VK physics for class XII

He passed radioactive rays through electric and magnetic fields and observed the deflection.. He also passed radioactive rays through different thicknesses of solids and gases and obser

conclusion that all elements heavier than lead are naturally radioactive

Alpha, Beta and Gamma Particles/Rays

Rutherford performed a series of experiments to study the nature of @

radioactive rays He passed radioactive rays through electric and magnetic

fields and observed the deflection He also passed radioactive rays through

different thicknesses of solids and gases and observed the relative absorption

On the basis of these observations he concluded that the radioactive rays are

of three types; classified as a-rays, fi-rays and y-rays

Rutherford found that one of the radiation was readily absorbed by even

a thick sheet of paper These less penetrating radiation are called @-rays The

other type of radiation was relatively more penetrating; which could pass

through 5 mm thick aluminium or 1 mm thick lead; these radiations were

named as -rays The third type of radiations were most penetarting which

could pass through even 30 cm of iron and were named as y-rays \

Later experiments revealed that a and f-rays are actually formed of Yc nnxaavệ particles; therefore it is better to call them œ-particles and {-particles Fig 19.3

Effect of Magnetic field on Becquerel Rays : A small quantity of a l

Pry tae

Effect of Magnetic field on Becquerel Rays : A small quantity of a

radioactive substance (say radium bromide) is placed within a thin © P lead block containing a small hole so that some of the rays could pass

through the hole as shown in Fig 19.3 A strong magnetic field is

applied at right angles and into the plane of the paper It is found that

a-rays are deflected to the left, showing, according to Fleming's left -

hand rule, that they are positively charged Morever the deflection of

a@-rays is very small, indicating that they possess particles of

relatively large mass, f}-rays are deflected to the right, showing that _

they are negatively charged Moreover the deflection of f-rays is

relatively large, indicating that they are extremely light particles —

Y-rays are undeflected indicating that they carry no charge Also each

set of particles is deflected by different amounts showing that their

Effect of Electric Field on Becquerel Rays : Becquerel rays can

also be separated by electric field If Becquerel rays are subjected to an

electric field (Fig 19.4) between two plates P and Q, Pbeingathigher _

potential relative to plate Q, then B-particles are deflected towards the

Composition of Nucleus and Radioactivity 1211

I[lustratiue AXatnples

Formulae used:

A

A constant, T = half life; t= mean

N =Noe”“* number of half-lines 1 =

HH

Fraction undecayed, aN =( 1 )

my No 2 Concept Used : In formula N = Noe~ At Nis the number of undecayed nuclei;

Number of nuclei decayed = (Ng — N)

Ex.5 Aftera certain lapse of time, the fraction of radioactive polonium undecayed is found to be 12-5% of its initial quantity What is the duration of this time lapse, if the half life of polonium is 138

Solution.Given = 12.5% = 23-1) 72138 days

m 1 a

1 (1Ý 71/7 (1Ý

(3 (3) (8) 8 \2 2 2

t

t= 3T = 3 x 138 = 414 days

Ex 6 The half life of su against «-decay is 4-5 x 10°? years Calculate the activity of 1 g sample

of ‘g2U (CBSE (F) 2006; CBSE AI 2005)

Solution.Given —- T = 4-5 x 10” years

=4-5x10? x3-16x10s — {sincel y= 3-16 10" s)

=1-42x10!7 s

Number of atoms in 238 g of Tư Xuyên = Avogadro number

= 6-023 x 107

“ Number of atoms in 1 g of U-238 is

_ 6-023 x 1073

=

Activity R = AN = (=) N

_ 0.6921

1-42 x 1017

=2-53x 107! atoms

x 2-53 x 107! =1.23x 104 becquerel

(I'roton) This carbon isotope is oxidised to carbon di-oxide ca) Oy, which is mixed with ordinary carbon-dioxide of atmosphere The ratio of CM toc? in atmosphere is 1] : 1012,

In nature all living plants take carbon-dioxide from the atmosphere So in living plants the ratio of

C™ to C’* remains 1:10°° When a plant dies, it stops taking carbon-dioxide from atmosphere; due to which the amount of stable carbon C!? remains unchanged while the radio-carbon C4 begins to decay Therefore by measuring the ratio of C'* to Cc’? ina dead plant, the age of old log may be estimated Let

Ry be the ratio of cM toc? in living plant and after time f of its death, suppose this ratio becomes R If half-life period of C'’ be T, then

_¢at/T

7 (2)

