1. RADIOACTIVITY

It was observed by Henri Becquerel in i896 that some minerals like pitchblende emit radiation spontaneously, and this radiation can blacken photographic plates if photographic plate is wrapped in light proof paper. He called this phenomenon radioactivity. It was observed that radioactivity was not affected by temperature, pressure or chemical combination, it could therefore be a property that was intrinsic to the atom-more specifically to its nucleus. Detailed studies of radioactivity resulted in the discovery that radiation emerging from radioactive substances were of three types :

a

-rays,

b

-rays and

g

-rays.

a

-rays were found to be positively charged,

b

-rays (mostly) negatively charged, and

g

- rays uncharged. Some of the properties of

,

ab

and

g

rays, are explain below:

(i)Alpha Particles: An (-particle is a helium nucleus, i.e. a helium atom which has lost two electrons. It has a mass about four times that of a hydrogen atom and carries a charge +2e. The velocity of (-particles ranges from 5 to 7 percent of the velocity of light. These have very little penetrating power into a material medium but have a very high ionising power.

(ii)Beta Particles: (-particles are electrons moving at high speeds. These have greater (compared to (-particles) penetrating power but less ionising power. Their emission velocity is almost the velocity of light. Unlike (-particles, they have a spectrum of energy, i.e., beta particles possess energy from a certain minimum to a certain maximum value.

(iii)Gamma Rays: (-rays are electromagnetic waves of wavelength of the order of ~ 1012 m. These have the maximum penetrating power (even more than Xray) and the least ionising power. These are emitted due to transition of excited nucleus from higher energy state to lower energy state.

Property

a

b

g

NaturePositive chargedNegatively charged Uncharged

particles, 2He4particles (electrons).

~0.01A,

l

o

nucleuselectromagneticradiation

Charge+2ee0

Mass6.6466 1027 kg9.1091031kg0

Range~10 cm in air, can beUpto a few m inSeveral m in air

stopped by 1mm of A

l

can be stopped stopped by ~ cm of

~cm of A

l

Pb

Natural By natural radioiso-By radioisotopesExcited nuclei formed

sourcestopes e.g. 92U236e.g. 29Co68as a result of

a

,

b

decay

The other simple laws of radioactivity that were discovered by were the Rutherford-Soddy displacement laws:

(a)During a

a

decay, the daughter element was always two position below the parent in the periodic table; the mass number of the daughter nucleus was 4 units smaller than that of the parent nucleus.

-

-

=+a

bb4

aa2

XY

(b)During

b

-decay, the daughter nucleus had the same mass number as the parent while its atomic number was greater than that of the parent by 1 (smaller by 1 unit in

+

b

decay).

+-

+b

bb

aa1

XY

(c)

g

emission did not result in change of atomic number or mass number.

Radioactivity is observed to be a random process, it cannot be predicted when a particular nucleus will decay. One can only discuss the probability of its decay.

2. RADIOACTIVE DECAY LAW

The number of atoms disintegrating per second

dN

dt

is directly proportional to the number of atoms (N) present at that instant. For most radioisotopes,

l

is very small in the SI system it take a large number N(~ Avogadro number, 1023) to get any significant activity.

dN

N

dt

=-

l

.(i)

Where ( is the radioactivity decay constant.

If No is the number of radioactive atoms present at a time t = 0, and N is the number at the end of time t, then

-

=

0

NNe

t

The term

-

dN

dt

is called the activity of a radioactive substance and is denoted by 'A'.

Units:1 Becquerel (Bq)= 1 disintegration per second (dps)

1 curie (Ci)= 3.7 x 1010 dps

1 Rutherford=106 dps

The changes in N, due to radioactivity, are very small in 1 s compared with the value of N 'its self.

Therefore. N may be approximated by means of a continuous function N(t).

For the case considered in equation (i), we can write,

A = rate of decrease of N =

N

l

or

dN

N

dt

-=l

.....(ii)

Note the equation (ii) is valid only for large value of N, typically 1018 or so.

If the number of atoms at time t = 0 (initial) is N = N0, then one can solve equation (ii):

0

Nt

N0

dN

dt

N

=-l

orIn

0

N

t

N

=-l

orN(t) = N0

t

e

-l

......(iii)

The activity A (t) =

N

l

(t) and is given by

A (t) =

t

0

Ne

-l

l

= A0

t

e

-l

taking (A0 =

l

N0)....(iv)

The variation of N is represented below:

In N

Straight Line

t

N

Expontential decay

t

av

1t2oN2

N

o

.31N

0

Graphs of A vs t or In A vs t are similar to the above graphs

2.1 HALF-LIFE

A significant feature of the above graph is the time in which the number of active nuclei (N) is halved. This is independent of the starring value, N = N0. Let us compute this time:

r'

0

0

N

Ne

2

-l

=

\

t'

e

-l

=

1

2

or,

t'

l

= In 2 = 0.693 (approx)

or,t' =

1/2

T

=

0.693

l

This half-life represents the time in which the number of radioactive nuclei falls to

1

2

of its starting value. Activity, being proportional to the number of active nuclei, also has the same half-life.

