Modern Physics for IITJEE

8/3/2019 Modern Physics for IITJEE

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Modern PhysicsIITJEE Syllabus:

Alpha, beta and gamma radiations, law of radioactive decay, decay constant, half-life and meanlife. Photoelectric effect. De Broglie wavelength. Bohrs theory of hydrogen like atoms.Production of characteristic and continuous X-rays, properties of X-rays. Atomic nucleus, bindingenergy and its calculation. Fission and fusion processes, energy calculations in these processes.

1. STRUCTURE OF ATOMAll matter is made up of tiny particles known as atoms. There are only about 105 different kindsof atoms, and they combine with each other in different ways to form groups called molecules.All matter has been found to be composed of atoms or molecules, and some knowledge of howatoms are made will give us valuable information about the behaviour of matter.

Thomson's Atomic ModelOn the basis of his experiments J.J. Thomson proposed a model of internal atomic structureaccording to which atom consisted of positively charged substance (+ve electric fluid) distributeduniformly over the entire body of the atom, with negative electrons embedded in this continuouspositive charge like seeds in a watermelon. It was a good effort to reveal mystery of an atom but itwas not the true picture of an atom.

Rutherford's Atomic ModelThe correct description of the distribution ofpositive and negative charges within an atom wasmade in 1911 by a New Zealander when working at Manchester University in England. This was

Ernest Rutherford, who was later made Lord Rutherford for his many scientific achievements. He

entered into physics during that crucial period of its development when the phenomenon ofnatural radioactivity had just been discovered, and he was first to realize that radioactivityrepresents a spontaneous disintegration of heavy unstable atoms.

Rutherford realized that important information about the inner structure of atoms could beobtained by the study of collisions between on rushing particles and the atoms of variousmaterials forming the target.

Particle Source GoldFoil

Zinc SulphateScreen Microscope

Fig.(1) Schematic diagram of the experimental set up used by Rutherford.

Collimator


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The basic idea of the experimental arrangement used by Rutherford in his studies was explainedas follows:

a speck of - emitting radioactive material; a lead shield with a hole that allowed a narrow beamof the particles to pass through; a thin metal foil to deflect or scatter them; and a pivoted

flourescent screen with a magnifier through which the tiny flashes of light were observedwhenever an -particle struck the screen. The whole apparatus was evacuated, so that the particleswould not collide with air molecules.

ObservationMost of the -particles penetrated the foil withvery little deflection. An appreciable fraction ofthem were deflected through large angles - a fewwere turned back almost as though they had beenreflected from the foil. This was a deflection ofnearly 180 and a completely impossiblephenomenon according to the Thomson's model.

Such large deflections required strong forces tobe acting, such as those between very smallercharged particles very close together. This wouldbe possible, Rutherford reasoned, if all thepositive charge, along with most of the atomicmass, were concentrated in a very small centralregion which Rutherford called the atomicnucleus.

Nucleus

Fig.(2) According to Rutherfords nuclear model, the alphaparticles were scattered by the coulomb force of atiny particle (the nucleus) rather than a largesphere, as in Thomsons model of the atom, Eachalpha particle experienced a single string collision.

Rutherford, knowing the kinetic energy of the -particles, calculated that they would be withinabout 10-12cm from its centre if - particles were to be turned back in the direction from whichthey came.Because there would be a Coulomb force of attraction between the positive nucleus and thenegative electrons, the two would be down together and the atom would vanish unless someprovisions were made to prevent it. It was suggested that the electrons might be orbiting rapidlyaround the nucleus, so that the electrostatic attraction would merely provide the necessarycentripetal force.

Drawbacks

(i) Rutherford's atomic model was unable to make any predictions about the light that an atomwould emit

(ii) More serious than this was its conflict with the accepted laws of electromagnetic theory. Anelectron revolving rapidly around a nucleus must have a continual centripetal acceleration, and

this acceleration would cause a continuous loss of energy by radiation. Bohr calculated thatthis emission of radiation would cause the electrons in an atom to lose all their energy and fallinto the nucleus within a hundred - millionth of a second. Since matter composed of atomsexists permanently, as far as we know, there was obviously something wrong here. Bohr'sconclusion was that the conventional classicallaws of physics must be wrong, at least whenapplied to the motion of electrons within an atom.


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Bohrs Theory

Bohr in defiance of the well - establishedlaws of classical mechanics andelectrodynamics, proposed that the followingrules must hold1. Of all the infinite number of

mechanically possible orbits for anelectron revolving around a nucleus, onlya few are permitted. These are the orbitsin which the angular momentum of theelectron is an integral multiple of h/2.

2. While circling around these permittedorbits, the electrons do not emit anyelectromagnetic radiation, even though

conventional electrodynamics holds thatthey should.

Fig.(3) Planetary model of the hydrogen atom.A massive positive charge is nearly stationary in thecenter. An electron travels in a circular orbit. The electronis held in its orbit by the Coulombs law attraction of thepositive nucleus and the negative electron.

v

r

mv2

3. Electrons may jump from one orbit to another, in which case the difference in energy betweenthe two states of motion is radiated as a photon whose frequency is determined by thequantum rule E= hf.

Bohr's Orbits

For an electron orbiting in a hydrogen atom, the necessary centripetal force is the electrostaticattraction between the negative electron and the massive, positively-charged proton, that is thenucleus.

Thus, k 2

2

r

e

= r

mv2

or r= 2

2

mv

ke

(1)According to Bohr's theory

mvr=2

nhwhere, n = 1, 2, 3,.

r=mv

nh

2(2)

From (1) and (2)

v =nh

ke22

22

22

4 kme

hn

r where h = 6.63 10-34

J s (1)


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Electron energiesKinetic energy

K=2

1mv2 =

2

m222

nh

ke=

22

4222

hn

mek (2)

Potential energy

U= r

ke2

= ke2

22

224

hn

kme=

22

4224

hn

mek

(3)Total energy

E=K+ U= 2

1

n

2

4222

h

mek(4)

Putting the values ofk= 9 109 Nm2/C2

e = 1.6 10-19 C and h = 6.63 10-34 Js,we get

E= 2

1

n(2.18 10-18) J =

2

6.13

n

eV (5)

E

Free Electron

Bound Electronn = 5n = 4n = 3

n = 2

n = 1

0 eV

-13.6eV

-3.40eV

-1.51eV

Fig.(4) Energy level diagram for the Bohr model of

the hydrogen atom. The vertical axisrepresents energy. The (arbitrary) zero ofenergy is taken as the energy of a stationaryelectron, infinitely far from the positivenucleus. The lowest energy level (n = 1) isknown as the ground state.