R

RS ' / Rọ

Ro

R Taking log, we get log yp > = log tg 2

R

rs

x | y —

logia2 0:3010 R

As half life T, the ratios R and Rg are known, therefore the age of old specimen of wood may be estimated, Using this method Liby estimated the age of various old species This method is called radio-carbon dating

iilustrative Example

r=“N ean

dt ` R_(1Ÿ

Rạ (5)

t= 3-222T logo =

li= 3 OIF —== 3 c,[= Ji =3XS4QAVSE (<Qa\*

T

Ao

Ex 4 The graph shows how the activity of a sample of |

radon-220 changes with time Use the grapit to determine its half

life Calculate the value of decay constant of radon-220 TÊN

(CBSE Dethi 2004)

Solution Key idea : Half life of a radioactive sample is the Án

time in which its activity is reduced to half of its initial value

From graph the activity becomes half (reduces from Ay to “| in

16 days; so half life of radon-20 is 16 days Decay constant, Am x6 day : —

Fig 19.10

Ex.5 Two radioactive nuclei X and Y initially contain an equal number of atoms Their half lives

are Lh and 2h respectively Calculate the ratio of their rates of disintegration after two hours,

(CBSE 2002}

Solution For nuclide X, half life T, = 1h, time? = 2/1

Number of half-lives 7, = = =—=2

%~(3]*z(3] =‡

For nuclide Y, half life Tz = 2h, timef = 2h

9 Number of half lives, 1) = = - el

2

Ratio, XX = 4/41

Ny 1/2 2

fhe half life of radon is 3-3 days What will be the mass of radon left after 19 days in a

Ex 6,

sample of 25 mg ?

Solution Given T = 3-8 days, ? = 19 days, my = 25mg

f 19

T 3:8

=-(1Ï-(1Ï-‡

m= 5 = 35 mg = 0-78 mg

Ex.13 The ratio of amounts of two isotopes U-” and U~ > in natural uranium is 99-3 and 0-7 If their half-lives are respectively 4-5 x 10” years and? x 10° years, then find the contributions of activity

of each isotope in total activity of uranium

Solution Decay constant A =

If mt) and m3 are masses of U-"” and U~” respectively and total mass is m1, then

my + My =m

Given mị : ty = 99: 3: 0-7

If N be Avogadro number and the number of uranium isotopes (having atomic weights A; and A>)

be N; and N> respectively, then

ny = MON and N, = UEN -Á3)

From (1) and (2)

NeX

Activity, R= AN = “82 =

If R, and R, are activities of u*"8 and U**’, then

_ Hg Ms

RyRy TT TT 1/1 242

( 99.3 0-7

100 | 100 238x4-5x10? 235x7x10°

99.3 OF

~238x4-5x10 235x7

= 92-7x10°4 : 4-255x10°4

Total activity, R = (92-7 + 4-255) x 1074 = 96-955 x 107 4

Ratio of activities in percentage

Ry: Rp = cig ; BEB 100

96-955 x 10” Í _ 96.955 x 10” Ì

= S:-b6l°%u : 4-391%

Radiocarbon Dating

Ex.14 The half life of radioactive carbon (cl )is 5700 years, In a sample the ratio of cM toc” is

reduced to 5 times of its initial equilibrium value Find the age of the sample

Solution Lett be the age of sample

We have Given T = 5700 years R

t = 3-222T logtg 7

R 1 Ro 3

f = 3-222 x 5700 log yp (2)°

= 3-222 x 5700 x 3 logy 2

= 3-222 x 5700 x 3 x 0- 3010

= 17,100 years

Prob, 1 Two stable isotopes of lithium : Li and Shi have respective abundances of 7.5% and

92.5% These isotopes have masses 6.01512 u and 7 -01600 wu respectively Find the atomic weight of

lithium

Solution Masses of isotopes are

tr, = 6-1512 u, My =7 -01600 u

Percentage of isotopes are P; =7 -5%, Py = 92-5%

Pym, + Pos

"PP

- 7:5 x 6:01512 + 92:5 x7 -01600

7-‹55+92.5

Average atomic mass =

= 6-941 u Prob 2 Boron has two isotopes, 1p Band 1 B Their respective masss are 10-01294 4 and 11 -00931

nand the atomic weight of boron ts 10-811 nu, Find the abundances of Ẳ Band 0 B,

Solution If P; and P; are percentage abundances of lp and Mạ,

_ Pụm + Pama

M

ad Pị + P›

10-811 = PLX10-01294 + Py x 11-0931

100

1081 -1 = 10-01294P, + (100 — P,) x 1100931

or 1081-1 = 10-01294P + 1100-931 — 11-00931P,

— 19-831 _ 49.9%,

0-99637

1 = 100 = P = 80-1%

That abundances of WB and lB are 19-9%, and 80: 1%

Prob 3 The three stable isotopes of neon Ne®, Ne*! and Ne** have respective abundances of