Illustration 1: A count rate-meter is used to measure the activity of a given sample. At one instant the meter shows 4750 counts per minute. Five minutes later it shows 2700 counts per minute. Find:

(a)decay constant

(b)the half life of the sample.

Solution:Initial activity = A0 = dN/dt at t = 0

Final activity = At = dN/dt at t = t

0t

t0t5

dNdN

N&N

dtdt

==

=l=l

0

t

N

4750

2700N

=

0

t

N

Using t2.303log

N

l=

(

)

1

47502.3034750

52.303loglog0.1129min

270052700

-

l==

1/2

0.693

t6.14min

0.1129

==

2.2 MEAN-LIFE

A very significant quantity that can be measured directly for small numbers of atoms is the mean lifetime. If there are n active nuclei, (atoms) (of the same type, of course), the mean life is

122

.......

n

t+t++t

t=

Where

1

t

,

2

t

,.......

n

t

represent the observed lifetime of the individual nuclei and n is a very large number. It can also be calculated as a weighted average:

1122nn

1n

NN.......N

N.....N

t+t++t

t=

++

Where N1 nuclei live for time

1

t

,

N2 nuclei live for time

2

,

t

....... and so on.

This quantity may be related with

l

.

Using calculus (vi) may be rewritten as:

tdN

dN

t=

Where |dN| is the number of nuclei decaying between t, t + dt; the modulus sign is required to ensure that it is positive.

dN =

t

0

Nedt

-l

-l

and |dN| =

t

0

Nedt

-l

l

\

t

0

0

t

0

0

tNedt

Nedt

-l

-l

l

t=

l

=

t

0

t

0

tedt

edt

-l

-l

=

2

1

1

l

l

(integrating the numerator by parts)

or

1

t=

l

Illustration 2:Cascade Decays: Frequently, a nucleus A decays into another nucleus B, which again decays into C, and so on until the series ends in a stable nuclide. These are known as cascade decays. We will consider a simple example of a cascade decay where a nuclide a decays into B, and then, B decays into C:

AB

ABC

Suppose that there are N0 atoms of A to begin with the problem is to find the numbers 0f atoms of A, B, C respectively as a function of time t.

Solution:Suppose that, at time t, there are NA atoms of A. Nb atoms of B. and We write the equations:

A

dN

dt

=

activity of A =

AA

N

-l

.....(i)

B

dN

dt

= + (activity of A) (activity of B)

=

AABB

NN,

l-l

....(ii)

Since each decay of A increases the number of atoms of B

c

BB

dN

N

dt

=l

....(iii)

Solving equation (i), we get

NA (t) = N0

A

t

e

-l

....(iv)

Substituting this in equation (ii), we get. the solution,

NB (t) =

A

A

t

t

A0

1

AB

Ne

Ce

-l

-l

l

+

l+l

......(v)

Putting the initial value (t = 0) of NB, we get C1, and substituting this value of C1 we get the expression for NB :

NB (t) =

AB

tt

A0

ABAB

N

1e1e

-l-l

l

--

-

-l+lll

.......(vi)

Substitution the value of NB in equation (iii) and solving, we get

NC =

AB

tt

AB0

ABAB

N

1e1e

-l-l

ll

--

-

-l+lll

......(vii)

Using these values of NA, NB and NC. the activity can be calculated, if required.

3. The Nucleus

It exists at the centre of an atom, containing entire positive charge and almost whole of mass. The electron revolve around the nucleus to form an atom. The nucleus consists of protons (+ve charge) and neutrons.

(i)A proton has positive charge equal in magnitude to that of an electron (+1.6 x 1019 C) and a mass equal to 1840 times that of an electron.

(ii)A neutron has no charge and mass is approximately equal to that of proton.

(iii)The number of protons in a nucleus of an atom is called as the atomic number (Z) of that atom. The number of protons plus neutrons (called as Nucleons) in a nucleus of an atom is called as mass number (A) of that atom.

(iv)A particular set of nucleons forming an atom is called as nuclide. It is represented as ZXA.

(v)The nuclides having same number of protons (Z), but different number of nucleons (A) are called as isotopes.

(vi)The nuclide having same number of nucleons (A), but different number of protons (Z) are called as isobars.

(vii)The nuclide having same number of neutrons (A - Z) are called as isotones.

3.1 MASS DEFECT & BINDING ENERGY

The nucleons are bound together in a nucleus and the energy has to be supplied in order to break apart the constituents into free nucleons. The energy with which nucleons are bounded together in a nucleus is called as Binding Energy (B.E.). In order to free nucleons from a bounded nucleus this much of energy (= B.E.) is to be supplied.