Radiation and Energy Levels E = hf E =E2 E1

Using equation (5)

E= 2.18 10-18

2221

11

nn

Applying Plancks Law,

f=h

E= 3.29 1015

22

21

11

nnHz (6)

Dividing the above equation by c = 3 108 m/s, we get

22

21

8

15 11

103

1029.31

nnm-1

or

2

221

111

nnR

m-1 (7)

where R = Rydberg constant = 1.097 107 m-1 .


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0 eV

-1.51eV-3.40eV

-13.6 eV Lyman

BalmerPaschen

Bracket

n = n = 5n = 4n = 3n = 2

n = 1

Fig.(5) Light is emitted from the hydrogen atom only when the electron makestransitions between stationary orbits. The Balmer series of spectral lines,for instance, results when electrons from higher energy levels fall into then = 2 level, releasing their energy as a single photon.

Successes and Limitations

Bohr showed that Planck's quantum idea were a necessary part of the atom and its innermechanism; he introduced the idea of quantized energy levels and explained the emission orabsorption of radiation as being due to the transition of an electron from one level to another. As amodel for even multielectron atoms, the Bohr picture is still useful. It leads to a good, simple,rational ordering of the electrons in larger atoms and qualitatively helps to predict a great dealabout chemical behaviour and spectral details.

Bohr's theory is unable to explain the following facts

1. Thespectral lines of hydrogen atom is not a single line but a collection of several lines veryclose together.

2. The structure of multielectron atoms is not explained.3. No explanation for using the principle ofquantisation of angular momentum.4. No explanation forZeeman effect

If a substance which gives a line emission spectrum is placed in a magnetic field, the lines ofthe spectrum get split up into a number of closely spaced lines.

This phenomenon is known asZeeman effect.

Conclusion

The atom consists of a heavy positively charged nucleus and negatively charged electronsmoving around it. The electron is an elementary particle having a massme 9.1 10

31 kg and a charge e, e being an elementary charge approximately equal to1.60 10-19C.


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The nuclear charge is equal to +Ze, whereZis the atomic number. The atom contains Zelectrons,their total charge being Ze. Consequently, the atom is an electrically neutral system. The size ofthe nucleus varies depending onZfrom 10-13 cm to 10-12 cm. The size of the atom is a quantity ofthe order of 10-8 cm.

The energy of the atom is quantized. This means that it can assume only discrete (i.e. separatedby finite gaps) values: E1, E2, E3,, which are called the energy levels of the atom(E1


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4. Wavelength of photon emitted for a transition fromn2 ton1

2

221

2 111

nnZR

(12)

whereR= 1.096 107 m-1 (for a stationary nucleus)

If nucleus is not considered to be stationaryR =

M

m

R

1(13)

where m is the mass of electron and Mis the mass of nucleus.

5. Wavelength () of a photon of energyE (eV) is given by

=)eV(

12400E

(14)

6. Momentum of a photon of energyE

p =c

E (15)Example: 1

A single electron orbits around a stationary nucleus of charge +Ze, where Zis a constantand e is the magnitude of electronic charge. It requires 47.2 eV to excite the electron from

secondBohr orbit to the thirdBohr orbit.(a) Find the value ofZ(b) Find the energy required to excite the electron from n = 3 to n = 4(c) Find the wavelength of radiation required to remove electron from first Bohrs Orbit to

infinity.(d) Find the kinetic energy, potential energy and angular momentum of the electron in the first

Bohr orbit.

Solution

(a) Given E23 = 47.2 eV

Since E= 13.6Z2

22

21

11

nneV

47.2 = 13.6Z2

22 3

1

2

1 Z= 5

(b) To find E34; n1 = 3; n2 = 4

E= 13.6Z2

22

21

11

nn

eV

E= 13.6 52

22 4

1

3

1= 16.53 eV

(c) Ionization energy is the energy required to excite the electron from n = 1 to n =

Thus, E= 13.6 52

22

111

= 340 eV


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The respective wavelength is

=E

hc

=

340

1240012400

E= 36.47

(d) K= -E= +340 eVU= 2E= -680 eV

L =2

h=

21063.6 34 = 1.056 10-34 J-s

Example 2Find the quantum number n corresponding to excited state of He+ ion if on transition to theground state, the ion emits two photons in succession with wavelengths 108.5 nm and30.4 nm. The ionization energy of H atom is 13.6 eV.

SolutionThe energy transitions for the given wavelengths are

E1 = eV43.111085

1240012400

1

E2 =304

1240012400

2

= 40.79 eV

Total energy emitted E= E1 + E2 = 52.22 eV

Now E= 13.6 Z2

22

21

11

nneV E= energy emitted

or 52.34 = 13.6 22

22

1

1

1

n

Thus, n = 5

Example 3An isolated hydrogen atom emits a photon of 10.2 eV.

(a) Determine the momentum of photon emitted(b) Calculate the recoil momentum of the atom(c) Find the kinetic energy of the recoil atom.

[Mass of proton, mP= 1.67 10-27 kg]

Solution(a) Momentum of the photon is

p1 =c

E=

8

19

103

106.12.10

= 5.44 10-27 kg m/s

(b) Applying the momentum conservation

p2 atm photon p1

p2 =p1 = 5.44 10-27 kg m/s

(c) K=21

mv2 (v = recoil speed of atom, m = mass of hydrogen atom)


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or K=m

p

m

pm

22

1 22

Substituting the value of the momentum of atom, we get

K=

27

227

1067.12

1044.5

= 8.86 10

-27

J

Example 4A hydrogen atom in a state of binding energy 0.85 eV makes a transition to a state ofexcitation energy of 10.2 eV. Find the energy and wavelength of photon emitted.