90-51%, 0-27% and 93-22%, The atomic masses of the three isotopes are 19:99 u, 20-99 u and 21-99 u

respectively Obtain the average atomic mass of neon

Solution Average atomic mass of neon

Anwon - xịn + xa4¿ + X3A3

Xp, eX +X

Here Ay = 19-99 u, Ap = 20-99 u, Az = 21-99 u, x) + Xp + X53 = 100,

Xy = 90-51, Xp = 0-27, x4 =9-22 90-51 x 19-99 + 0-27 x 20-99 + 9.22 x 21-99

_ 1809-29 +5-67 + 202-75

=20:1Su

Prob 7 Obtain the amount of Co necessary to provide a radioactive source of 8-0 mCi strength

The half life of $2 Co is 3-3 years

Avogadro Number = 6-02 x 10^ per g-atom

Given R = 8-0 mCi = 8-0 107 3 Ci

=8-0 x10” Ÿ x 3-7 x 1019 s~Ï = 29-6 x 107 s7 Ï

he 0-6931 _ 0-6931 _ 0-6931 s^!=4-15x 109 s71

T S-3years 5.3X4/15XI0

From equation (1)

Number of undecayed nuclei, N = + 29-6 x10 =7 -13 x 1019

Rh 415x107?

The mass of 602x107 atoms is 60 grams, so mass of N=7-13™x 10'® atoms is

60

= 7-13 x1016 —nx | prams

| 602x105 |”

16

Required mass of Co, 1 = mex ng =-7.1x10"”g

6-02 x 10

Prob 8 The half life of aq Sr is 28 years What is the disintegration rate of 15 mg of titis isotope ?

Disintegration constant A= a .(2)

Here T = 28 years = 28 x 3-154 x 10” seconds

90 g of ™Sr contain 6-023 x 1079 atoms

Example of a-particle :a-particle is nucleus of helium atom (He) Its mass is 4-00150 amu and it

contains two protons and two neutrons

Mass of 2-protons (= 2nr,) = 2 x 1.00727 amu

= 2-01454 amu mass of 2-neutrons = 2m, = 2 x 1-00865

= 2-01730 amu Total mass of nucleons of a-particle

= 2m, + 2m„ = 2-01454 + 2-01730

= 4:03184 amu mass of a-particle, My; = 4:00150 amu

Mass defect, Ant = 4-03184 -— 4-00150

= 0-03034 amu

Binding energy E = 0-03034 x 931 MeV = 28-25 MeV Therefore to break an a-particle into 2-protons and 2 neutrons 28-25 MeV energy is required + Binding energy of a-particle per nucleon

EF 28-25

A 4 =7-06 MeV / nucleon

SOME IMPORTANT TRIGONOMETRIC FORMULAE

1 Addition and Substraction Formulae

(1) sin(A+8)=sin Acos Btcos Asin B

(2) cos (A + 8)=cos Acos 8#sin 4 sin 8

tan A ‡ tan 8

(3) tan (4 £B)= Fn dAtanB

2 Conversion of a Sum into Product and Vice-Versa Formulae

(1) sin (A+B8)+sin (A-—B)=2sin Acos B

(2) sin (A +8)-sin (A-8)=2cos Asin B

(3) cos (A+ B)+cos (A-B8)=2cos A cos B

(4) cos (A + B)-cos (A —B)=-2sin Asin B

In inverse form, above formulae can be written as :

(5) 2sin Ccos D=sin (C+D)+sin (C-D)

(6) 2cosC sin Ð =sin (C+ Ð)-sin (C—Ð)

(7) 2cos Ccos Ð =cos (C + D)+cos (C—D)

(8) 2sin C sin D =cos (C-D)-cos (C + D)

3 Multiple Formulae

(1) sin 2A =2sin Acos A

(2) cos 2A =cos* A-sin*? A

(3) cos 24 =1-2sin? A =2cos” A-1

2tan A

1—tan? 4

4 The Sine and Cosine Formulae for a Triangle

For a triangle ABC, with sides a,b,c and angles A,B and C the following formulae hold good : (1)