It is observed that the mass of a nucleus is always less than the mass of constituent (free) nucleons. This difference in mass is called as mass defect and is denoted as (m.

Ifmn : mass of a neutron;mp : mass of a proton

M (Z, A) : mass of bounded nucleus

Then, (m = Z . mp + (A Z). mn M (Z, A)

This mass-defect is in form of energy and is responsible for binding the nucleons together. From Einstein's law of inter-conversion of mass into energy:

E = mc2 (c : speed of light; m : mass)

binding energy = (m . c2

Generally, (m is measured in amu units. So let us calculate the energy equivalent to 1 amu. It is calculated in eV (electron volts; 1 eV = 1.6 x 1019 J)

(

)

2

278

6

19

1.6710310

E1amueV93110eV931MeV

1.610

-

-

===

(

)

B.E.m931MeV

=D

There is another quantity which is very useful in predicting the stability of a nucleus called as Binding energy per nucleons.

(

)

m931

B.E. per nucleonsMeV

A

D

=

02060180A

mass number

B.E.A

(In Mev)

8.5 Mev

2

4

6

8

c

o

Be

From the plot of B.E./nucleons Vs mass number (A), we observe that:

(i)B.E./nucleons increases on an average and reaches a maximum of about 8.7 MeV for A ( 50 80.

(ii)For more heavy nuclei, B.E./nucleons decreases slowly as A increases. For the heaviest natural element U238 it drops to about 7.5 MeV.

(iii)From above observation, it follows that nuclei in the region of atomic masses 50-80 are most stable.

Illustration 2:If mass of proton = 1.008 amu and mass of neutron = 1.009 amu, then the binding energy per nucleon for 4Be9 (mass = 9.012 amu) will be

(A) 0.0672 MeV(B) 0.672 MeV

(C) 6.72 MeV(D) 67.2 MeV

Solution:Mass defect

m(41.00851.009)9.012

D=+-

9.0779.0120.065amu

=-=

BE0.065931

6.72MeV

A9

\==

Illustration 4:The energy released in the following (-decay process will be -

110

011

nPev

-

++

Given that

27

n

27

p

27

e

m1.674710kg

m1.672510kg

m0.0009110kg

-

-

-

=

=

=

(A) 0.931 MeV(B) 0.731 MeV

(C) 0.511 MeV(D) 0.271 MeV

Solution:Mass defect

27

m(1.67471.67250.0091)10

-

D=--

27

0.001210kg

-

=

2782

13

0.001210(310)

E0.731MeV

1.610

-

-

\D==

Illustration 5:The binding energy per nucleon for 3Li7 will be, if the mass of 3Li7 is 7.01653 amu.

(A) 5.6 MeV(B) 39.25 MeV

(C) 1 MeV(D) zero.

Solution:

Em931

EMeV

AA

DD

==

7

pn

m(3m4m)massofLi

D=+-

(31.0075941.008898)7.01653

0.04216amu

=+-

=

0.0421693139.25

E5.6MeV

77

D===

Illustration 6:How much energy is released in the following reaction?

224

112

HHHe

+=

If the B.E./Nucleon of 1H2 and 2He4 are 1.123 MeV and 7.2 MeV respectively.

(A) 12 MeV(B) 24.3 MeV

(C) 36 MeV(D) zero

Solution:B. E. of 1H2

E1.125

EAE

E21.125

2.25MeV

D=

=D

=

=

2

1

d

B.E.oftwoH2.25

E4.5MeV

\=

=

B.E. of an ( -particle = 4 ( 7.2

E( = 28.8

( energy released ER = E( Ed

ER = 28.8 4.5 = 24.3 MeV

3.2 NUCLEAR FORCES

The protons and neutrons are held together by the strong attractive forces inside the nucleus. These forces are called as nuclear forces.

(i)Nuclear forces are short-ranged. They exist in small region (of diameter 1015 m = 1 fm). The nuclear force between two nucleons decrease rapidly as the separation between them increases and becomes negligible at separation more than 10 fm.

(ii)Nuclear force are much stronger than electromagnetic force or gravitational attractive forces.

(iii)Nuclear force are independent of charge. The nuclear force between two proton is same as that between two neutrons or between a neutron and proton. This is known as charge independent character of nuclear forces.

In a typical nuclear reaction

(i)In nuclear reactions, sum of masses before reaction is greater than the sum of masses after the reaction. The difference in masses appears in form of energy following the Law of inter-conversion of mass & energy. The energy released in a nuclear reaction is called as Q Value of a reaction and is given as follows.

If difference in mass before and after the reaction is (m amu

((m = mass of reactants minus mass of products)

thenQ value = (m (931) MeV

(ii)Law of conservation of momentum is also followed.

(iii)Total number of protons and neutrons should also remain same on both sides of a nuclear reaction.

