Crackiitjee.in.Phy.ch27

Table of Contents

Properties of X-rays ... 3

X-ray Spectrum ... 5

Continuous X-ray Spectrum ... 5

Characteristic Spectrum ... 7

Uses of X-rays ... 8

Moseley’s Law ... 9

de-Broglie Waves or Matter Waves ... 13

Properties of Matter Waves ... 14

Davisson – Germer Experiment ... 15

Applications of de-Broglie Wave Hypothesis ... 16

Radiation Pressure/Force ... 19

Laws of photoelectric effect (Experimental Observation) ... 25

First Law ... 25 Second Law ... 25 Third Law ... 25 Fourth Law ... 26 Fifth Law ... 26 Photoelectric Effect ... 26 Work Function ... 27 Intensity of Radiation ... 28

Two Theories of Radiation ... 30

Experimental Study of Photoelectric Effect ... 31

Einstein’s Explanation for Photoelectric Effect ... 32

Photoelectric Effect and Wave Theory ... 32

Summary of Photoelectric Effect ... 32

Graphical Variation ... 33

Atomic Excitation ... 39

Ways of Atomic Excitation ... 39

Atomic Models ... 45

Dalton’s Atomic Model... 45

J.J. Thompson’s Atomic Model ... 45

Special Points ... 46

Results from Rutherford’s Experiment ... 46

Failure of Thompson’s Model ... 47

Rutherford’s Atomic Model ... 47

Rutherford’s - scattering experiment ... 47

Failure of Rutherford’s Atomic Model ... 50

Bohr’s Atomic Model ... 51

Energy of Electrons ... 52

Hydrogen Spectrum ... 55

Special Points ... 56

Properties of Electron in nth Orbit ... 56

Summary ... 58

Limitations of Bohr’s Atomic Model ... 58

Nuclear Motion ... 63

Special Points ... 66

Intensity of X-rays

Properties of X-rays

 X-rays are invisible to human eyes

 X-rays affect photographic plate similar to visible radiation

 X-rays, under suitable condition cause photoelectric effect

 X-rays exhibit interference, diffraction and polarisation under suitable condition

 X-rays cause ionization of the gas through which they pass

 X-rays produce fluorescence and phosphorescence

 X-rays cause genetic mutation which leads to cancer

 X-rays are absorbed by the material through which they pass

Due to interaction of X-ray photons with electrons of atoms of material Io = Intensity of incident radiation

x = Thickness penetrated I = Io e-x

 = Absorption coefficient

N

0

Depth of Penetration

(x)

 X-ray equipments are never operated continuously

 X-ray equipments are always operated intermittently

 The process of X-ray production can be looked upon as inverse photo-electric

effect

 The intensity of X-rays depends upon the number of electrons striking the target i.e.

rate of emission of electrons from filament. This can be controlled by varying the filament current by adjusting electron gun circuit parameter.

 the quality of X-rays which is measured by their penetrating power is a function of potential difference

The efficiency of an X-ray tube is given as Efficiency = 1.4 x 10-19 _{Z V}

Z = Atomic no. of target V = Potential difference For example:

Z = 74 (Tungsten as target) V = 100 kV

Efficiency  1 %

Hence, 99% of the energy of striking electrons is converted into heat. High Potential Difference

High Penetrating Power Hard X-rays

Low Potential Difference Low Penetrating Power

Soft X-rays

X-ray Spectrum

Continuous X-ray Spectrum

 These continuous X-rays are produced due to variable deceleration/retardation

experienced by electrons beam as it hits and penetrates target metal.

 This is called continuous as we get all possible wavelengths starting from minimum

wavelength called cut of wavelength.

   max min hc eV hf _min  hc eV

Continuous X-ray Spectrum Characteristics X-ray Spectrum

X-ray Spectrum

Intensity of

radiation

_{V = Voltage}



min

Intensity of

radiation

_{V = Voltage}



_min



_max

I

0



(wavelength)

 The continuous X-radiations are also known as Bremsstrahlung Radiation, a

german word meaning, literally breaking radiations. This word is describing the mechanism of production of X-rays.

 In practice, X-rays of wavelength min are not produced.

 There is wavelength (m) corresponding to which intensity of radiation is maximum

0



(wavelength)



_m

V



cons tant

Characteristic Spectrum

 These X-rays are produced due to knockout of electrons of target metal atom

 When electron from K-shell is knocked out and vacancy is filled by electron from L,

we get K radiation, if it is filled by electrons from M shell, we get K radiations.

Similarly K radiations. Hence, when electrons from shell is knocked out, we get

K-series.

  



hc

E hf

ΔE is Energy difference between two transition levels

ΔE is minimum for K

Hence, XK is longest. Also, first member of any series will have longest wavelength.

