56

1. INTRODUCTION

(1) The phenomenon of radioactivity wasdiscovered by Henry Becquerel in uraniumsalts. (1896)

(2) After the discovery of radioactivity in uranium,Piere curie & Madame curie discovered anew radioactive element called ‘radium’. It is106 times more radioactive than uranium.Radium was extracted from pitch blende(a kind of coal tar) for which curies werehonoured by nobel prize in 1903.

(3) Definition : -The self, spontaneous,disintegration (i.e. decay) of unstableradioactive nuclei’s is known as radioactivity& the nucleils exhibiting this phenomenonare known as radioactive nuclelis.

(4) Some example of radio active substance(or elements) are :Uranium, Radium, Thorium, Polonium,Neptumium etc.

(5) All elements having atomic number greaterthan 83 are radio active element.

(6) Lead (Z = 82) is the most stable element inhigh atomic number elements that’s why allradio active elements emits radio activeradiation till they are converted into Lead.

(7) Radioactivity is a nuclear event & not atomic.Hence electronic configuration of atom don’thave any relationship with radio activity.

(8) Decay processes are random. Here onesimply knows that in a radioactive element,radioactivity is taking place or it is definitethat a certain no. of atoms will decay in agiven time interval but one never knows thatwhich particular radioactive nuclei will decaywhen. It is just a matter of chance i.e.probability which is explained by quantummechanics. It is statistical in nature.

(9) This process is spontaneous i.e. it can neitherbe started, stopped, accelerated nor retardedby any physical (i.e. by changing temperaturepressure, force) or chemical change.

(10) In this phenomenon parent-daughter chaincontinues. Original radio active element arecalled parent element or nucleus but newelement is called Daughter element ordisintegrated nucleus.

92U238 90Th234 + 24

(Parent) (Daughter)

90Th234 91Pa234 + -10 +

(Parent) (Daughter) (Antinentrino)

(11) In this phenomenon, particles and -raysare emitted. For a given nuclei at a particulartime emission of either or takes place,never both at time.

(12) rays are emitted in when an excited nucleimakes transition to any lower or groundenergy state the form of photon.

92U238 *90Th234 + 24

*90Th234 90Th234 + -ray(higher (lower energy nuclei)energy nuclei)(lower stability) (higher stability)excited nuclei and normal nuclei are knownas radioactive isomers.

(13) Due to emission of during radioactivedecay, mass of parent nuclei goes ondecreasing.

(14) Deflection of radioactive radiations wasmeasured by rutherford in electric field & it isas below.

(+)ve plate

(-)ve plate

or

-ray

Radioactive Source

57

(15) Defiection in magnetic field.

(16) Remember : Proton is not emitted duringradioactive decay.

2. LAWS OF RADIOACTIVE DISINTEGRATION

(1) If N = No. of active nuclei at time tN – dN = No. of active nuclei after timeinterval of dt.

Rate of decay = dtdN

(2) According to the Rutherford and Soddy lawfor radio active decay, "At any instant therate of decay of radioactive atom isproportional to the number of atoms presentat that instant " Rate of decay N

dtdN N

or dtdN = N

is called decay constant. (3) If Ni = No of active nuclei at time ti.

Nf = No of active nuclei left at time tf.

f

i

N

NNdN

= – f

i

t

t

dt

f

i

NNnn – f

i

ttt

ni

f

NN

= – [tf – ti]

i

f

NN

= if tte

If tf = t & Nf = N& ti = 0 Ni = N0

then0N

Ν = te N = N0

te ....... (1)

(4) eq. (1) is known as Ruther ford & soddy’sexponential equation of radioactivity.

(5) If t = then N = 0Therefore, for a complete decay of aradioactive nuclei, it takes infinite time.

(6) Half life (T1/2) Half life of a radioactive element is the timeor time interval is which 50% of radioactivenuclei will disiintegrates.

or when t = T1/2 then N =2

N0

From N = N0te

2N0 = N0

2/1Te

T1/2 =

2n

=693.0

(7) Physical sighificance of Half life.If half life of a radio active nuclei is more thenin a given time interval probability of itsavailability is also more. Half life of a elementdoes not depend on physical & chemicalcircumstances.

