10$

UNITS OF RADIOACTIVITY

The standard unit of radioactivity is curie (C) which is defined as that amount of any radioactive material which gives 3.7 x 10¹⁰ disintegrations per second (dps). i.e., 1C = 3.7 x 10¹⁰ dps.

Now a - days, the unit curie is replaced by Rutherford (rd) which is defined as the amount of a radioactive substance which undergoes 10⁶ dps; i.e., 1 rd = 10⁶ dps.

1 C = 3.7 x 10¹⁰ dps = 37 x 10³ rd

1 mC = 3.7 x 10⁷ dps = 37 rd

However, in SI system the unit of radioactivity is becquerel (Bq)

1Bq = 1 disintegration per second

10⁶ Bq = 1 rd

3.7 x 10¹⁰ Bq = 1 C

Illustration 8. Define one curie?

Solution: One curie (Ci) is defined as the amount of radioactive isotope that gives 3.7 × 10¹⁰ distintegrations per second.

Radioactive Equilibrium

According to the theory of radioactive disintegration, the product of a radioactive disintegration may also be radioactive and the rate of disintegration of the daughter element depends upon the amount of it present at different time. When steady state is reached,

${\left(\frac{dN}{dt}\right)}_{parent}={\left(\frac{dN}{dt}\right)}_{daughter}$

Or ${\lambda }_1{N}_1={\lambda }_2{N}_2$ Where ${\lambda }_1$ and ${\lambda }_2$ are the disintegration constants for the decay of parent and daughter radioactive species and N₁ and N₂ are the number of atoms disintegrating at a given time.

Thus, $\frac{N_1}{N_2}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{t_1(half\text{ life)}}{t_2(half\text{ life)}}$

Activity: Activity is the rate at which the radioactive substance decays

Activity = $\lambda$xN =$\frac{0.693}{t_{1/2}}\times \frac{mass}{at\overrightarrow{\leftarrow }mas}\times N_A$

Specific Activity: Activity of the unit mass of radio isotope.

Illustration 9. ⁹⁰Sr shows ${\beta }_1$ activity and its half- life period is 28 years. What is the activity of sample containing 1 gm of ⁹⁰Sr?

Solution: Activity = $\lambda$N

= $\frac{0.693}{28\times 365\times 24\times 60\times 60}\times \frac{1}{90}\times 6.023\times {10}^{23\overrightarrow{\leftarrow }}=147.89\overrightarrow{\leftarrow }curie$

Parallel Path: Let the radioactive element 'A' decay to B and C in two parallel paths.

Decay constant of 'A' = Decay constant of B + Decay constant of C

${\lambda }_A=({\lambda }_B+{\lambda }_C)$

(Frication of B) = $\frac{{\lambda }_B}{({\lambda }_A+{\lambda }_B)}=\frac{{\lambda }_B}{{\lambda }_A}$

(Frication of C) = $\frac{{\lambda }_C}{({\lambda }_A+{\lambda }_B)}=\frac{{\lambda }_C}{{\lambda }_A}$

Maximum yield of daughter element

Let A $\to$ B (daughter element)

If ${\lambda }_A$ and ${\lambda }_B$ are decay constants of 'A' and 'B'. Maximum activity time of daughter element can be calculated as:

t_max $\frac{2.303}{({\lambda }_B-{\lambda }_C)}\log \left(\frac{{\lambda }_B}{{\lambda }_A}\right)$

Number of atoms of daughter element B after time t:

[N_B] =$\frac{N_{\circ }{\lambda }_B}{\left({\lambda }_B-{\lambda }_A\right)}\times (e^{-{\lambda }_{A}t}-e^{-{\lambda }_{B}t})$

Nuclear Fission: It is a nuclear reaction in which a heavy nucleus splits into lighter nuclei and energy in released.

$\bullet$ Californium - 252 decays both by alpha emission (97%) and by spontaneous fission (3%). During spontaneous fission, the nucleus splits into two stabler nuclei plus several neutrons.