The last member will have shortest wavelength.

 All members of K-series will have wavelength lesser than member of series and

L-series members will have wavelength lesser than M-L-series.

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 Characteristic X-ray radiations are so called as these radiations are characteristic target used in X-ray production.

 When target is changed say from tungsten to molybdenum, the positions of peaks

will change.

 When target is not changed, only min is affected.

 fk = fk + fL

Uses of X-rays

 Surgery  Radiotherapy  Industry  Defective Departments  Scientific Research  Crystallography  Metallography

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Slope = b

a

Moseley’s Law

This states that square root of frequency of characteristic of spectral line is directly proportional to the atomic number of target used in X-ray production



f

b (Z – a)

a = screening constant

b = A constant which depends on series

0

Depth of Penetration

Z

 For K-series a = 1

 For other series, it is determined experimentally

 The value of b is worked using atomic model

 From Bohr’s atomic model







1

RZ2



_









21 22



1

1

n

n

 f = RCZ2













12 22



1

1

n

n

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 The value of b for different members of K-series K Kβ Kγ Kδ 3 RC 2 8_RC 9 15 RC 16 24_RC 25

Example: A X-ray tube is operating at 12 kV and 5 mA. Calculate the number of electrons striking the target per second and the speed of striking electrons.

Solution: We know that



Q



ne

I

t

t

Where, n is the number of electrons striking the target per second.

  





3



16 19

I.t

5x10 x1

n

3.125x10 / s

e

1.6x10

Ans. Also,

 

2eV

m

= (2x1.6x10

-19

x12x10

3

/9.1x10

-31

)

1/2 = 6.49x107 _{m/s Ans.}

Example: The wavelength of a certain line in the X-ray spectrum for tungsten (Z=74) is

200 Å. What would be the wavelength of the same line for platinum (Z=78)? the constant 

is unity. Solution:

Using Moseley’s law, we get

 



 

1 1 2 2

f

b(Z

)

b(Z

)

f

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

 





 

2 1 1 2 2 2

c

(Z

)

c

(Z

)

2 = 1

 

 

2 1 2 2

(Z

)

(Z

)

= 200 x (74-1)2_/(78-1)2 _{= 179.76 Å Ans.}

Example: The mass absorption coefficient for aluminium for X-rays having λ = 0.32 Å is 0.6 cm2_{/g. If density of aluminium is 2.7 g/cm}2_{, find}

(i) the linear absorption coefficient of aluminium, (ii) half value thickness, and

(iii) the thickness of the absorber needed to cut down the intensity of beam to 1/20 of initial value.

Sol. (i) The mass absorption coefficient m and linear absorption coefficient  are related as

m =

where  is the density of material

 =  × m = 2.7 × 0.6

= 1.68 cm-1 _Ans.

(ii) Half value thickness = =

= 0.428 Ans.

(iii) We know that I = I0 e-x

According to problem = or = e -1.62x _{or 20 = e}1.62x or x = 1.85 cm Ans.

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Example: The wavelength of the characteristic X-ray K line emitted by a hydrogen like

element is 0.32 Å. Calculate the wavelength Kβ line emitted by the same element.

Sol. For hydrogen like element

= R For K line, = R For Kβ line, = R ∴

=

or = = 0.27 Å Ans.

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de-Broglie Waves or Matter Waves

 The waves associated with moving bodies are referred as de-Broglie waves or matter waves.

 This concept of matter waves was given by de-Broglie.

 De-Broglie had following two things in mind while giving this hypothesis.

………. (1)

Energy Planck’s Constant Frequency of Radiation

……….. (2)

Momentum

Using above two relations, we get P =



or

de-Broglie relation λ = de-Broglie wavelength P = momentum of moving body

Dual Nature of Radiation Nature of Lone Symmetry

E = hf

P =

λ =

We know KE =

 P =

Properties of Matter Waves

 de-Broglie waves are different from electromagnetic waves.  Speed of matter wave is more than the speed of light.

v = speed of wave = f λ = = =

As speed of particle can never exceed speed of light. ∴

 In ordinary situation, de-Broglie wavelength is very small and wave nature of matter can be ignored.

λ = =  4.8 × 10-34 _m

 The wave and particle aspects of moving bodies can never be observed at the same time i.e. the two natures are mutually exclusive.

λ =

v > c

Davisson – Germer Experiment

(Proof of Matter Wave)

In this experiment, high energy electron beam was made to impinge on a Nickel crystal and electron beam was found to be diffracted.

One experimental observation indicated that the first order maxima of electrons, accelerated through a potential difference of 54 V, was obtained when both the incident beam and the

detector made 65 angle with a particular family of crystal planes.