Illustration 7: Calculate the Q-value of the nuclear reaction:

0C12

10Ne20 + 2He4

The following data are given:

m(6C12) = 12.000000 u

m(10Ne20) = 12.000000 u

m(2Ne4) = 4.002603 u

Solution:The Q-value of this reaction may be easily calculated may be easily calculated from the masses of the individual nuclei.

Q = [2m (6C12) { m (10Ne20) + m (2He4)}]e2

= 24.000000 (19.992439 + 4.002603)} u c2

= 4.618 MeV

Illustration 8:Binding energy of the calculate the mass defect and the binding energy of an

a

-particle given the following date:

mn = 1.0086665 u

mp = 1.007825 u

m

= 4.0026 u

(1u = 931.5 MeV/c2)

Solution: The mass defect of an

a

-particle is given by

m

D

= (2 1.008665 + 2 1.007825) 4.0026)u

= 0.03038u

The Binding energy is related to the mass defect by the reaction.

B.E. =

2

m.c

D

= 0.03038 931.5 MeV

= 28.3 MeV (approximately)

3.3 NUCLEAR FISSION

The breaking of a heavy nucleus into two or more fragments of comparable masses, with the release of tremendous energy is called as nuclear fission. The most typical fission reaction occurs when slow moving neutrons strike 92U235. The following nuclear reaction takes place.

2351141921

92056360

UnBaKr3n200MeV

++++

If more than one of the neutrons produced in the above fission reaction are capable of inducing a fission reaction (provided U235 is available), then the number of fissions taking place at successive stages goes increasing at a very brisk rate and this generates a series of fissions. This is known as chain reaction. The chain reaction takes place only if the size of the fissionable material (U235) is greater than a certain size called the critical size.

If the number of fission in a given interval of time goes on increasing continuously, then a condition of explosion is created. In such cases, the chain reaction is known as uncontrolled chain reaction. This forms the basis of atomic bomb.

In a chain reaction, the fast moving neutrons are absorbed by certain substances known as moderators (like heavy water), then the number of fissions can be controlled and the chain reaction is such cases is known as controlled chain reaction. This forms the basis of a nuclear reactor.

Illustration 9:When a beta particle is emitted from a nucleus the effect on its neutron-proton ratio is

(A) increased(B) decreased

(C) remains same(D)first (1) then (2)

Solution:

AA0

ZZ11

XYe()v

+-

+b+

110

n

011

p

N

nPe()visdecreased

N

-

+b+\

Illustration 10:In which sequence, the radioactive radiations are emitted in the following nuclear reactions?

AAA4A4

ZZ1Z1Z1

XYKK

--

+--

(A) (, ( and ((B) (, ( and (

(C) (, ( and ((D) (, ( and (

Solution:We know that on emission of (-particle from the nucleus, atomic number reduces by 2 and mass number reduces by 4. And on emission of (-particle from the nucleus, the atomic number increases by 1. Similarly on emission of (-particle from the nucleus, there is no change in atomic number of mass number.

3.4 NUCLEAR FUSION

The process in which two or more light nuclei are combined into a single nucleus with the release of tremendous amount of energy is called as nuclear fusion. Like a fission reaction, the sum of masses before the fusion (i.e. of light nuclei) is more than the sum of masses after the fusion (i.e. of bigger nucleus) and this difference appears as the fusion energy. The most typical fusion reaction is the fusion of two deuterium nuclei into helium.

124

112

HHHe21.6MeV

++

For the fusion reaction to occur, the light nuclei are brought closer to each other (with a distance of 1014 m). This is possible only at very high temperature to counter the repulsive force between nuclei. Due to this reason, the fusion reaction is very difficult to perform. The inner core of sun is at very high temperature, and is suitable for fusion, in fact the source of sun's and other star's energy is the nuclear fusion reaction

4. assignment

1.The mass number of a nucleus is

(A) always less than atomic number

(B) always more than atomic number

(C)equal to atomic number

(D) some times more than and some times equal to atomic number.

Solution- (D) Mass number of a nucleus represents the number of nucleons (neutrons + protons). When no neutrons are present in the nucleus, mass number equals to the atomic number.

2.In which of the following decays, the element does not change?

(A) decay(B)+ decay

(C) decay(D)y decay

Solution- (D) (-ray has no charge and no mass during (-decay the element will not have any change in atomic number or mass number.

3.In the reaction

144171

7281

NHeOH

++

the minimum energy of the

a-

particle is

(A)1.21 MeV(B)1.62 MeV

(C)1.89 MeV(D)1.96 MeV.