(4) tan2A=

(2) c? =a? +b? —2abcosC

(3) a* =b* +c* —2becos A

(4) b* =c* +a? -2accos B

(5) Area of triangle ABC =A=,/s(s—a)(s—b)(s—c) where s=(a+b+c)/2

10,

12 [sosx+«= sin x

13 ƒ cos ar dr = =

14, | sec” xdx= tan x

| sn đv ẩv = ~

STANDARD FORMULAE FOR INTEGRATION

[ eude=ef war

| (ut vt w)de= f wdr+ | vá + | wdc

+1

, provided n¥-1

n+]

ƒ c&«== where c is 4 constant

+1

[ aM drwc

a+

[+ !&=loez

[ Saree

[ are at

loge a

nat b)

a

| sin x dx = — cosx

cos ax

[sin axde= -

Vo Us

sin ax

a

l5 f cosec® rar =~ cot x

16 Ị sec x tân x dx = s€€C X

17 | cosec x cot x dx = — cosec x

19, ƒ cos (ax + ice et?

_ (ax+ b)"* Ì

20 j (ax+ 6)" d=

provided n#—1

a

" [tO aa

a

ght * 4 plogea

` b)dr< 6=

23 [az*?&=

waar uy

24 | sec* (ax+ 6) dr=

25 [ cosee? (ax+ b) dx = — S0 (4x + 8)

26 j sec (ax + b) tan (ax+ b) dv~ SE (+ 6)

cosec (ax + b)

27 f cosec (ax + 6) cot (ax + 6) dr =

-1 Integration by Substitution

Though derivative of any function can be easily obtained, but this is not generally true for integration However, a few functions which cannot be integrated directly by applying the standard formulae of integration, can be changed into a standard integrand by making a suitable substitution and then can be integrated directly In order to illustrate the method, we take here few examples :

(1) Calculate : f ai

1

(ax +b)

Let 1={

Substitute z=ax+b, so that a x

a

r=Ị 1$)-zí @ lings „1 log (ax + 5)

(2) Evaluate : | cos? xsin x dx

Substitute z=cosx, so that dz=—sin xdx

z! 1 I=-Í z3 dđ=—^—— =——cos` x

cot x

log sin x

(3) Calculate : j dx

2 Integration by Parts

This method corresponds to the rule for differentiating a product of two functions Let u and y be two functions of x If we find out the derivative of product uv w.r.t x, then

— (uv)=u— +¥—

| wt as =u (#)+

Suppose, oa

then v=f w dx

[ (uve) de =u J wde-f £)I ws

Above equation represents the rule for integration by parts In words, it can be written as follows :

Integral of the product of two functions

= first function x integral of the second — integral of (derivative of the first function

x integral of the second function) While applying above rule, following points should be taken care of :

(1) Out of the two functions (whose product is to be integrated) one should select that function as the

second function whose integral is known

(2) The function sclected as the first function should be such that its derivative is simpler than the

function itself

In order to illustrate the method, we take here few cxample :

(1) Calculate : | x? sin x de

(1) Calculate : [ sin x ức

Let r=Í x” sin x dx

I=Í x sin xdv=x? ƒ sin xả&~ | FG )x f sin va

=x? (—cos x)~ƒ 2x (-—cos x) dx

=- x" cos x+ 2Í xcos x dr

=-x? sex cos x dx ~ f lᜠx Jos s| a

=-x* cos x + 2[xsin x - Ix sin x dx]

=x? cos x + 2[xsin x —(—cos x)]

=~ x? cos x+2x sin x+2 cos x

(2) Evaluate : | x" log x dx

., I=Í x" log x dx =log x [ x" dx — | 4 toe x" as|as

=log x wl)! yt! deed Re l fx" de

n+1

-x”" logx x

n+l (n1)?

DEFINITE INTEGRAL

When integration is carried out between definite limits, it is called as definite integration Suppose / (x)

be a function of x and J I(x) de=F (x) then F(6)-F(a) is called as the definite integral of / (x) between the limits a and ở and is written as

[2 fedde=F(x)|? =F(O)-F(a)

Here a is called the lower limit and 6 the upper limit of integration The definite integral thus represents the difference between the values of the indefinite integral F(x) at the upper and lower limits

The basic reason for calling this integral as the definite integral is that the indefinite constant of integration does not appear in the final result Suppose that instead of F(x) we write F(x)+c for

j f(x) dx, then

Hence we find that the arbitrary constant c does not appear in the process and the value of definite

integral is the same as on considering F(x) alone

Examples :

(1) Evaluate : [2° sin xe0s x dx