For lower and higher voltages, the peak is suppressed. According to Bragg’s law, for maxima,

2d sin  = n λ

where ‘λ’ is the wavelength of electron.

In this case, d = 0.91 × 10-10 _m

 = 65

n = 1

∴ 2 × 0.91 × 10-10 _{sin 65 = 1 × λ}

or λ = 1.65 Å

V = potential difference (in volt) between two electrodes

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∴ mv2 _{= eV}

or mv2 _{= 2 eV}

or mv =

If ‘λ’ is the wavelength of wave associated with the electron λ = or λ = we know that, h = 6.626 × 10-14 _Js m = 9.1 × 10-31 _kg e = 1.6 × 10-19 _C ∴ λ = or λ = Å For V = 54 V λ = Å = 1.67 Å

Applications of de-Broglie Wave Hypothesis

1. Electron Microscope

2. Quantization of Angular Momentum 2  r = n λ = n =



=



=

 de- Broglie wavelength of a charged particle q = charge

mvr =



V = potential difference through which charged particle is accelerated.



charged particle

=

For electron, q = 1.6 × m = 9.1 × _kg

 de- Broglie wavelength of a gas molecule K.E.gas molecule = KT

T= Absolute temperature

K= Boltzmann’s constant

This relation is valid for any gas irrespective of type of gas molecule (mono atomic, diatomic, poly atomic).

 de- Broglie wavelength of a thermal neutron Thermal neutrons behave like gas molecule

KE = KT KE = q v



=



gas molecule

=



thermal neutron

=

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Example: Find the ratio of de-Broglie wavelength of proton and alpha particle which has been accelerated through same particle difference.

Solution: We know that



=



=

 

=

Alpha particle = doubly ionized helium particle

= 4 = 2



 = = = Ans.

Example: Find the ratio of de- Broglie wavelength of molecule of hydrogen and helium which are at 27°C and 127°C respectively.

Solution:

For gas molecule, we know that



=

 

=

=

= Ans.

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Radiation Pressure/Force

 The force/pressure experienced by any surface exposed to radiation is called radiation force/pressure as the case may be

 Radiation Pressure =

c = speed of light

Case (I) : Surface is perfectly reflecting I = Intensity of radiation

E = Energy received by surface per sec = I A

N = No. of photons falling per sec = =

| | = change in momentum of one photon due to reflection = 2

We know that the force is rate of change of momentum, F = N. Δ P

= 2. ∴

Calculation of Radiation Pressure/Force

=

Particle nature of Photons

E = hf

E = P.c

F =

Also, Radiation Pressure (P) = =

Case (II) : Surface is perfectly absorbing | | =

∴

Case (III) : Surface is partially reflecting Reflection coefficient = 0.7

= +

Due to reflected Due to absorbed photons

photons

and are in same direction, F = F1 + F2 = 0.7 + 0.3

F =

F =

=

F =

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Radiation is falling obliquely E = (I) (A cos )

For Perfectly Surface ∆ pone photon = 2 cos θ

For perfectly absorbing surface

∆ p =

 In case of perfectly reflecting surface, the surface will experience force is downward direction and is care of perfectly absorbing surface, it will experience force in the direction of incident radiation.

 For partially reflecting surface

F1

= force due to reflecting photons

=

F₂

= force due to absorbed photons

=

f = Resultant force =

 Whenever exposed surface behaves like a black body

Radiation force (F)=

(

projected area of surface in thedirection of

radiation)

F =

F =

Projected Area A = × = RH Projected Area A = = 2 RH

Example: A plank of mass ‘m’ is lying on a rough surface having coefficient of friction as ‘’ in situation as shown in figure. Find the acceleration of plank assuming that it slips and surface of body exposed to radiation is black body.

Solution: E = Energy received by surface per second = I.A. cos  = Iab cos 

N = number of photons received by surface per second = =  F=

F =

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Δ Pone photon =

Fradiation = Δ Pone photon .N =



Let us draw the FBD of plank

F sin  - N = m.a

N = mg + F cos 

From above two equations, we get

a =    Ans.

Remark:

If surface is perfectly reflecting then acceleration will be zero because surface will experience radiation force perpendicular to surface in downward direction.

Example: A metallic sphere of radius R is kept in the path of a parallel beam of light beam of intensity I. Find the force exerted by the beam on the sphere.

Solution: Let us consider an elementary ring as shown in figure.

r = Radius of elementary ring = R sin 

Δ A = Area of elementary ring = (2r) (R d) = 2R2 _{sin  d}

Further, let us imagine a very small portion of this elementary ring of area Δ A.