(MN = 14.00307 amu, MHe = 4.00260 amu and MO

= 16.99914 amu, MH = 1.00783 amu and 1 amu = 931 MeV)

Solution- (A) 7N14 + 2He4

8O17 + 1H1

Total mass of reactants = 18.00567 amu

Tota mass of products = 18.00697 amu

Mass defect = 18.00697 18.00567 = 0.0013 amu

\

Energy (E) = 931 (0.0013) = 1.2103 MeV

4.In the carbon cycle of fusion

(A) Four 1H1 fuse to form 2He4 and two positrons

(B) Four 1H1 fuse to form 2He4 and two electrons

(C) Two 1H2 fuse to form 2He4

(D) Two 1H2 fuse to form 2He4 and two neutrons

Solution- (A) 41H1

2He4 + 2. +1e0 + Energy.

5.In each fission of U235, 200 MeV of energy is released. If a reactor produces 100 MW power, the rate of fission in it will be

(A) 3.125 1018 per minute >(B) 3.125 1017 per second

(C) 3.125 1017 per minute>(D) 3.125 1018 per second

Solution- (D) We know

nE

P

t

=

6

13

nP10010

tE200[1.610]

\==

= 3.125 1018 /sec

6. To generate a power of 3.2 MW, the number of fissions of U235 per minute is (Energy released per fission = 200 MeV, 1 eV = 1.6 ( 1019 J)

(A)61018(B)61017(C)1017(D)61016

Solution- (A) Power of reactor

nE

P

t

=

Where 'n' is number of fissions 't' is time 'E' is energy released per fission

619

6

n(20010)(1.610)

3.210

60

-

\=

18

n610

=

7.If in nuclear reactor using U235 as fuel, the power output is 4.8 MW, the number of fissions per second is (Energy released per fission of U235 = 200 MeV watts, e eV = 1.6 ( 1019 J)

(A)1.51017(B)31019(C)1.5025(D)31025

Solution- (A) P = 4.8 MW = 4.8106 W

Power of a nuclear reactor

nE

P

t

=

6

13

nP4.810

tE(200)(1.610)

-

\==

= 1.5 1017

8.In the following nuclear reaction

1111

65

CBX

+b+

. What does X stand for ?

(A) A neutron (B)A neutrino (C)an electron(D) A proton

Solution-In -decay process a neutrino is accompanied with the emission of a positron.

9.If 200 MeV energy is released in the fission of a single U235 nucleus, the number of fissions required per second to produce 1 kilowatt power shall be (given 1 eV = 1.6 ( 1019 J)

(A)3.1251013(B)3.1251014

(C)3.1251015(D)3.1251016

Solution- (A) Energy released in the fission

E = 200 MeV = 2001061.61019 = 3.21011 Joule

nE

P

t

=

3

11

nP10

tE3.210

-

==

= 3.1251013

10.MP = 1.008 a.m.u., MN = 1.009 a.m.u. and 2He4 = 4.003 a.m.u. then the binding energy of (-particle is

(A)21.4 MeV(B)8.2 MeV(C)34 MeV(D)1.35 A/m.

Solution-(D) BE = 2Mp + 2Mn Ma

= 21.008 + 2 1.009 4.003

= 0.031 amu

= 0.031 931 = 28.8 MeV

11.Energy released in fusion of 1 kg of deuterium nuclei

(A) 91013 J(B) 61027 J

(C) 2107 KwH(D) 81023 MeV

Solution-(A) Fusion reaction of deuterium

1H2 + 1H2

2He3 + 0n1+3.27 MeV

23313

6.0210103.271.610

E

22

-

=

13

E910J

12.The energy produced in the sun is due to

(A)fission reaction (B)fusion reaction

(C)chemical reaction (D)motion of electrons and ions

Solution-(A) We know that source of the huge solar energy is the fusion of lighter nuclei. Fusion of hydrogen nuclei into helium nuclei is continuously taking place in the plasma, with the continuous liberation of energy. Therefore energy produced in the sun is due to fusion reaction.

13.Atomic number of a nucleus is Z, while its mass number is M. What will be the number of neutrons in its nucleus ?

(A)M(B)Z(C)(MZ)(D)(M+Z)

Solution-(C) M = Z + N

NMZ

\=-

14.In uncontrolled chain reaction, the quantity of energy released, is

(A)very high (B)very low

(C)normal (D)first (A) to (B)

Solution-(A) We know that in the uncontrolled chain reaction, more than one of the neutrons produced in a particular fission cause further fissions so that the number of fissions increases very rapidly. Thus this is a very fast reaction and the whole substance is fissioned in a few moments. A huge quantity of energy is released.

15.The net force between two nucleons 1 fm apart is F1 if both are protons, F2 if both are neutrons, and F3 if one is a neutron and the other is a proton.

(A)F1 < F2 < F3(B)F2 < F1 < F3

(C)F1 < F2 = F3(D)F1 = F2 < F3

Solution:(C) The nuclear force of interaction between any pair of nucleons is identical i.e. force between two neutrons (F2) equals the force between neutron and proton (F3). However, between two protons net force is equal to the resultant of nuclear force between them (attractive in nature) and electrostatic force between them (repulsive in nature). Hence F1 is a value lesser than F2 and F3. So F1 < F2 = F3.