SolutionSince the binding energy is always negative, therefore,

Ei = -0.85 eVLet ni be the initial binding state of the electron, then

En = 2

2

6.13in

Z

or -0.85 = -13.622

n

Z

or ni = 4Binding energy =En = -13.6 Z

2/n2 0.85 eV = -13.6(1)2/n2

2 n2 = 4Let the electron now goes to an energy level n whose excitation energy is 10.2 eV. Sincethe excitation energy Eis defined with respect to ground state, therefore

E= 13.6Z2

22

21

11

nneV

or 10.2 = 13.6 12

221

1

1

fn

thus nf= 2So the electron makes a transition from energy level ni = 4 to nf= 2.Thus, the energy released is E=E4 E2

or E= 13.6

22 4

1

2

1= 2.55 eV

Since =E

hc

= 5511

eV25.212400

Example 5

A particle of charge equal to that of an electron, -e and mass 208 times the mass ofelectron (called a -meson) moves in a circular orbit around a nucleus of charge +3e (takethe mass of the nucleus to be infinite). Assuming that the Bohr model of the atom isapplicable to this system:

(i) Calculate the radius ofnth Bohr orbit(ii) Find the value ofn, for which the radius of orbit is approximately the same as that of first

Bohr orbit for the hydrogen atom;


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(iii)Find the wavelength of radiation emitted when the -meson jumps from the third orbit tofirst orbit(Rydbergs constant = 1.097 107/m)

Solution(i) Radius of the nth Bohr orbit for hydrogen atom is

rn = 0.53Z

n2

Since rm

1

Radius of nth orbit for -meson is

rn = Z

n

208

53.0 2

or rn = (8.5 10-4)n

2

(ii) (8.5 10-4)n2 = 0.53 n2 = 623

or n 25(iii)In case of hydrogen like atom,

E=E3 E1 = 13.6Z2

23

11 = 12.08 eV

since E m -meson, E= (12.8)(208) = 22.6 keV

Thus =3106.22

1240012400

E= 0.548

2. de BROGLIE WAVES

The first step in the understanding of the hidden meaning of Bohr's quantum orbits was made by aFrenchman, Louis de Broglie, who tried to draw an analogy between the sets of discrete energylevels that characterise the inner state of atoms and the discrete sets of mechanical vibrations thatare observed in the case of violin strings, organ pipes etc.

de Broglie asked himself,"Could it not be that the optical properties of atoms are due to some kind of standing wavesenclosed within themselves?"

As a result of these considerations, de Broglie came out with his hypothesis that the motion ofelectron within the atom is associated with a peculiar kind of waves which he called "pilot

waves".In order to have n complete wavelengths (n) fit into the circumference of the nth orbit, thefollowing relation must be true:

nn = 2rnFrom Bohr's theory of the hydrogen atom,

rn = 22

22

4 kme

hn


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or nn = 2

22

2 kme

hn

or n = 2

2

2 kme

nh

=

m

h22 ke

nh

n

nmv

h (16)

The wavelength of the wave associated with a moving particle is equal to Planck's quantum

constant divided by the momentum of the particle.

Example 6What is the energy and wavelength of a thermal neutron ?

SolutionBy definition, a thermal neutron is a free neutron in a neutron gas at about 20o C (293 K).Since it has three degrees of freedom, therefore

K= 2123

1007.62931038.12

3

2

3

kT J

= 2127

34

1007.61067.12

1063.6

2

Km

h

p

h

o

= 0.147 nm

Example 7A particle of mass m is confined to a narrow tube of lengthL.

(a) Find the wavelengths of the de-Broglie waves which will resonate in the tube,(b) Calculate the corresponding particle momenta, and(c) Calculate the corresponding energies.

Solution

(a) The de Broglie waves will resonate with a node at each end of the tube.A few of the possible resonance forms are as follows :

n =n

L2; n = 1, 2, 3, ......

(b) Since de-Broglie wavelengths are

n =np

h

pn =L

nhh

n 2

n = 1, 2, 3....

(c) The kinetic energies of the particles are

Kn =mL

hn

m

pn2

222

82 ,

n = 1, 2, 3, ........

L

L = 3(/2)

L = 2(/2)

L = (/2)

NN

N

N

N

N

AAA

A A

A

N

N

N


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3. X RAYSWhen a heavy metal target is bombardedwith high-energy (30 50keV) electrons, itemits X rays. The radiation involves both a

continuous and a line spectrum, as shownin the fig.(6). The continuous spectrum,which starts at some minimum wavelengtho, arises from the rapid deceleration of theelectrons when they enter the target it iscalledbremsstrahlung, or braking radiation.The existence of minimum wavelength (ormaximum frequency) is further evidence infavour of the photon concept. The highestfrequency photon is emitted when anelectron loses all its energy in one step. By

equating the energy of the electron (eV) tothe energy of the photon (hfo), we findhfo = eV

or o =eV

hc(17)

min Fig.(6) The X-ray emission when stream of fast moving

electrons strike a target of heavy element.

Intensity

The minimum wavelength depends on the electron energy, but not on the target material.

W

Mo

Cr

WavelengthO

Intensity

Fig.(7) Different targets of tungsten (W), Molybdenum(Mo) and Chromium (Cr) are used and thekinetic energies of the incident electrons are keptconstant. It is observed that the minimumwavelength o is independent of target material.

K3

Wavelength3O

Intensity

Fig.(8) Electron beam of different energies K1, K2 andK3 are incident on a target of same material. Theminimum wavelength is inversely proportionalto the kinetic energy.IfK3 >K2 >K1 , then 3 < 2 < 1.

K2K1

2 1

Fig.(9) Energy level diagram for an electron in an atom.The arrows indicate the transitions that give riseto the different series ofX rays.