Δ E = Energy received per sec = (I) (Δ A cos )

Δ N = Number of photon received per sec =

=



Δ F = Force experienced = ΔN ΔPone photon

= 



=



The direction of this force is radial as shown in figure.

∴ Force experienced by elementary ring is given by D =

The components of ΔF in the plane of elementary ring will cancel out.

∴ dF = cos 

=  =  ΔA =  2 R2 _{sin  d}

The force on the entire sphere is given by =

or F = cos3_{ d = Ans.}

I

_S

= Saturation

Photocurrent

Remark:

Had the sphere been perfectly absorbing, the expression of radiation force experienced would be same.

The existence of radiation force supports quantum theory or photon nature of light.

Laws of photoelectric effect (Experimental Observation)

First Law

Saturation photocurrent is directly proportional to intensity of radiation

0

I (Intensity)

Second Law

There is no time lag between photon absorption and photoelectron emission (photoelectric effect)

Third Law

 Photoelectric effect takes place when frequency of incident radiation is more than a

certaib minimum value known as threshold frequency (fo)

 If frequency of incident radiation is less than threshold frequency, then photoelectric

effect does not take place, however intense is the radiation.

 Threshold wavelength (o) is the minimum wavelength which will cause photoelectric

effect.

 For photoelectric effect:

i) o

ii) f  fo

Fourth Law

Maximum Kinetic Energy of emitted photoelectron does not depend on intensity of radiation. It depends on frequency of wavelength of radiation.

Fifth Law

Photoelectrons emitted in photoelectric effect have a range of kinetic energies (K.E.)min = 0

(K.E.)max = eVs

Photoelectric Effect

The phenomenon of ejection of electrons, when metallic surfaces are exposed to certain energetic radiations, is called photoelectric effect.

The electrons which The current constituted The minimum

are ejected as a when randomly ejected energy required

result of photoelectric photo electrons are to free an

effect are known as made to flow in one electron from

photo electrons. direction by suitable metal bondage is

application of electric called work

field. function.

Work Function

Work function is a property of metallic surface.

Ionization energy should not be confused with work function. Ionization energy is the energy required to remove an electron from outer most shell in isolated state. Theoretically, we can expect that work function will be lesser than ionization energy.

Experimentally, it is found

Basic Terms

Photo electrons

Photo Current

_{Work Function}

Work function () =

P

Metal Work Function

Cesium 1.9 eV Potassium 2.2 eV Sodium 2.3 eV Lithium 2.5 eV Copper 3.2 eV Silver 4.5 eV Platinum 5.6 eV

Table: Work functions for Some Photosensitive Metals

Work Function can be overcome by many ways: Thermionic Emission: By application of heat.

Field Emission: By application of high electric field.

Secondary Emission: By impinging a beam of high speed electrons.

Intensity of Radiation

This is defined as energy flowing per second per unit area normally. Case I : Point Source

For a point source I

=

PS = Power of

Source

I

0

r

Case II : Linear Source

I =

Remark : In general, we can say, as we move away from source, intensity decreases. How intensity will be function of distance, will depend on the geometry of source.

I



I



l

r P

Two Theories of Radiation

 We assume, energy is  We assume, energy is

propagated in the form emitted in the form of

of waves. energy packets known as

 We assume, the energy photons. Energy of photon is

from the source is emitted given by

continuously.

E = energy of photon h = Planck’s constant f = frequency of radiation * Energy from source is not emitted continuously.

Photoelectric effect supports photon theory of radiation. A photon is a charge-less and mass-less particle.

A photon is bound to move with speed of light.

P = momentum of photon c = speed of light

Whenever photon interacts with mater, it transfers energy and momentum.

Wave theory

Photon Theory

E = hf

Photocurrent

(I)

I

_S

V

S

V

_P

Voltage (V)

Experimental Study of Photoelectric Effect

Variation of photoelectric current with anode potential

(a) Saturation Current (Is): The maximum value of photocurrent in a given situation is

called saturation point.

(b) Stopping Potential (Vs): The negative collector plate potential at which corresponding

photocurrent is zero, is called stopping potential. This is also known as cut off voltage. (c) Pinch of Voltage (VP): The positive plate potential at which photocurrent saturates, is

called pinch of voltage. If plate voltage is increased beyond pinch off voltage, there is no increase in photocurrent.

Einstein’s Explanation for Photoelectric Effect

 Einstein explained photoelectric effect using Planck’s quantum theory of radiation

 Einstein says that the entire energy of incident photon is absorbed by electron

E =  + (K.E.)max

or hf =  + (K.E.)max

 represents work function

 It is very clear, photo electric effect will take place only if energy of incident photon is greater than work function

 = hfo = hc/o

Hence photon theory of radiation beautifully explains the existence of threshold wavelength or frequency.