16.Which of the following process represents a

g

-decay?

(A)

A

Z

X+

EMBED Equation.DSMT4

A

Z-1

X+a+b

(B)

A1A-3

Z0z-2

X+n

X+c

(C)

AA

Zz-1

X

X+f

(D)

A

Z-1z-1

X+e

AX+g

Solution:(C) In gamma decay neither the proton number (Z) nor the neutron number (A-Z) changes. Only the quantum states of the nucleons change.

17.The activity of a sample of radioactive material is R1 at time t1 and R2 at time t2 (t2 > t1). If mean life of the radioactive sample is T, then:

(A)R1t1 = R2t2(B)

12

21

R-R

=constant

t-t

(C)

12

21

t-t

R=Rexp

T

(D)

1

21

2

t

R=Rexp

Tt

Solution:(C)

If R0 be the initial activity of the sample, then

1

t

10

RRe

l

=

and

2

t

20

RRe

-l

=

Where

1

T

l=

1

MeanlifeT

==

l

Q

(

)

2

12

1

t

tt

2

t

1

R

e

e

R

e

-l

l-

-l

==

12

21

tt

RRexp

T

-

=

18.When 15P30 decays to become 14Si30, the particle released is

(A)electron (B)(-particle

(C)neutron (D)positron

Sol:(D)

30030

15114

PeSi

++

19.Among electron, proton, neutron and (-particle the maximum penetration capacity is for

(A)electron (B)proton

(C)neutron (D)(-particle

Sol:(C)

20.Plutonium decays with a half-life of 24000 years. If plutonium is stored for 72000 years, the fraction of it that remains is

(A)

1

2

(B)

1

3

(C)

1

4

(D)

1

8

Sol(D) Number of half-life periods

72000

3

24000

==

( The fraction that remains

3

11

28

==

21.The decay constant (() and the half-life (T) of a radioactive isotope are related as:

(A)

e

1

log2T

l=

(B)

e

T

log2

l=

(C)

2

T

l=

(D)

e

log2

T

l=

Sol(D) Half-life

e

log2

0.693

T

T

=l=

l

22.The decay constant of a radioactive sample is (. The half-time and the average-life of the sample will be respectively

(A)

(

)

ln2

1

and

ll

(B)

(

)

ln2

1

and

ll

(C)

(

)

1

ln2and

l

l

(D)

(

)

1

and

ln2

l

l

Sol(B) Because ln 2 = 0.301 ( 2.303 = 0.693

23.In the uranium radioactive series, the initial nucleus is 92U238 and the final nucleus is 82Pb206. When the uranium nucleus decays to lead, the number of (-particle emitted will be

(A)1(B)2(C)4(D)8

Sol(D)

12

n

A_A

23820632

8

444

-

a====

24.The activity of a radioactive sample is 1.6 curie, and its half-life is 2.5 days. Its activity after 10 days will be:

(A)0.8 curie(B)0.4 curie(C)0.1 curie(D)0.16 curie

Sol(C)

10

n4

2.5

=

n4

0

A11

A1.6

A22

==

A = 0.1 Curie.

25.In a mean life of a radioactive sample:

(A)about 1/3 of substance disintegrates

(B)about 2/3 of the substance disintegrates

(C)about 90% of the substance disintegrates

(D)almost all the substance disintegrates.

Sol(B)

0

N

1

tlog

N

=

l

Given

1

t

=t=

l

00

NN

11

loglog1

NN

==

ll

00

00

NN

1

2.7N0.37NN

N2.73

===

26.Concept of neutrino was given by

(A)Sommerfield (B)Pauli

(C)P-Fermy (D)J. Chadwick

Sol(B)

27.Half-life of an element is 30 days. How much part will remain after 90 days

(A)1/4th part (B)1/16 part

(C)1/8th part (D)1/3rd part

Sol(C)

(1/T)(90/30)

0

N111

N228

===

28.In the radioactive decay

234222

9287

XY

number of emitted ( and ( particles emitted are

(A)3, 3(B)3, 1(C)5, 3(D)3, 5

Sol(B)

12

AA

234222

n3

44

a

-

-

===

21

nZ2nZ876921

ba

=+-=+-=

29.After five half lives what will be the fraction of initial substance remaining undecayed

(A)

10

1

2

(B)

5

1

2

(C)

4

1

2

(D)

1

2

Sol(B) After 5 half-lives

5

0

N1

N2

=

30.If half-life of a substance is 3.8 days & its initial quantity is 10.38 gm, then quantity of substance reaming after 19 days will be:

(A)0.151 gm(B)0.32 gm

(C)1.51 gm (D)0.16 gm

Sol(B) No. of half periods

19

5

3.8

==

Quantity remaining after n half lives

n5

1110.38

No.10.380.32gm

2232

====

For Test

1.1 atomic mass unit is equal to

(A)1/12 (mass of one C-atom)(B)1/16 (mass of O2-molecule)

(C)1/14 (mass of N2-moleucle)(D)1/25 (mass of F2-moleucle)

Solution-(A) We know that the atomic mass unit is defined as

1

12

th of the mass of one 6C12 atom.