0NN

= T/t)2(1

.

(8) Graph between no. of active nuclei left & timeof decay

N0

N0

N0

N

/2

/4

(9) Average life or mean life () (i) The time, for which a radioactive material

remains active, is defined as mean life ofthat material.

58

(ii) = presentnucleiof.noTotals'nucleialloflivesofSum

=0

0N

dNt

(iii) The average time taken in decaying by thenuclei of an element is defined as its meanlife .

(iv) =1

(v) n0N

N = – t

The magnitude of inverse of slope of n0N

N & t

curve is known as mean life.

(vi) In N = N0te

If t =1

then N = N0 e–1

= eN0 = 718.2

N0= 0.3676 N0 0.37 N0

Therefore if = t =1

then N = eN0

0.37 N0

Mean life of a radio active nuclei is equal to that

time is which no of nuclei left becomes either e1

times the original nuclei or approximately 37% ofthe original nuclei.or, Number of decayed nuclei in mean life

= N0 – N = N0 – eN0

= N0

e11 0.63 N0

Mean life is equal to that time in which no. of

decayed nuclei becomes either

e

1e times

original nuclei or apporoximately 63% of originalnuclei.

(vii) Fraction of active nuclei left =0N

N

(probability of survival)(viii) Fraction of decayed nuclei

= 1 –0N

N =

0

0

NNN

(Probability of decay)

(ix) When decay process is too slowthen N = N0 [e–t]

N = N0 [1– t]N = [– N0] t + N0

On comparingy = m x + c

(x) Half life, mean life & decay constant for aparticular nuclei always remains constant.

(xi) Decay constant () :

(i) =NdtdN)(

= instantthatatnucleiactiveofNo.tantinsgivenaatdecayofRate

(ii) =

dtNdN

= Probability of decay in unit time.

(iii) If = t–1 then N = eN0 = 0.3676

N0 0.37 N0Decay constant is equal to inverage of thattime in which no. of active nuclei left becomes

either e1

times or approx 37% original

nuclei.(iv) N = N0

–t

can be written in the form of mass as belowm = m0e–t

59

When m = mass of radio active nuclei attime t & m0 = mass of radioactive nuclei attime t = 0

0mm

= T/t2

1

(v) Decay constant is equal to inverse of thetime in which 63% of initial atoms (N0) isbeing decayed.

(vi) Unit :- (second)–1 or (minute)–1 or (year)–1

3. ACTIVITY (A or R)

(1) Activity of any element shows instability ofthe element at any time (t).

(2) Activity is defined as rate of decay at giveninstant.

A = – dtdN

= N

A N

(3) Original activityA0 = N0

0A

A =

0NN

= te = T/t)2(1

(4) Units of Activity(a) 1 disintegration / sec = 1 becquerd (Bq)(b) 1 curie = 3.7 × 1010 dps(c) 1 Ruther ford = 106 dps

(5) 1 mille curie = 37 Rutherford

(6) Specific Activity Activity per gram of asubstance is known as specific activity. Thespecific activity for 1 gm of radium 226 is1 curie.

4. PAIR PRODUCTION & PAIR-ANNIHILATION

Collision of -ray photon by a nucleus &production of electron positron pair is known aspair production.

The rest mass of each of the electron & thepositron is 9.1 x 10–31 kg. so, the rest massenergy of each of them is

E0 = m0c2= (9.1 × 10–31) (3 × 108)2

= 8.2 × 10–14 Joule = 0.51 MeV

Hence for pair-production, it is essential that theenergy of -photon must be at least 2 × 0.51= 1.02 MeV.