 The typical penetration of radiation is 10-8 cm. Hence, electrons are emitted out not

only from surface but also from subsurface layer also.

Photoelectric Effect and Wave Theory

 According to wave theory, energy radiation is bound to cause photoelectric effect.

The more the intensity of radiation, the sooner is the photoelectric effect.

 According to wave theory, (K.E.)max will depend on the intensity of radiation.

 Wave theory predicts appreciable time lag between energy absorption and

photoelectron ejection.

Summary of Photoelectric Effect

1. hf =  + (K.E.)max

2. (K.E.)max = eVs

3.  = hfo = hc/o

 Frequency of radiation is frequency of oscillation of electric field or magnetic field

 Einstein received Nobel Prize for explaining photoelectric effect not for his famous

theory of relativity.

K.E.

_MAX

0

_fo

₁

_f

Metal 1 Metal 2 Metal 3 Slope = h

fo

2

fo

3

-



Graphical Variation

1. Maximum Kinetic Energy Vs Frequency hf =  + (K.E.)max

or (K.E.)max = hf - 

V

S

0

_fo

₁

_f

Metal 1 Metal 2 Metal 3 Slope = h/e

fo

2

fo

3

-



/e

2. Stopping Potential Vs Frequency eVs = hf -  or



 

_{ }





 

s

h

V

f

e

e

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V

S

0

_fo

₁

_f

Metal 1 Metal 2 Metal 3 Slope = h/e

fo

2

fo

3

-



/e

3. Stopping Potential Vs



1







_

_

 







s

hc

eV

or





_

  

_{  }









  

s

hc 1

V

e

e

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V

S

0

_f

₀

_f

4. Maximum speed of photoelectron Vs frequency (K.E.)max = hf - 



2 max

1

_m

2

= hf - 

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Example: The stopping potential for photoelectrons emitted from surface illuminated by light wavelength 5893 Å is 0.36 volt. Calculate the maximum kinetic energy of

photoelectrons, the work function of the surface and the threshold energy. Solution: We know that (K.E.)max = hf -  =





  









hc

 





max

hc

_K.E.

Also, K.E.max = eVs = 0.36eV Ans.

   

 

34 8



19 10

(6.62x10 )(3x10 )

_0.36x1.6x10

5893x10

= 1.746 eV Ans.

The threshold frequency is given by:   







19



14 o 34

2.794x10

f

4.22x10 hertz

h

6.62x10

Example : A beam of light has three wavelengths 4144 Å, 4972 Å and 6261 Å with a total

intensity of 3.6 × 10-3 Wm-2 _{equally distributed amongst the three wavelengths. The beam}

falls normally on an area 1.0 cm2 _{of a clean metallic surface of work function 2.3 eV.}

Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectron liberated in two seconds.

Solution : We know that threshold wavelength (λ0) =

 ∴ λ0 = = 5.404 × 10-7 _{m = 5404 Å}

Thus, wavelength 4144 Å and 4972 Å will emit electrons from the metal surface. Energy incident on surface for each wavelength

= Intensity of each wavelength × Area of the surface = × (1.0 cm2_{) = 1.2 × 10}-7 _watt

Energy incident on surface for each wavelength in two second E = (1.2 × 10-7_{) × (2) = 2.4 × 10}-7 _Joule

Number of photons n1 due to wavelength 4144 Å = 4144 × 10-10 m

n1 =

= 0.5 × 10

12

Number of photons n2 due to wavelength 4972 Å

n2 = = 0.575 × 1012 N = n1 + n2 = 0.5 × 1012 + 0.575 × 1012 = 1.075 × 1012 _Ans.

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Atomic Excitation

Ways of Atomic Excitation

Two ways of Atomic Excitation

An atom will not Whenever an atom is

absorb photon of any excited by collision,

arbitrary energy. collision must be

inelastic.

For a photon to be If loss of K.E. as

absorbed by an atom, it permitted by

must have energy equal conservation of linear

to difference of energy momentum is less than

of ground state and any the minimum

excited state excitation energy, then

collision will be elastic. By Photon

Absorption

By Collision

Example: A hydrogen atom is projected towards a stationary hydrogen atom in ground state. Find the minimum kinetic energy of projected hydrogen atom so that after collision, one of the hydrogen atom is capable of emitting photon. Given that ionization energy of hydrogen atom = 13.6 eV.

Solution:

v………. v’

Before collision After collision

Conservation of linear momentum gives mv = 2 mv’



Loss of K.E. = mv2 _{- m(v/2)}2 ₌

=

= 50 % of initial kinetic energy

∴ KEmin = (10.2 eV) × 2 = 20.4 eV Ans.