2.For the construction of nuclear-bomb which of the following substances is taken ?

(A)U-235(B)Pu-239(C)F-14(D)both (A) and (B)

Solution-(D) We know that uranium (U235) and plutonium (Pu) are used as fissionable substances in a nuclear bomb. They can be fissioned easily by neutrons.

3.Energy obtained when 1 mg mass is completely converted to energy is

(A)3108 J(B)31010 J(C)91013 J(D)91015 J

Solution-(C) (E) = mc2 = (1 103) (3108)2 = 9 1013 J

4.1 MeV energy is

(A)1.61019 J(B)1.61016 J

(C)1.61013 J(D)1.61011 J

Solution-(C) 1eV = 1.6 1019 J

196

1MeV1.61010

\=

= 1.6 1013 J

5.Nuclear fission and fusion can be explained on the basis of

(A)conversion of energy principle

(B)Einstein mass-energy equivalence relation

(C)binding energy per nucleon variation with nucleon number

(D)variation of mass with increasing atomic nucleus

Solution- (B) We know that is nuclear fission and fusion, the mass is converted into the energy. Therefore nuclear fission and fusion can be explained on the basis of Einstein mass-energy equivalence relation.

6.If 1 gm hydrogen is converted into 0.993 gm of helium, in a thermonuclear reaction, the energy released in the reaction is

(A)63107 J(B)631010 J

(C)631014 J(D)631020 J

Solution-(B)

22

12

(E)mc(mm)c

=D=-

= {(1103) (0.993103)} (3 108)2

= (7 106) (9 1016) = 63 1010 J

7.In the given particles, which of the following is stable ?

(A)electron (B)proton (C)positron (D)neutron

Solution:(D) We know that neutrons do not have any charge. Therefore a neutron is more stable than proton, electron or position.

8.1 a.m.u. is equal to

(A)1 gm(B)4.81010 esu

(C)6.0231023 gm(D)1.661027 kg

Solution:(D) We know that one atomic mass unit (a.m.u.) is defined as

1

th

12

of the mass of the isotope of carbon atom (6C12) and is numerically equal to = 1.66 1027 kg.

9.Hydrogen bomb is based on the phenomenon of

(A)nuclear fission (B)nuclear fusion

(C)radioactive decay (D) none of these

Solution:(B) We know that in a hydrogen bomb, a uranium fission bomb is exploded, first, which produce high temperature of pressure. Due to this the heavy hydrogen nuclei come extremely close and fuse together, liberating huge energy. Thus the massive energy liberated is due to nuclear fusion.

10.When two deuterium nuclei fuse together in addition to tritium, we get a

(A)neutron (B)proton (C)(-particle(D) deuteron

Solution:(B) 1H2 + 1H2

1H3 + 1H1 + Q

11.In nuclear reaction there is conservation of

(A)energy only (B)mass, energy and momentum

(C)mass only (D)momentum only

Solution:(B) Mass, energy as well as momentum are conserved in nuclear reaction.

12.An atomic power reactor furnace can deliver 300 MW. The energy released due to fission of each of uranium atom U238 is 170 MeV. The number of uranium atoms fissioned per hour will be

(A)51015(B)101020(C)401021(D)301025

Solution:(C) No. of Atoms fissioned per sec

820

12

310310

27.21027.2

==

No. of Atoms fissioned per hour

20

2222

3360010336

10410

27.227.2

===

13.r1 and r2 are the radii of atomic nuclei of mass numbers 64 and 27 respectively. The ratio

1

2

r

r

is

(A)

64

27

(B)

27

64

(C)

4

3

(D)1

Solution:(A) 1H3 + 1P1

2He4

14.Two radioactive materials. X1 and X2 have decay constants 10

l

respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of X1 to that of X2. will be 1/e after a time

(A)

1

10

l

(B)

1

11

l

(C)

11

10

l

(D)

1

9

l

Solution:(D)

N1 = N0

10t

e

-l

N2 = N0

t

e

-l

9

t

l

= 1, or t =

1

9

l

15.The wave length of the characteristic X-ray

K

a

line emitted by a hydrogen like atom is 0.32 . The wave length of

K

b

line emitted by the same element is:

(A) 0.18 (B) 0.48

(C) 0.27(D) 0.38A

Solution:(C) The wavelength of x-rays emitted for

K

a

line and

K

b

line are given by

a

xL

hc

EE

l=

-

and

KM

hc

EE

b

l=

-

\

2

KM

KL

2

1

1

EE

32

13

11

EE27

12

a

b

-

l

-

===

l-

-

\

o

2727

0.320.27A

3232

ba

l=l==

16.The ratio of the mean life of a radionuclide to its half-life is

(A)1(B)1.44(C)2(D)3.12

Sol(B) Mean-life (t)mean

1

=

l

and half-life

1/2

0.693

(t)

=

l

mean

1/2

(t)

1/1

1.44

(t)0.693/0.693

l

===

l

17.The half-life of radioactive substance is 48 hr. How much time it will take to disintegrate to its 1/16th part.