5. FUNDAMENTAL PARTICLES & THEIRANTIPARTICLES

The particles which are not constituted by anyother particles i.e., which have no structure, arecalled ‘fundamental particles’

(1) Electron :It is the first fundamental particle whichwas discovered by Thomson in 1897. It revolvesaround the nucleus of an atom indifferent orbits.Electron plays an important role in explaining thephysical and chemical properties of substances.Its charge is – 1.6 × 10–19 coulomb and massis 9.1 × 10–31 kg. Its symbol is e¯ (or –1

0).(2) Proton :It was discovered by Rutherford in 1919

in artificial nuclear disintegration. It has a positivecharge (+ 1.6 x 10–19 coulomb) equal to theelectronic charge and its mass is (1.673 × 10–27kg)1836 times the electronic mass. In free state,the proton is a stable particle. Its symbol is p+.It is also written as 1H1.

60

(3) Neutron :It was discovered by Chadwick in 1932.It carries no charge. Its mass is 1839 times theelectronic mass (1.675 x 10–27 kg). In free statethe neutron is unable (its mean life is about17 minutes), but it constitutes a stable nucleusalong with proton. Its symbol is n or 0n1.

(4) Positron :It was also discovered in 1932 byAnderson. Its charge and mass are same asthose of electron, the only difference being thatis positively-charged whereas the electron isnegatively-charged. Its symbol is e+ (or +1

0).

(5) Antiproton :It was discovered in 1955. Itscharged and mass are same as those of proton,the only difference being that it is negativelycharged. Its symbol is p¯.

(6) Antineutron : It was discovered in 1956. It hasno charge and its mass is equal to the mass ofneutron. The only difference between neutron andantineutron is that if they spin in the samedirection, their magnetic moment will be inopposite directions. The symbol for antineutronis n¯.

(7) Neutrino and Anti-neutrino : The existence ofthese particles was predicted in 1930 by Pauliwhile explaining the emission of -particles fromradio-active nuclei, but they were observedexperimentally in 1956. Their rest-mass andcharge are both zero but they have energy andmomentum. Both neutrino and anti-neutrino arestable particles. The only difference between themis that their spins are in opposite directions.Their symbols are v and v respectively..

(8) Pi-mesons :The existence of these particles waspredicted by Yukawa in 1935 as originator ofexchange-forces between the nucleons, but theywere actually discovered in 1947 in cosmic rays.Pi-mesons are of three types.(i) Positive pi-meson : It is a positively chargedparticle whose charge is equal to the electroniccharge and whose mass is 274 times theelectronic mass. It is an unstable particle. Itsmean life is of the order of 10–8 second. Itssymbol is +.(ii) Negative pi-meson : It is a negativelycharged particle whose charge is equal to theelectronic charge and whose mass is 274 timethe electronic mass. It mean life is also of theorder of 10–8 second. Its symbol is ¯.

(iii) Neutral pi-meson : This particle has nocharge. Its mass is nearly 264 times theelectronic mass. Its mean life is of the order of10–15 second. Its symbol is 0. On disintegration,it forms two -photons:0 .

(9) Photon :These are the bundles ofelectromagnetic energy and travel with the speedof light. If the frequency of waves be , then theenergy of a photon is h and momentum is h/ c.Its symbol is .

Name ofparticle

Symbol Antiparticle

Mass(in comparision

to mass ofelectron)

Average life(in second)

Photon ( ) 0 StableElectron e-1 e+1 1 StableProton p+ p– 1836 StableNeutron n n– 1839 1010NeutrinoCharged v v– 0 Stablepi-meson

Uncharged + – 274 2.6 × 10–8

pi-meson 0 (0) 264 0.9 × 10–16

6. CHARACTERISTICS OF RADIOACTIVERADIATIONS

6.1 Characteristics of - decay(1) - particle are two times ionised Helium

atoms.(2) - particle carry 2- proton and 2 neutrons.(3) 2He4 (Nucleus) = -particle(4) In general - decay is given by

ZXA decayZ-2Y A-4 + 2He4 + (Energy),

(-particle)atomic number decreases by 2 & mass no.decreases by 4.

(5) Mass of - particle = (2p + 2n) = 6.68 × 10-27 kg

Charge of - particle = + 2e= 2 × 1.6 × 10-19 coulomb, (+) Ve.