Example: A neutron of kinetic energy 65 eV collides in-elastically with a singly ionized helium atom at rest. It is scattered at an angle of 90 with respect to its original direction.

(i) Find the allowed values of the energy of neutron and that of the atom after the

collision.

(ii) If the atoms get de-excited subsequently by emitting radiation, find the

frequencies of the emitted radiation.

Solution: Applying the law of conservation of momentum in X and Y directions, We have, m = 4 m2 cos  m1 = 42 sin  Eliminating ‘’, we get v’ = v/2

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2 _{+ } 12 = 16 22 [sin2 + cos2 ] = 16 22 22 = 2 2 1

16

  

If ΔE represents the loss of energy in the collision, then COE gives

65 eV = 1 2m 1 2 ₊

1

2

(4m) 2 2 _{+ ΔE} = 1 2m 1 2 _{+ 1} 2(4m) + 2 2 1

u

16



 











+ ΔE =

1

2

m 1 2 ₊

1

8

u 2 _{+ 1} 8 m 1 2 _{+ ΔE} = 1 2m 1 2 ₊ 1 4 2 1_mu 2       + 1 8 m 1 2 _{+ ΔE} = 5 8 m 1 2 ₊ 1 4 (65 eV) + ΔE or

65

3

4













eV =

5

8

m 1 2 _{+ ΔE} We know that, En = 2 2

13.6Z

n



Here, Z = 4 ∴ E1 = - 54.4 eV E2 = - 13.6 eV

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E3 = - 6.04 eV E4 = - 3.4 eV

E5 = - 2.17 eV

Now, loss of energy in the collision process must have been used in exciting the atom. There may be different possibilities regarding the loss of energy in the collision process.

First Possibility:

Let ΔE = 54.4 – 13.6 = 40.8 eV

If we make substitution for this value of ΔE, we get

5 8m 1 2 _{= 65 ×}

3

4

- ΔE =

3

65

40.8

4



_{ }











eV = (48.75 – 40.8) eV = 7.95 eV ∴ Kinetic energy of scattered neutron is

=

1

2

m 1 2 _{+ 4}

5× 7.95 eV = 4 × 1.59 eV = 6.36 eV

Hence 1st _{allowed values of the energy of neutron}

after collision = 6.36 2 _{+ } 12 = 16 22 or 1 2m  2 ₊ 1 2m 1 2 ₌ 1 2m × 16 2 2 or 1 2m  2 ₊ 1 2m 1 2 _{= 4 ×} 1 2 (4m) 2 2 or 65 eV + 6.36 eV = 4 × K.E. of atom ∴ K.E. of atom =

71.36

4

eV = 17.84 eV Ans. Second Possibility: Let ΔE = 54.4 – 6.04 = 48.36 eV In this case, 5 8m 1 2 _{= 65 ×} 3 4 - ΔE = 3 65 48.36 4  _{ }      eV = (48.75 – 48.36) eV = 0.39 eV

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∴ 1

2m 1 2 ₌

4

5

× 0.39 eV = 2.565 eV = 0.312 eV

Hence 2nd allowed value of the energy of neutron after collision = 0.312 eV 2 _{+ } 12 = 16 22 or 1 2m  2 ₊ 1 2m 1 2 _{= 4 ×} 1 2 (4m) 2 2 or 65 eV + 0.312 eV = 4 × K.E. of atom ∴ K.E. of atom =

65.312

4

eV = 16.328 eV Ans.

Proceeding as above, if ΔE = 54.4 – 3.4 = 51 eV Kinetic energy of neutron is

1 2m 1

2 ₌

4

5

× (48.75 - 51) = - 45 × 2.25 eV = - 4 × 0.45 eV = - 1.80 eV

This gives negative energy. Hence, further values of energy are not allowed. Thus, allowed values of energies of neutron are 6.36 eV, 0.312 eV while the allowed values of energies of He-atom are 17.84 eV and 16.328 eV.

(ii) It is clear from first part that the helium atom is excited to third state or second state.

Hence, there can be three possible emission transitions i.e. 3  1, 3  2, and 2  1. The

frequencies of radiations emitted in these transitions can be calculated as follows: (a) 1 1  = Z 2 _R 2 2 1 1 1 3  _      = 4 × 1.097 × 8 9 × 10 7 f1 = 1

c



= (1.097 × 10 7_{) × 4 × 8} 9 × (3 × 10 8₎ = 11.7 × 1015 _{Hz Ans.} (b) 2

1



= Z 2 _R 2 2 1 1 2 3  _      = 4 × 1.097 × 10 7 _{× 5} 36 f2 = 2

c



= (1.097 × 10 7_{) × 4 ×} 5 36 × (3 × 10 8₎ = 1.827 × 1015 _{Hz Ans.}

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(c) 3 1  = Z 2 _R 2 2

1

1

1

2



_











= 4 × 1.097 × 10 7 _× 3 4 f3 = 3

c



= (1.097 × 10 7_{) × 4 ×}

3

4

× (3 × 10 8₎ = 9.85 × 1015 _{Hz Ans.}

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Atomic Models

Any explanation regarding atomic structure is called atomic model. Daltons atomic model.