O

Fig.(10) A plot of the square root of the frequency versusatomic numberZ.

N

M

L

K

L L

K K K

f

Atomic numberZ


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The line spectrum depends on the element used as target. These characteristic X rays areproduced when an electron knocks out an atomic electron from one of the inner levels. The ejectedelectron leaves a vacancy, which is then filled by an electron falling from a higher level. In theprocess a high energyphoton is emitted. If the transitions are to the n = 1 level, the Xrays arelabeledK,K.If they are to the n = 2 level, they are labeledL,L

In 1913, Moseley noted that the characteristic lines shifted systematically as the target materialwas changed. He plotted thesquare rootof thefrequency of theK line versus the atomic number

Zfor many elements. Thestraight line he obtained is shown in the fig.(10).

Moseleys plot did not pass through the origin. Let us see, why?. Once one of the two electrons inthe n = 1 level is ejected, an electron in the next highest level will drop to the lower state to fill thevacancy and in the process it emits theK frequency. For this electron the electric field due to thenucleus is screened by the remaining electron in the n = 1 level. Moseley estimated that theeffective nuclear charge for theK transition is (Z 1)e. Thus Moseleys law for the frequency oftheK line is

Kf = a(Z 1) (18)

when a = RC4

3whereR is the Rydberg constant c is the speed of light.

The wavelength ofK lines is given by

22 111

1

nZ

where n = 2, 3, 4, (19)

Example 8Find the cut-off wavelength of the X-rays emitted by an X-ray tube operating at 30 kV.

SolutionFor minimum wavelength, the total kinetic energy should be converted into an X-rayphoton.Thus,

=31030

1240012400

EE

hc= 0.41

Example 9Show that the frequency ofK X-ray of a material equals to the sum of frequencies ofKandL X-rays of the same material.

Solution

L

K K

M

K

L

The energy level diagram of an atom with one electron knocked out is shown above.Energy ofK X-ray isEK =EL -EK

ofKX-ray is K

E EM-EK


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and, of L X-ray isEL =EM -ELthus,

KE EK +EL

or LKK

fff

4. RADIOACTIVITYRadioactive decay is a random process: Each decay is an independent event, and one cannot tellwhen a particular nucleus will decay. When a given nucleus decays, it is tranformed anothernuclide, which may or may not be radioactive. When there is a very large number of nuclei in asample, the rate of decay is proportional to the number of nuclei,N, that are present

Ndt

dN (20)

Where is called the decay constant. This equation may be expressed in the form dtN

dN

and integrated

tN

N

dtN

dN

o 0

to yield

tN

N

o

ln

where No is the initial number of parent nuclei at t = 0. The number that survive at time t istherefore

N=Noe-t (21)

This function is plotted in Fig.(11)

N

No

0.5No0.37No

T1/2 1/Fig.(11)The number of radioactive nuclei in a sample

as a function of time. The half-life is thetime required for the number to fall to 50%of any initial value (not just t= 0).

t

The time requiredfor the number of parent nuclei to fall to 50% is called thehalf-life, T, and maybe related to as follows. Since

0.5No =NoT

e


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we have T= ln|2| = 0.693. Therefore

T=693.0

(22)

It takes one half-life to drop to 50% of any starting value. The half-life for the decay of the freeneutron is 12.8 min. Other half-lives range from about 10-20 s to 1016 years.Since the number of atoms is not directly measurable, we measure thedecay rate oractivity (A)

A = -dt

dN. On taking the derivative of equation we find

A =N = Noe-t = Roe

-t (23)where A = Nis the initial activity. The activity is characterized by the same half-life. The SI unitfor the activity is the becquerel(Bq), but the curie (Ci) is often used in practice.

1 becquerel (Bq) = 1 disintegration per second (dps)1 curie = 3.7 1010 dps1 rutherford = 106 dps

Mean life of a radioactive sample is defined as the average of the lives of all nucleus.

Tav =693.0

10 T

N

dteN

o

to

(24)

Example 10The half-life of Cobalt - 60 is 5.25 years. How long after its activity have decreased toabout one-eigth of its original value ?

SolutionThe activity is proportional to the number of undecayed atoms.In each half-life, half the remaining sample decays.

Since12

12

12

18

`, therefore, three half-lives or 15.75 years are required for the

sample to decay to 18

th its original strength.

Example 11A count rate-meter is used to measure the activity of a given sample. At one instant themeter shows 4750 counts per minute. Five minutes later it shows 2700 counts per minute.

(a) Find the decay constant(b) Also, find the half life of the sample

Solution

Initial velocity Ai = 47500

ot

Ndt

dN (i)

Final velocity Af= 27005 Ndt

dN

t (ii)Dividing (i) by (ii), we get

t

o

N

N

2700

4750(iii)


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The decay constant is given by

=t

o

N

N

tlog

303.2

or = log

5

303.2

2700

4750= 0.113 min-1

Half life of the sample is

T= min14.6113.0693.0693.0

Example 12The mean lives of a radio active substance are 1600 and 400 years for - emission and - emission respectively. Find out the time during which three fourth of a sample willdecay if it is decaying both by - emission and - emission simultaneously.

Solution

When an substance decays by and emission simultaneously, the average disintegrationconstant

avis given by

av = + where = disintegration constant for - emission only

= disintegration constant for - emission onlyMean life is given by Tm = 1/

av = + or4001

16001111

TTTm

= 3.12 10-3

avt= 2.303 logt

o

N

N

(3.12 10-3)t= 2.303 log25

100

t= 2.303 4log1012.3

13

= 443.5 years

There are two types of radioactivity, natural radioactivity observed in unstable elements in natureand artificial radioactivity observed in artificially obtained isotopes.

Example 13The half-life of radium is 1620 years. How many radium atoms decay in 1s in a 1g sampleof radium. The atomic weight of radium is 226 kg/mol.

SolutionNumber of atoms in 1 g sample is

N= 0001226

6 02 10 2 66 1026 21. . .

atoms.