(6) In - particle emission an element goes twocolumn backward in periodic table.

(7) Energy in - particle emission

= (Mx– My – 42HeM ) C2

Energy in - particle emission varies from4.5 MeV to 11 MeV. Through calculation- Particle has to crop potential barrier of21 MeV for their emission. Hence -particleemission can not be explained on the basisof classical theory.

61

(No.

of-p

artic

le)

n

n0

0.15 MeV E1.17 MeV(End point energy)

(8) Emission of -particle can be explained onthe basis of quantum mechanics (Tunneleffect).

(9) Energy spectrum of - particle is linespectrum.

(10) Energy spectrum of -particle also has microdetails

(11) Energy spectrum of -particle shows that anucleus also has energy levels like atomshave.

(12) Energy of -particle emitted from a singlenucleus are not same. These emit in variousenergy groups.

(13) Range of - particle (velocity)3

R v3

(14) When -particle emission takes place,-rays are also emitted.

(15) R E3/2 or R = 0.318 E3/2

E : Energy of - particle(16) (Geiger's and Nuttal law)

Relation between decay constant of a elementand range of -particle as follows:-

log = A + B log RA and B are constant. B has equal valuewhile A have different values for radioactiveseries.

6.2 Characteristics of - decay(1) -particle are high energy electron or positron.

(a)- or -1eº (electron)(b) + or +1e º (Positron)

(2) Resultant charge on -particle= ± 1.6 × 10-19 coulomb

(3) Rest mass of -particle are equal to massof electron.

(4) -particle emission can be represented byfollowing reactions:-

ZxA particle Z+1YA + -1e0 + Q

ZxA particleZ-1YA + +1e0 + Q

where Q = Energy (5) In - particle , atomic no. increase by one

and in +emission, atomic no. decreases byone.

(6) Mass number does not changes in-emission.

(7) Emission of particle can be explained byconversion of nuetron in proton & vice verseain the nucleus.

1p1 0n1 + 1e0 (+)or 0n1 1p1 + -1e0 (-)

(8) To explain energy conservation, linearmomentum conservation and angularmomentum conservation, a hypotheticalneutrino was considered

(9) Neutrino was first given Pauli. (10) According to neutrino hypothesis, some

particle also emits with -emission, which iscalled neutrino.

(11) Rest mass and charge of neutrino are bothzero and angular or spin momentum ofneutrino is ± ½ (h /2). It travel with speedof light and it's spin value is ± ½.

(12) So by neutrino hypothesis , emission of particle reaction

0n1 1p1 + -1e0 + (Anti-Neutrino)

1p1 0n1 + 1e0 + (Neutrino)Hence reaction ZXA Z-1YA + +1e0 + + Q

ZXA Z+1YA + -1e0 + + Q

(13) Existance of nutrino is practically explainedby Rein's Collin.

(14) Energy spectrum of -emission iscontinuous.

6.3 Characteristic of spectrum(1) Energy range of emitted - particle has all

possible energy's -particle.

(2) Maximum value of energy of -particle iscalled end point energy.

(3) During -emission the decreases in energyof parent nucleus is equal to end point energyof -particle, latter it is shared by -particleand neutrino.

6.4 Characteristics of - rays:-(1) - rays are electromagnetic waves of short

wavelength (10-12m)

(2) They emit from nucleus.

(3) They travel with speed of light(3 × 108 m/sec)

(4) These are high energy rays (of photons)

62

(5) When or particle emission takes place,nucleus come in excited state and duringcoming back to normal state radiationemission takes place

27Co60 28Ni60(exicited state) + -1e0

28Ni60(exicited state) 28N60 (ground state) + Q (gama rays)

(6) In -decay atomic no. and mass no does notchanges.

(7) Energy spectrum of -rays is line spectrum.

(8) This spectrum verifies that same energy levelsare found in nucleus as that of in atomoutside the nucelus.

(9) It affects photographic plate.

(10) It has ionising power.