Thompson’s atomic model. Rutherford’s atomic model.

Dalton’s Atomic Model

According to this model, matter is made up of very tiny particles called atom. An atom is in divisible, i.e. it can’t be divided by any physical or chemical process.

J.J. Thompson’s Atomic Model

(Water- Melon Model)

J. J. Thompson gave the first idea regarding structure of atom. The model is known after him as Thompson’s atomic model.

According to this model, whole of positive charge is distributed uniformly in the form of a sphere. Negatively charged electrons are arranged within this sphere here and there. The model is popularly known as plum pudding model.

Every electron is attracted towards the centre of uniformly charged sphere while they exert a force of repulsion upon each other. The electrons get themselves arranged in such a way that the force of repulsion is exactly balanced.

e e e e e e e e Electron s Positively Charged Matter e e e

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Special Points

(i) As the gold foil is very thin, it can be assumed that - particles will not suffer more than one collision during their passage through gold foil. Hence , single nucleus computation of - particles trajectory is sufficient.

(ii) The nucleus of gold is about 50 times heavier than an - particle, therefore gold nucleus remains stationary throughout the scattering process and produces a large deflection in - particle.

(iii) The formula that Rutherford obtained for - particle scattering by a thin foil on the basis of the nuclear model, of the atom and Coulomb’s law is

N =

Where, N = Number of alpha particles per unit area that reach the screen at a

scattering angle of .

Ni = Total number of alpha particles that reach the screen

n = Number of atoms per unit volume in the foil Z = Atomic number of the foil atoms

r = Distance of the screen from the foil KE = Kinetic energy of the alpha particles t = Foil thickness.

The - particles on striking the atoms of the foil, get scattered in different directions. By rotating the chamber, the number of particles scattered along different directions can be recorded by observing the scintillations on the fluorescent screen.

Results from Rutherford’s Experiment

Most of the - particles, either passed straight through the metallic foil, or suffered only

small deflections. This could be explained by Thompson’s atomic model.

A few particles were deflected through angles which were less than or equal to 90.

Very few particles were found to be deflected at greater than 90. It was observed that

only one in 2000 - particles was found to be deflected through 180. In other words, it

was sent back in the same direction from where it came. The target could not be explained by Thompson’s atomic model. It was one of the main reasons for rejecting Thompson’s atomic model.

N

Scattering angle (θ) = 180°

r

The graph between N and  was found to be as shown in figure.

If ‘t’ is the thickness of the foil and ‘N’ is the number of - particles scattered in a particular direction ( = constant) it was observed that

= Constant

When distributed, electrons vibrate to and fro within the atom and causes emission of visible, infrared and ultraviolet light.

Failure of Thompson’s Model

According to this model, hydrogen can give rise to a single spectral line. Experimentally, hydrogen is found to give several spectral lines.

Rutherford’s Atomic Model

This atomic model is based on Rutherford’s - scattering experiment.

Rutherford’s



- scattering experiment

Rutherford performed experiments on the scattering of alpha particles by extremely thin

metal foils. A radioactive source (radon) of - particles was placed in a lead box having a

narrow opening as shown in figure. This source emits - particles in all possible directions.

However, only a narrow beam of alpha particles emerged from the lead box, the rest being

absorbed by the lead box. This beam of - particles is made incident on a gold foil, the -

particle is made incident on a gold foil whose thickness is only one micron, i.e. 10-6 _{m. When}

passing through the metal foil, the - particles get scattered through different angles.

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These particles fall on a fluorescent screen, producing a tiny flash of light on the screen. This can be easily viewed by a low power microscope in a dark room.

Conclusion

(i) The fact that most of the - particles passed undeviated led to the conclusion

that an atom has a lot of empty space in it.

(ii) - particles are heavy particles having high initial speeds. These could be deflected through large angles only by a nearly the entire mass of the atom were concentrated in a tiny central core. Rutherford named this core as nucleus.

(iii) The scattering of - particles by the nucleus was found to be in accordance to coulomb’s law which proved that coulomb’s law hold for atomic distances also.