The decay constant is

=

117

2/11035.1

1016.31620

693.0693.0

T

s-1

Taking one year = 3.16 107 s


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Now, 102111 106.31066.21035.1 N

t

N s-1

Thus, 3.6 1010 nuclei decay in one second.

Natural radioactive processes are of two kinds:

(1) decay associated with the emission of particles, viz. nuclei He42 of helium. Alphaparticles are heavy positively charged particles having a massm 4 amu and a charge q = +2e. The velocity of -particles is relativelylow: v = (c/30 - c/15), where c is the velocity of light.

(2) decay (beta-minus-decay) associated with the emission of electrons formed at theinstant of decay.Both processes are accompanied by -radiation, i.e. the flow of photons

having a very small wavelength, andhence a very high energy. Like otherelectromagnetic waves, -rays propagateat a velocity of light. The penetrabilityof-rays is 0-100 times higher than thepenetrability of -rays and 1000-10000times higher than the penetrability of-rays. It also exceeds the penetrabilityofX-rays.In a magnetic field, a beam of -, -,and -rays splits into three parts.

Lead

Fig.(12) In a magnetic field, rays areundeviated and - particles arethe most deviated.

Nuclei possessing theartificial radioactivity are obtained by bombarding stable nuclei ofheavy elements by -particles, neutrons, or (sometimes) protons and other particles.Nuclear transformations occur in two stages in this case. First a particle hits a targetnucleus and causes its transformation into another, unstable (radioactive), nucleus. Thisnewly formed nucleus spontaneously emits a particle and is transformed either into astable nucleus or into a new radioactive nucleus. Artificial radioactivity obeys the samelaws as natural radioactivity.Radioactive processes occur in accordance with the laws of conservation of energy,momentum, angular conservation, electric charge, and mass number (amount ofnucleons).In -decay, the mass number of the nucleus decreases by four and the charge decreases bytwo units, as a result of which two electrons are removed from the atomic shell. The

element transforms into another element with the atomic number which is two units lower.In --decay, a neutron in the nucleus transforms into a proton. Such a transformation of theneutral neutron into the positive proton is accompanied by the birth of an electron, i.e. by-radiation. The mass number of the nucleus does not change in this process, while thecharge increases by +e and atomic number increases by one.


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5. ATOMIC NUCLEUS

The atomic nucleus consists of two types of elementary particles, viz. Protons and neutrons.These particles are called nucleons.

The proton (denoted by p) has a charge +e and a mass mp 1.6726 10-27 kg, which isapproximately 1840 times larger than the electron mass. The proton is the nucleus of the simplestatom withZ = 1, viz. the hydrogen atom.

The neutron (denoted by n) is an electrically neutral particle (its charge is zero). The neutronmass mn 1.6749 10

-27 kg. The fact that the neutron mass exceeds the proton mass by about 2.5electron masses is of essential importance. It follows from this that the neutron in free state(outside the nucleus) is unstable (radioactive). During the time equal on the average to 12 min, theneutron spontaneously transforms to the proton by emitting an electron (e-) and a particle calledthe antineutrino ( v~ ). This process can be schematically written as follows:

n p + e- + v~ .

The most important characteristics of the nucleus are the charge numberZ (coinciding with the atomic number of the element) and the mass numberA. The chargenumber Z is equal to the number of protons in the nucleus, and hence it determines the nuclearcharge equal toZe. The mass numberA is equal to the number of nucleons in the nucleus (i.e. tothe total number of protons and neutrons).

Nuclei are symbolically designated asAZX or ZX

A

where X stands for the symbol of a chemical element. For example, the nucleus of the oxygenatom is symbolically written as 188O or 8O

18.Most of the chemical element have several types of atoms differing in the number of neutrons in

their nuclei. These varieties are called isotopes. For example, oxygen has three stable isotopes:168O ,

178O and

188O . In addition to stable isotopes, there also exist unstable (radioactive) isotopes.

Atomic masses are specified in terms of the atomic mass unit or unified mass unit (u). The massof a neutral atom of the carbon isotope 6C

12 is defined to be exactly 12 u.1u = 1.66056 10-27 kg = 931.5 MeV

Example 14(a) Calculate the value of 1 u from Avogadros number.(b) Determine the energy equivalent of 1u.

Solution

(a) One mole of C12

has a mass of 12 g and contains Avogadros number,NA, of atoms.By definition, each C12 has a mass of 12 u.Thus, 12 g corresponds to 12NA u which means

1u =2310022045.6

1g1

AN

or 1u = 1.66056 10-27 kg


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(b) From Einstein relation E= mc2

E= (1.66056 10-27) (3 108)2 = 1.4924 10-10 JSince 1eV = 1.6 10-19 J E= 931.5 MeVHence 1u = 931.5 MeV

The shape of nucleus is approximately spherical and its radius is approximately related tothe mass number by

R 1.2A1/3 fmwhere 1 fermi (fm) = 10-15 m

Example 15 Find the mass density of the oxygen nucleus 8O

16.Solution

Volume V= AR 33 2.134

34

= 1.16 10-43 m3

Mass of oxygen atoms (A = 16) is approximately 16 u.Therefore, density is =

v

m

or =

43

27

1016.1

1066.116

= 2.3 1017 kg/m3

This is 1014 times the density of water.

Binding Energy

The rest mass of the nucleus is smaller than the sum of the rest masses of nucleonsconstituting it. This is due to the fact that when nucleons combine to form a nucleus, the

binding energy of nucleons is liberated. The binding energy is equal to the work that mustbe done to split the nucleus into particles constituting it.