(11) It also has penetration power.

(12) It is not affected by electric & magnetic field.

(13) Intensity of -rays after travelling x- distanceis I = I0 e x [same as x-rays]I0-Initial intensity, - Absorption coefficient.I-Intensity of -rays after x- distance

(14) -rays also shows diffraction by crystal gratinglike x- rays.

(15) -rays radiations after entering into substanceare absorbed in three process dependingupon energy(A) Photoelectric effect (B) Compton effect(C) Pair production.

7. RADIOACTIVE SERIES

There are mainly four radio active series. Threeare natural and one is artificial.

(1) Thorium series (4n series)

90Th232 Irregular

decay ,

82Pb208 (stable element)

(2) Neptunium series (4n + 1 series)This is artificial series

93Np237 Irregular

decay , 83Bi209 (stable)

(3) Uranium series (4n + 2 series ) :-

92U238 Irregular

decay , 82Pb206 (stable)

(4) Actinium series (4n+3 series) :-

92U235Irregular

decay , 82Pb207

Last element of radio active series is stableand decay constant of that element has valueequal to zero.

8. RADIOACTIVE EQUILIBRIUM

NA A = NB B = ...........

orA

A

TN

=B

B

TN

= ..............

9. USES OF RADIOACTIVE ISOTOPES

(a) In Medicine -(i) For testing blood-Chromium - 51(ii) For testing blood circulation - Sodium - 24(iii) For detecting brain tumor-Radio mercury-203(iv) For detecting fault in thyroid gland-Radio

iodine-131(v) For cancer - cobalt - 60(vi) For blood-Gold-189(vii) For skin diseases-Phospohorous-31

(b) In Archaeology -(i) For determining age of archaeological sample

(Carbon dating) C14

(ii) For determining age of meteorities-K40

(iii) For determining age of earth-Lead isotopes (c) In Agriculture -

(i) For protecting potato crop from earthworm -CO60

(ii) For artificial rains-Ag(iii) As fertilizers-P32

(d) As tracers - (Tracer) -Very small quantity of radioisotopes presentin a mixture is known as Tracer

(i) Tracer technique is used for studyingbiochemical reaction in tracer and animals.

(e) In industries -(i) For detecting leakage in oil or water pipe

lines.(ii) For determining the age of planets

63

10. POINTS OF REMEMBER

(1) Radioactive decay is a nuclear process. (2) The decay constant of the end product of a

radioactive series is zero. (3) Radioactivity is not an atomic process. (4) - rays are never emitted directly by the

nuclei. (5) The time taken by any radioactive material to

decay completely is infinity irrespective of itsmass and decay constant.

(6) Radioactivity is not associated with theelectron configuration in atom.

(7) The spectrum of -rays is a discrete linespectrum.

(8) The charge on neutrino is zero. (9) Activity depends on the quantity & nature of

that material. (10) Radioactivity remains unaffected due to the

physical and chemical changes of thematerial.

(11) 1 curie = 3.7 x 1010 disintegration/sec. (12) 1 Rutherford = 106 disintegrations/sec. (13) decay is explained on the basis of tunnel

effect.

(14) Size of nucleus decreases by -emission. (15) Cloud chamber is used for detecting

radioactive radiations and for determining theirpaths, range and energy.

(16) Geiger-Muller counter is used for detecting-and -particles and measuring half life.

(17) Coefficient of absorption depends on thewavelength of -rays and the nature ofabsorbing material.

(18) The mass number and the charge numberremain unchanged by -decay

(19) When a -particle is emitted by a nucleusthen its mass number remains unchangedwhere as the charge number increases ordecreases.

(20) When an -particle is emitted by a nucleusthen its atomic number decreases by 2 andmass number decreases by 4.

(21) There are five types of radioactivedisintegration.

(22) Radiation dose is measured in sieverts (Sv).(23) The interaction responsible for beta decay is

called weak interaction.