${}_{98}^{252}Cf$$\to$ ${}_{56}^{142}Ba$ + ${}_{42}^{106}Mo$ + $4_0^{1}n$$\bullet$ ;;When a neutron strikes ${}_{92}^{235}U$;nucleus, the nucleus splits into roughly equal parts, giving off several neutrons ${}_0^{1}n{+}_{92}^{235}U$$\to$${}_{54}^{142}Xe$;;+;;;;;;;;;${}_{38}^{90}Sr$;;+;;;;;;;;;$4_0^{1}n$or, ${}_{56}^{139}Ba$ +;;;${}_{36}^{96}Kr$;;+;;;$4_0^{1}n$or,;${}_{55}^{144}Cs$ +;;;${}_{37}^{90}Rb$;+$2_0^{1}n$$\bullet$;;;If the neutrons from each nuclear fission are absorbed by other uranium - 235 nuclei, these nuclei split and release even more neutrons. Thus a chain reaction can occur.
Spallation: Bombardment with high energy charged particles can break some target nuclei into several smaller nuclei with the emission of a large number of nucleus (10 to 20 or more). These are called spallation reaction.${}_{33}^{75}As{+}_1^{2}H{\to }_{25}^{56}Mn+9H+12_0^{1}n$${}_{92}^{238}U{+}_2^{4}He{\to }_{74}^{187}W20_1^{1}H+35_0^{1}n$
Nuclear fusion: In this type of nuclear reaction, certain light nuclei may fuse together with the liberation of tremendous amount of energy. To achieve this, the colliding nuclei must possess enough kinetic energy to overcome the initial repulsion between the positively charged cores. This energy may be made available by raising the temperature of the reacting system to several million degress. Such reactions are therefore also known as thermonuclear reactions.${}_1^{2}H{+}_1^{3}H{\to }_2^{4}He{+}_0^{1}n+17.6Mev$The energy of a fusion process is due to ma.6ss defect (converted to Binding Energy). The high temperature required to initiate such reactions may be attainted initially through a fission process.
Artificial Radioactivity: Bombardment of stable elements with high energy a-particles, protons, neutrons, deutrons or $\gamma$-rays produce radioactive nuclide. These radionuclide do not occur naturally and may be called man - made or artificial. The radioactivity exhibited by these artificial radio nuclide is referred to as artificial radioactivity or as induced radioactivity.${}_{13}^{27}Al{+}_2^{4}He\to {}_{15}^{30}P{+}_{15}^{30}P{+}_0^{1}n$${}_{11}^{23}Na{+}_0^{1}n\to {}_{12}^{27}Mg+{}_{-1}^{0}e$;Binding Energy and Packing fractionThe nearest integer to the mass of a nuclide is called the mass number of that isotope. The difference between the actual isotopic mass and mass number is termal as mass defect of the nuclide.Thus Dm (mass defect) = isotopic mass - mass number Aston introduced a term "Packing fraction" for each nuclide to compare their mass excess, it was defined as:Packing fraction (f)= ;;$\frac{\Delta m}{A}\times 10^4$Binding energy $\left(B\right)=\Delta m\times c^2$Binding energy of the nucleus is equal to the energy required to split the nucleus into its component nucleons. Nuclear binding energy is generally expressed in MeV (mega - electron volt).Since 1 a.m.u. = 1.66 x 10^–27 Kg$\therefore$(B) for one a.m.u. = 1.66 x 10^–27 x (3x10⁸)²J$\frac{=1.66\times {10}^{-27}\times {\left(3\times {10}^8\right)}^{2}J}{1.60\times {10}^{-19}J/eV}$ = 931 x 10⁶ eV = 931 MeV$\therefore \left(B\right)=\Delta M$ (in a.m.u.);x 931 MeVBinding energy per nucleon $\left(\bar{B}\right)=\frac{B}{A}=\frac{total\text{ binding energy}}{total\text{ number of nucleons}}$
Applications of Radioactivity(i);;Radio Carbon Dating: The application of the C - 14 dating is based on the fundamental assumption that the intensity of cosmic ray and hence of ¹⁴C in the atmosphere has been remaining constant over many thousands of years. This gives the initial activity of;C - 14 corresponding to the time when the plant or animal died and assimilation of radio active carbon ceased to continue.;;;;;${}_7^{14}N+{}_0^{1}N\to {}_6^{14}C+{}_1^{1}H$;;;;;${}_6^{14}C\to {}_7^{14}N+{}_{-1}^{0}e$;;;;;Age = $\frac{2.30}{\lambda }\frac{\log original\text{ activity (}{\text{N}}_{\text{o}})}{final\text{ activity (N)}}$;;;;;or N =;$N_0$${\left(\frac{1}{2}\right)}^y$

where y = t/t_1/2

(ii) Rock Dating: It is based on the kinetics of radioactive decay. It is assumed that no lead was initially in the sample and the whole of it came from the uranium.

Initial radioactivity = [U] = N₀ in terms of gm atoms

Final radioactivity = [U] = N

We have assumed that due to high value of t_x/2 of uranium its amount remains unchanged.

$\frac{N_o}{N}=\frac{[V]+[Pb]}{[V]}=1+\frac{[Pb]}{[V]}$

Thus t = $\frac{2.303}{t_{1/2}}\log \left(\frac{N_0}{N}\right)=\frac{2.303}{t_{1/2}}\log \left[1+\frac{[Pb]}{[U]}\right]$

or $\left[1+\frac{[Pb]}{[v]}\right]={(2)}^y$

Where y =$\frac{t}{t_{1/2}}$, hence t is calculated.

Illustration 10. The amount \begin{align}  _{6}^{14}C \\   \\ \end{align} isotope in a pice of wood is found to be one fifth of that present in fresh piece of wood. Calculate the age of wood (t_1/2 = 5577 year.

Solution: t = $\frac{2.303}{K}\log \frac{No}{N}$ $\Rightarrow$ t =

$\frac{2.303\times t_{1/2}}{0.693}\log \frac{No}{N}\left(K=\frac{0.693}{t_{1/2}}\right)$

Given N = $\frac{No}{5}$ $\therefore$ t = $\frac{2.303\times 5577}{0.693}\log 5$

=$\frac{2.303\times 5577}{0.693}\times 0.6989=12953years$