(A)12 hr.(B)16 hr.(C)48 hr.(D)192 hr.

Sol(D)

nn

0

0

N

111

NNor

162162

===

n4

11

ororn4484192hr.

22

====

18.Radioactive substance emits

(A)(-rays (B)(-rays

(C)(-rays (D)all of these.

Sol(D) We know that when one (-particle is emitted, the mass (A) is reduced by 4 and atomic number (Z) is reduced by 2. Therefore for new nucleus, mass no. is (A 4) and atomic number is (Z 2).

19.A radioactive substance has a half-life of 1 year. The fraction of this material, that would remain after 5 years will be

(A)

1

32

(B)

1

5

(C)

1

2

(D)

4

5

Sol(A)

n5

0

0

N

1111

orNN

16223232

====

20.Half-life of a substance is 20 minutes. What is the time between 33% decay and 67% decay?

(A)40 minutes (B)20 minutes

(C)30 minutes(D)25 minutes

Sol(B) Let N0 be the number of nuclei at the beginning.

Number of undecayed nuclei after 33% decay = 0.67N0

Number of undecayed nuclei after 67% decay = 0.33N0

Also

0

0

0.67N

0.33N.

2

And in one half-life the number of undecayed nuclei becomes half.

21.The half-life of a radioactive substance is 3.6 days. How much of 20 mg of that radioactive substance will remain after 40 days?

(A)2.68 ( 103 mg (B)4.31 ( 102 mg

(C)6.20 ( 103 mg(D)9.76 ( 103 mg

Sol(D)

3

0

t/T40/3.611

m

2020

m9.7610mg

222

-

===

22.A substance reduces to 1/16th of its original mass in 2 hours. The half-life period of the substance will be:

(A)15 min(B)30 min(C)60 min(D)120 min

Sol(B)

t/T4

0

N111

N2162

===

tt120

4T30min

T44

\====

23.Which of the following processes represents a gamma-decay?

(A)

AA

ZrZ1

XXab

+-

++

(B)

A13

Z0Z2

XnZC

-

+

(C)

AA

ZZ

XXf

+

(D)

AA

Z1Z1

XeXg

--

++

Sol(C)

24.The half-life of 215At is 100 (s. The time taken for the radioactivity of a sample of 215At to decay to (1/16) th of its initial value is

(A)400 (s(B)6.3 (s

(C)40 (s(D)300 (s

Sol(A) Number of half lives = 4

Total time =

1/2

4t4100s400s

=m=m

25.A sample of a radioactive substance contains 2828 atoms. If its half-life is 2 days, how many atoms will be left intact in the sample after one day?

(A)1414(B)1000(C)2000(D)707

Sol(C)

0

1/2

N

2828

N2000

21.4

===

26.If N0 is the original mass of the substance of half life period t1/2 = 5 years, then the amount of substance left after 15 years is

(A)N0 / 8(B)N0 / 16

(C)N0 / 2(D)N0 / 4

Sol(A)

(

)

(

)

000

t/T15/5

NNN

N

8

22

===

27.At a specific instant emission of radioactive compound is deflected in a magnetic field. The compound can emit

(i)electrons (ii)protons (iii)He2+(iv)neutrons

The emission at the instant can be

(A)i, ii, iii(B)i, ii, iii, iv (C)iv(D)ii, iii

Sol(A)

28.A radioactive sample at any instant has its disintegration rate 5000 disintegrations per minute. After 5 minutes, the rate is 1250 disintegrations per minute. Then, the decay constant (per minute) is

(A)19 : 81 (B)18 : 92

(C)81 : 19 (D)92 : 18

Sol(A)

0

10

10

N

0.693log

N

tlog2

l=

or

10

10

5000

0.693log

1250

5log2

l=

or

2ln2

0.4ln2

5

l==

29.A nucleus with Z = 92 emits the following in a sequence: (, (, (, (, (, (, (, ( ; (, (, (, (+, (+, (. The Z of the resulting nucleus is

(A)76(B)78(C)82(D)74

Sol(B)

R

ZZ2NNN

-+

a

bb

=-+-

922842

=-+-

9241878

=+-=

30.Which of the following cannot be emitted by radioactive substances during their decay?

(A)protons (B)neutrinos

(C)helium nuclei (D)electrons

Sol(A)

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