64

Ex.1 The half life of polonium is 140 days. In whattime will 15 gm of polonium be disintegratedout of its initial mass of 16 gm ?(A) 500 days (B) 480 days(C) 560 days (D) 280 days

Sol. Suppose the initial mass of a radio activeelement is N0. The mass of the element left

after n half-lives is given by N = N0

n

21

0N

N =

n

21

N0 = 16 gm. The mass of the disintegratedelement is 15 gm So, the mass of theelement left is N = 16 – 15 = 1 gm. Thus

161

=n

21

n = 4

The half life of polonium is 140 days. Hence,the time taken in the disintegration of150 gm polonium = half-life × no. of half lives= 140 × 4 = 560 days

Ex. 2 The half life of radium is 1600 years. In how

much time will its 1615

fraction disintegrate ?

(A) 6400 years (B) 3200 years(C) 1600 years (D) 8000 years

Sol. Ist time of Decay t = 2log)N/Nlog(T 0

t =2log

116log1600

= 2log2log16004

,

t = 6400 yearsEx. 3 The half-life of radium is 1600 years. After

how many years 25% of radium blockremains undecayed?(A) 3200 years (B) 4800 years(C) 7200 years (D) 9600 years

Sol. Suppose the initial quantity of radium is N0.Then the quantity left after n half-lives will be

N = N0

n

21

Here, N = 25 % of N0 =4

N0

SOLVED EXAMPLES

4

N0 = N0

n

21

,

41

=n

21

n

21

=

2

21

n = 2

time of disintegration = half life x number ofhalf lives = 1600 × 2 = 32 00 years

Ex.4 Find the half-life period of a radio-activematerial if its activity drops to 1/16 th of itsinitial value in 30 years.(A) 15 yr. (B) 7.5 yr.(C) 22.5 yr. (D) 120 yr.

Sol. If the initial mass of some radio-activeelement be N0, then the mass of the elementremaining after n half-lives is given by

N = N0

n

21

0N

N =

n

21

16

1=

n

21

n

21

=

4

21

n = 4

Half life of the substance

= liveshalfofnumbertiondisintegraoftime

=4

30 = 7.5 years.

Ex.5 If a radio active material contains 0.1 mg ofTh234, how much of it will remain unchangedafter 120 days ? Its half life is 24 days.(A) 0.0312 gm (B) 0.0312 mg(C) .00312 mg (D) .00312 gm

Sol. T = 24 days, t = 120 days

mM0 =

T2t

= 2120/24 = 25 = 32, N = 32N0 ,

Mass left = 32massOriginal

= 321.0

= .003125 mgEx.6 The half life of a radioactive material is 12.7 hr.

What fraction of the original active materialwould become inactive in 63.5 hr ?(A) 1/32 (B) 1/23(C) 31/32 (D) 23/32

Sol. t = 63.5 Hr, T = 12.7 Hr.

65

0NN

= T/t21

= )7.12/5.63(21

= 521

0N

N = 32

1

Inactive fraction

= 1 –0N

N = 1 – 32

1 = 32

31

Ex.7 The activity of a radioactive sample drops of1/32 of its initial value in 7.5 h. Find the halflife ?(A) 7.5 Hr. (B) 5 Hr.(C) 1.5 Hr. (D) None

Sol. Given,0R

R = 32

1, t = 7.5 h

0RR

=T/t

21

321

=T/5.7

21

5

21

=

T/5.7

21

,

5 = 7.5/T T = 55.7

= 1.5 hours

Ex.8 A radio active sample contains 106

radioactive nucleus. It's half life is 20 sec.Number of remaining nucleus after10 seconds.(A) 7.09 (B) 7.09 × 105

(C) 79 (D) 709

Sol. N = 106T/t

21

N = 10620/10

21

N = 106/ 2 = 106/1.41

N = 7.9 × 105

Ex.9 The count rate of a radio active source att = 0 was 1600 count/s and at t = 8 sec, itwas 100 counts/s. The count rate (in counts)at t = 6 sec will be-(A) 150 (B) 200(C) 300 (D) 400