The difference between the total mass of the nucleons and the mass of the nucleus is calledthemass defect of the nucleus:

m = [Zmp + (A Z)mn] mnuc.Multiplying the mass defect by the square of the velocity of light, we can find the bindingenergy of the nucleus:

BE = mc2 = [(Zmp + (A Z)mn) mnuc]c2 J (25)

If the masses are taken in atomic mass unit, the binding energy is given byBE = [ZmP + (A Z)mn mnuc]931.5 MeV (26)

Dividing the binding energy by the number A of nucleons in the nucleus, we obtain thebinding energy per nucleon. Fig.(13) shows the dependence of the binding energy pernucleon BE/A on the mass number A of the nucleus. Nucleons in nuclei with massnumbers from 50 to 60 have the highest binding energy. The binding energy per nucleonfor these nuclei amounts to 8.7 MeV and gradually decreases with increasingA. For theheaviest natural element uranium it amounts to 7.5 MeV. Figure shows that when aheavy nucleus (withA 240) splits into two nuclei withA = 120, the released energy is of


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the order of 1 MeV per nucleon i.e. 240 MeV per parent nucleus. It should be mentionedfor comparison that when a carbon atom is oxidized (burnt) to CO2, the energy of the orderof 5 eV is liberated, which is smaller than the energy released in fission of a uraniumnucleus by a factor of 50 millions.

56

Fe

1

3

5

7

9

8040 120 160 200 A

Fig.(13) The average binding energy per nucleon as

a function of atomic number A. Themaximum value occurs at Fe56.

240

(MeV)

nucleonBE 238U

It also follows from Fig(13) that the fusion (synthesis) of light nuclei into one should beaccompanied by the liberation of a huge energy. For example, the fusion of two nuclei ofheavy hydrogen 21 H (this nucleus is called a deuteron) into a helium nucleus

42 He

would yield an energy equal to 24 MeV.

The forces binding nucleons in a nucleus manifest themselves at distances < 10 -15 m.In order to bring together two positively charged deuterons to such a distance, theirCoulomb repulsion should be overcome. For this, the deuterons must have a kinetic energyequivalent to their mean energy of thermal motion at a temperature of the order of 109 K.

For this reason, the fusion reaction of nuclei is also called a thermonuclear reaction.Actually, some thermonuclear reactions may occur at a temperature of the order of 107K.This is due to the fact that there is always a certain number of nuclei whose energyconsiderably exceeds the mean value.

Example 16Find the binding energy of 6

12 C ? Also find the binding energy per nucleon.Solution

One atom of C126 consists of 6 protons, 6 electrons and 6 neutrons. The mass of the

uncombined protons and electrons is the same as that of six 11 H atoms (if we ignore the

very small binding energy of each proton-electron pair).

Mass of six 11

H atoms = 6 1.0078 = 6.0468 uMass of six neutrons = 6 1.0087 = 6.0522 uTotal mass of component particles = 12.0990 u

Mass of 612C atom = 12.00004

Mass defect = 0.0990 uBinding energy = (931)(0.099) = 92 MeV

Binding energy per nucleon = MeV66.712

92


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Example 17A neutron breaks into a proton and electron. Calculate the energy produced in this reactionin MeV. Mass of an electron = 9 10-31kg, Mass of proton = 1.6725 10-27kg, Mass ofneutron 1.6747 10-27kg. Speed of light = 3 108 m/s.

Solution

on1 1H1 + -1eoMass defect (m) = [Mass of neutron (mass of proton + mass of electron)]

= [1.6747 10-27 (1.6725 10-27 + 9 10-31)]= 0.0013 10-27 kg

Energy released Q = m c2

Q = (0.0013 10-27) (3 108)2 = 1.17 10-13 J

=19

13

106.1

1017.1

= 0.73 106eV = 0.73 MeV

NUCLEAR REACTION

A nuclear reaction in which a collision between particle a and nucleusXproduces Yand particle bis represented as a +X Y+ bThe reaction is sometimes expressed in the shorthand notationX(a, b)Y.Reactions are subjected to the restrictions imposed by the conservation of charge, energy,momentum and angular momentum.

Energy of A Reaction

K1m1

Initial

K2m2

a X

K3m3

Final

K4m4

Y b

Initial energy: Ei = m1c2 + m2c2 +K1 +K2Final energy: Ef= m3c

2 + m4c2 +K3 +K4

Since Ei =Ef (energy conservation) [(m1 + m2) - (m3 + m4)]c

2 = (K3 +K4) - (K1 +K2)The energy, that is releasedor absorbedin a nuclear reaction is called the Q - value or

disintegration energy of the reaction.Q = [(m1 + m2) - (m3 + m4)]c

2 J (27 a)or Q = [(m1 + m2) - m3 + m4)] 931.5 u (27 b)IfQ ispositive, rest mass energy is converted to kinetic mass energy, radiation mass -energy or both, and the reaction is exoergic.If Q is negative, the reaction is endoergic. The minimum amount of energy that a

bombarding particle must have in order to initiate an endoergic reaction, is calledThreshold Energy Eth .

Eth = -Q

1

2

1

m

m(28)

where m1 = mass of the projectilem2 = mass of the target.


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Example 18Neon - 23 beta decays in the following way :

eNaNeo

12311

2310

Find the minimum and maximum kinetic energy that the beta particle eo1 can have. The

atomic masses of23Ne and

23Na are 22.9945 u and 22.9898 u, respectively.

SolutionReactant Products

Ne2310 22.9945 - 10me Na2311 22.9898 - 11 me

1o e -me

Total 22.9945 10 me Total 22.9898 10 me

Mass defect = 22.9945 - 22.9898 = 0.0047 uQ = (0.0047)(931) = 4.4 MeV

The - particle and neutrino share this energy. Hence the energy of the -particle can

range from 0 to 4.4 MeV.

Example 19How much energy must a bombarding proton possess to cause the reaction.

nBeHLi 1074

11

73

SolutionSince the mass of an atom include the masses of the atomic electrons, the appropriatenumber of electron masses must be subtracted from the given values.

Reactants Products

Li73 7.01600 - 3 me Be

74 7.01693 - 4me

H11 1.0783 - 1 me no

1

1.0866

Total 8.02383 - 4me Total 8.02559 - 4meThe Q-value of the reaction

Q = -0.00176 u = 1.65 MeVThe energy is supplied as kinetic energy of the bombarding proton. The incident protonmust have more than this energy because the system must possess some kinetic energyeven after the reaction, so that momentum is conserved.With momentum conservation taken into account, the minimum kinetic energy that theincident particle can be found with the formula.