Sol. t =2logN

NlogT 0

8 =2log100

1600logT = 2log

16logT

8 = 2loglogT 4

2 8 = 4T 2log2log

T = 2sec

N = T/t0

2N

= 2/621600

N = 200 counts/s

Ex.10 1 milligram radium has 2.68 × 1018 atoms.Its half-life is 1620 years. How many radiumatoms will disintegrate from 1 milligram ofpure radium in 3240 years ?(A) 2.01 × 1018

(B) 0.75 × 1018

(C) cannot be predicted(D) None

Sol. If the initial quantity of a radio-active elementbe N0, then the quantity left after n half-lives

is given by N = N0

n

21

The half life of radium is 1620 years. Thenumber of half-lives in 3240 years is

n = 16203240

= 2

N = 1 ×2

21

( N0 = 1mg), N = 1/4 = 0.25 m mass of disintegrated radium

= 1– 0.25 = 0.75 mNumber of atoms in it = 0.75 × (2.86 × 1018)

= 2.01 × 1018

Ex.11 The mean lives of a radioactive material for and radiations are 1620 and 520 yearsrespectively. The material decayssimultaneously for and radiations. Thetime after which one fourth of the materialremains undecayed is-(A) 540 years (B) 324 years(C) 720 years (D) 840 years

Sol. =

= 5201620

5201620

= 394 years

Time of decay t = 2.303 log10 NN0

t = 394 × 2.303 log104 t = 394 × 2.303 × 0.602 t = 546 years

66

Ex.12 When 90Th228 gets converted into 83Bi212,then the number of -and -particles emittedwill respectively be-(A) 4 , 7 (B) 4 , 1 (C) 8 , 7 (D) 4 , 4

Sol. 90Th228 = 83Bi212 + x (2He4) + y(-1e0)According to law of conservation of charge90 = 83 + 2x – y2x – y = 7According to law of conservation of massnumber 228 = 212 + 4x4x = 16, x = 4, 2 × 4 – y = 7, y = 1Hence 4 and 1 will be emitted.

Ex.13 A radioactive Nucleus decays as follows:

A A1 A2

A3 A4

If the mass number and charge number of Aare 180 and 72 respectively, then for A4 thesevalues will respectively be-(A) 172, 69 (B) 108, 252(C) 108, 72 (D) None

Sol. 72A180 = 2He4 + 70A1176

70A1176 = -1e0 + 71A2

176

71A2176 = 2He4 + 69A3

172 , 69 A3172

= + 69 A4172

Ex.14 If the activity of radioactive sample drops to1/32 of its initial value in 7.5 Hr. Half life willbe-(A) 3 Hr (B) 4.5 Hr(C) 7.5 Hr (D) 1.5 Hr.

Sol. A = A0/32, t = 7.5 Hr, T = ?

A

A0 = 2t/T

32 = 2t/T 25 = 2t/T 5 = t/T T = t/5 T = 7.5/5 = 1.5 Hr.

Ex.15 In 420 days, the activity of a sample ofpolonium, Po fell to one lights of its initialvalue. Nuclear reaction is following

ba

cd

fePo Pb 82

206 Half life ofpolonium & value of a,b, c, d, e, f will be-(A) 140, 210, 84, 4, 2, 0, 0(B) 420, 208, 83, 2, 1, 0,0(C) 420, 210, 84, 2, 4, 0, 0(D) 140, 210, 84, 2, 4, 0, 0

Sol.N

N0 = 2t/T, 8 = 2t/T 23 = 2t/T

3 = t/T T = t/3 = 420/3 = 140 daysReaction will be,

84210

24

82206

00Po Pb

Ex.16 A freshly prepared radioactive sample, withhalf-life 2 hours, emits radiations whoseintensity is 64 times higher than its safelevel. The minimum time after which it will besafe to work with the sample will be-(A) 42 hr (B) 6 hr(C) 128 hr (D) 12 hr

Sol.0N

N =

T/t

21

= 64

1 64

1 =

2/t

21

6

21

=

T/t

21

t = 6T = 12 hr.