Eth = - 89.165.1

7

111

Q

M

mMeV

Example 20In a nuclear reactor, fission is produced in 1 g for U235(235.0439 u) in 24 hours by a slowneutron (1.0087 u). Assume that 35Kr

92 (91.8973 u) and 56Ba141 (140.9139 amu) are

produced in all reactions and no energy is lost.(a) Write the complete reaction(b) Calculate the total energy produced in kilowatt hour. Given 1 u = 931 MeV.


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Solution

The nuclear fission reaction is 92U235 + on

1 56Ba141 + 36Kr

92 + 3 on1

Mass defect m = [(mu + mn) (mBa + mKr+ 3mn)]m = 256.0526 235.8373 = 0.2153 u

Energy released = 0.2153 931 = 200 MeV

Number of atoms in 1g=235

1002.6 23 = 2.56 1021

Energy released in fission of 1 g ofU235 is Q = 200 2.56 1021 = 5.12 1023 MeV= (5.12 1023) (1.6 10-13) = 8.2 1010 J

= kWh106.3

102.86

10

= 2.28 104 kWh

Example 21It is proposed to use the nuclear fusion reaction:

1H2 + 1H

2 = 2He4

in a nuclear reactor of 200 MW rating. If the energy from above reaction is used with a25% efficiency in the reactor, how many grams of deuterium will be needed per day. (Themasses of1H

2 and 2He4 are 2.0141 and 4.0026 u respectively).

SolutionEnergy released in the nuclear fusion is

Q = mc2 = m(931)MeV Q = (2 2.0141 4.0026) 931MeV = 23.834 MeV = 23.834 106eVSince efficiency of reactor is 25%So effective energy used = 25 / 100 23.834 106 1.6 10-19 J = 9.534 10-13 JSince the two deuterium nucleus are involved in a fusion reaction, therefore, energy

released per deuterium is2

10534.9 13

For 200MWpower per day

number of deuterium nuclei required =13

6

102534.9

8640010200

= 3.624 1025

Since 2gof deuterium constitute 6 1023 nuclei, therefore amount of deuterium required is

g=23

25

106

10624.32

= 120.83 g/day

6. PHOTOELECTRIC EFFECT

The emission of electrons from a metallic surface when irradiated by electromagnetic radiation iscalled the phenomenon ofphotoelectric effect. The emitted electrons are called asphotoelectrons.


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For the investigation of the photoelectriceffect a schematic diagram of theapparatus as used by Lenord (1902) isshown in the fig.(14). Monochromaticlight from the lamp L illuminates a plate

P in an evacuated glass enclosure. Abattery maintains a potential differencebetweenPand a metal cylinder C, whichcollects the photoelectrons. The potentialC can be varied to be either positive ornegative relative toP. When thecollector is positive with respect to theplate, the electrons are attracted to it andthe ammeter (A) registers a current.Lenard studied the dependence ofphotoelectric current on the followingfactors.

`

+V

A

CeP

Fig.(14) Schematic diagram of apparatus used for theinvestigation of the photoelectric effect.L - lamp;P metal plate,C conducting cylinder

L

(i) Intensity of incident radiation(ii) Potential difference between the plate and the collector(iii) Frequency of the incident radiationThe result of observations are as follows:

Effect of the intensity of Incident Radiation

When the collector ispositive relative to the plate andthepotential difference is kept fixed, then for a givenfrequency of radiation, the photoelectric current is

proportional to the intensity of the light, as shown in

fig.(15). It shows that the number of emittedphotoelectrons is proportional to the light intensity.Furthermore,there is no threshold intensity. Intensity (I) Wm-2

i

(mA)

O

Fig.(15)

Effect of Potential Difference

When thefrequency and intensity of radiation are kept constant and the positive potentialof collector relative to plate is gradually increased, then the photoelectric current iincreases with the potential difference V. At some value of the potential difference, whenall the emitted electrons are collected, thus increasing potential difference has no effect on

the current. The current has reached its maximum value, called thesaturation current .

When the polarity of the battery is reversed, the electrons are repelled and only the mostenergetic ones reach the collector, so the current falls. When the retarding potentialdifference reaches a critical value, the current drops to zero. At this stopping potentialVo,only those electrons with the maximum kinetic energy are able to reach the collector.

eVo =2max2

1mv (29)


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For a given frequency of light, the saturation currentdepends on theintensity of light.Larger the intensity;

higher the saturation current.However, the stopping potential does

not change with the intensity. It isclearly shown in fig. (16).

Effect of frequency

For a given intensity of radiation, thestopping potential depends on thefrequency. Higher the frequency,

higher the value of stopping potential.

i

I2

I1

I2 >I1

VOVoFig.(16) At positive accelerating potential differences,

the maximum current is determined by theintensity of the radiation. However, thestopping potential does not change with theintensity.

The maximum kinetic energy of the electrons depends on the light source and the platematerial, but not on the intensity of the source. Certain combinations of light sources andplate materials exhibit no photoelectric effect.

2max2

1mv

SaturationCurrent

VOVo2Fig.(17) For a given intensity, stopping potentialdepends on frequency. Iff2 >f1, Vo2 > Vo1.

Vo1

f1

f2

(1)

V

Fig.(18) The maximum kinetic energy is proportionalto the frequency of light.fo1 - threshold frequency of metal (1)fo2 threshold frequency of metal (2)

fo1 fo2

(2)

O

For a metal plate there exists a minimum frequency calledthreshold frequency (fo) belowwhich no electron is emitted however large the light intensity may be.The threshold frequency is a characteristic of the metal plate.

Einsteins Theory of Photoelectric Effect

According to Einstein, the experimental results of photoelectric effect can be explained by

applying the quantum theory of light. He assumed that light of frequencyfcontain packetsor quanta of energyE = hf. On this basis, light consists of particles, and these are calledphotons. The number of photons per unit area of cross-section of the beam of light per second is proportional to its intensity. But the energy of photon is proportional to itsfrequency and is independentof the light intensity.


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In the process of photo emission a single photon gives up all its energy to a singleelectron. As a result, the electron is ejected instantaneously. Since the intensity of light isdetermined by the number of photons incident, therefore, increasing the intensity willincrease the number of ejected electrons.

The maximum possible kinetic energy

2maxmv2

1

of the photoelectrons is determined bythe energy of each photon (hf) according to the Einstein equation (30)

Whfmv 2max21

(30)

where the work function, (W), is the minimum energy needed to extract an electron fromthe surface of the material.In terms ofthreshold frequency , it is given by

W = hfo (31)Using equation (31), we may write equation (30) as

2

maxmv2

1= hf hfo = h(f fo) (32)

Also, in terms ofstopping potential,eVo = h(ffo) (33)

Example 22Ultraviolet light of wavelength 2000 causes photoemission from a surface. The stoppingpotential is 2V.

(a) Find the work function in eV(b) Find the maximum speed of the photoelectrons.

Solution(a) Using Einstein relation

W= oeVhc

or W= eV2.42200012400

(b) Since oeVmv 2max2

1

vmax =

31

19

101.9

2106.122

m

eVo

or vmax = 8.4 105 m/s

How to determine the photoelectric current?

Let P be the power of a point source of electromagnetic radiations, then intensity I ata distance rfrom the source is given by

I=24 r

P

(W/m2) (34)


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Objective Solved Examples

1. The intensity of X-rays from a Coolidge tube isplotted against wavelength as shown in the figure.

The minimum wavelength found is C and thewavelength of the K line is K. As the acceleratingvoltage is increased(a) K - C increases(b) K - C decreases(c) K increases(d) K increases

I

C KSolution

As the accelerating voltage is increased, C decreases while K remains the same. (a)

2. The half-life of215

At is 100 s. The time taken for the radioactivity of a sample of215

At todecay to 1/16th of its initial value is(a) 400 s (b) 6.3 s(c) 40 s (d) 300 s

SolutionFor the decay of 1/16th of initial value, four half lives are required. (a)

3. Which of the following processes represents a gamma-decay?(a) baXX Z

A

Z

A 1 (b) cXnX ZA

oZ

A

231

(c) fXX ZA

Z

A (d) gXeX ZA

Z

A 11

SolutionIn gamma-decay, the atomic and mass number do not change. (c)

4. The attractive potential for an atom is given byo

or

rVV ln , oV and or are constants and

ris the radius of orbit. The radius nr of the nth Bohr's orbit depends on principal quantumnumber n as

(a) nr n (b) nr 21

n

(c) nr 2n (d) nr n1

Solution

o

eor

rVV log

r

V

dr

dVF o


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r

V

r

mV o2

oVmv 2

2nhmvr

oV

hn

mV

rVm 222

2

222 4/

2r 2nr n

(a)

5. Volume V of a nucleus is related to the mass M as(a) V M3 (b) V 3 M

(c) V M (d) V M1

Solution

Radius of nucleus is given by orAr3/1

volume = 334

r v A M

(c)

6. The ratio of radius of 100Fm257 atom (assuming Bohr's model to be valid) to the Bohr

radius is(a) 257 (b) 100

(c) 41

(d) 4Solution

Since the outermost orbit number is 5 and

nr =Z

n2

Hr = Hr4

1

(c)

7. A stationary nucleus (mass number 220) decays by emitting an -particle. The totalenergy released is 5.5 MeV. The kinetic energy carried by the -particle is(a) 5.4 MeV (b) 5.6 MeV(c) 4.9 MeV (d) 6.5 MeV

SolutionApplying law of conservation of linear momentum and using the energy released, velocityof-particle and daughter nucleus can be calculated.Therefore, kinetic energy of particle

=220

4220 5.5 MeV = 5.4 MeV.

(a)


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8. After 280 days, the activity of a radioactive sample is 6000 dps. The activity reduces to3000 dps after another 140 days. The initial activity of the sample in dps is(a) 6000 (b) 9000(c) 3000 (d) 24000

Solution

The half-life of the substance is 140 days. In 420 days, there will be three half-lives. (d)

9. The energy of a photon is equal to the kinetic energy of a proton. The energy of the photonis E. Let 1 be the de-Broglie wavelength of the proton and 2 be the wavelength of thephoton. The ratio 1/2 is proportional to(a)Eo (b)E1/2

(c)E-1 (d)E-2

Solution

1 =mE

h

2and

E

hc2

(b)

10. The figure shows the variation ofphotocurrent with anode potential for aphoto-sensitive surface for three differentradiations. LetIa, Ib and Ic be the intensitiesand fa, fb and fc be the frequencies for thecurves a, b and c respectively.(a)fa =fb andIa Ib(b)fa =fc andIa =Ic(c)fa =fb andIa =Ib(d)fb =fc andIb =Ic

Photo current

O

ab

c

Anode potential

SolutionThe stopping potential for curves a and b is same. fa =fbAlso saturation current is proportional to intensity. Ia


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SolutionDue to 10.2 eV photon one photon of energy 10.2 eV will be detected.Due to 15 eV photon the electron will come out of the atom with energy

(15 - 13.6) = 1.4 eV. (c)

12. K wavelength emitted by an atom is given by an atom of atomic number Z = 11 is .Find the atomic number for an atom that emits K radiation with wavelength 4 (a) Z = 6 (b) Z = 4(c) Z = 11 (d) Z = 44

Solution

1111

1 a

vf

12

2

Zav

f

By dividing,

110

1

2

Z

1

1041

Z

Z = 6 (a)

13. If a star can convert all the He nuclei completely into oxygen nuclei. The energy released

per oxygen nuclei is[Mass of He nucleus is 4.0026 amu and mass of Oxygen nucleus is 15.9994 amu](a) 7.6 MeV (b) 56.12 MeV(c) 10.24 MeV (d) 23.9 MeV

Solution

Mass defect = 4 4.0026 - 15.9994 = 0.011Energy released by oxygen nuclei = 0.011 931 = 10.24 MeV (c)

Modern Physics for IITJEE

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