(iv) The difference in deflection of various particles can be explained as follows:

- particles which pass at greater distance away from the nucleus, shown as 2 and 8 in

below figure., suffer a small deflection due to smaller repulsion exerted by the nucleus upon them. The particles like 3 and 7 which pass close to the nucleus experience a comparatively greater force and hence get deflected through greater angles.

A particle 5 which travels directly towards the nucleus is first slowed down by the repulsion force. Such a particle finally stops and is repelled along the direction of its approach. Thus, it gets repelled back after suffering a deviation of 180.

Rutherford’s Atomic Model:

(i) An atom consists of equal amounts of positive and negative charge so that

atom, as a whole, is electrically neutral.

(ii) The whole of positive charge of the atom and practically whole of its mass is concentrated in a small region which forms the core of the atom, called the nucleus.

(iii) The negative charge, which is contained in the atom of electrons, is distributed all around the nucleus, but separated from it.

(iv) In order to explain the stability of electron at a certain distance from the nucleus, it was proposed by ‘Rutherford’ that the electrons revolve round the nucleus in circular orbits. The electrostatic force of attraction between the nucleus and the electrons provides the centripetal force.

(v) The nuclear diameter is of the order of 10-14 _m.

This can be shown as follows:

Let us consider an - particle projected towards a nucleus having charge +Ze with velocity

v0 as shown in figure.

Now, at a distance of closest approach (r0), - particle will come to rest instantly.

From conservation of energy, we get mv02 =



r0 = _

In one of the experiments, - particles of velocity 2 × 107 _ms-1 _{was bomdarded upon gold} foil. Here, Z = 79 e = 1.59 × 10-19 _C m = 4 × 1.67 × 10-27 _kg v = 2 × 107 _ms-1 r0 = 4 × 9 × 109 × = 2.69 × 10-14 _m

This gives the radius of the nucleus.

Failure of Rutherford’s Atomic Model

(i) According to electromagnetic theory, a charged particle in accelerated

motion must radiate energy in the form of electromagnetic radiation. As the electron revolves in a circular orbit, it is constantly subjected to

centripetal acceleration . So, it must radiate energy continuously. As a

result of this, there should be a gradual decrease in the energy of electron. The electron should follow a spiral path and ultimately fall into the nucleus.

Electron spiraling inwards as it radiates energy due to its acceleration}

Thus, the whole atomic structure should collapse. This is contrary to the actual fact that atom is very stable.

(ii) According to Rutherford’s model, electrons can revolve in any orbit. If so, it must emit continuous radiations of all frequencies. But atoms emit spectral lines of only definite frequencies.

Bohr’s Atomic Model

(i) Law of circular orbit: According to Bohr, electron revolve around nucleus in circular orbit and the necessary centripetal force is provided by electrical force of attraction between nucleus and electron

∴

=

(ii) Law of stationary orbit: The electrons revolve around nucleus in orbits known as stationary orbits. When electron revolve in stationary orbit, they are permitted to disobey the law of electrodynamics i.e. electrons do not radiate energy.

(iii) Law of quantization of angular momentum: When electron revolves around nucleus in stationary orbit, its angular momentum is quantized and is given by

mvr =

(iv) Law of emission of radiation: When electron makes transition from higher orbit to lower orbit the different energy is emitted, as

electromagnetic radiation emitting one or more photons.

(v) Law of absorption of radiation: An electron can not absorb photon of arbitrary energy value. It absorbs photon of energy equal to difference of energy between ground state and any excited state. When an electron absorbs, photon is moved to higher orbit. In fact, probability of

absorption of photon having energy not corresponding to difference of energy between ground state and any excited state.

Energy of Electrons

All laws of classical mechanics are valid when an electron revolves around nucleus and are not valid during transition. During transition, quantum physics is applicable.

The energy En of an electron in orbit having principle quantum number n is the sum

of kinetic and potential energies En = KE + PE

PE = Electrostatic potential energy + Gravitational Potential energy ∴ En = mv2 - We can write mv2 ₌ ∴ En = - Now, we get r = 

This is the equation for the radius of the permitted orbits. Now, we get

En =

On substituting the value of different constants, we get En =

eV

Special Points

(i) In the derivation of above expression for the energy of the electron in the nth

orbit, following assumptions are taken:

a) Gravitational potential energy has been neglected and for writing electrostatic

potential energy, infinity is taken as zero potential position.

b) Nucleus is assumed to be infinitely massive i.e. nucleus is at rest.

(ii) The quantity

is known as Rydberg constant ‘R’ and its value is

1.09 × 107 _m-1_.

Also, Rhc = 13.6 eV

(iii) Total energy, (TE) = = - KE (iv) For Hydrogen atom, Z = 1

∴ En =

For sake of convenience and fast calculation, it is better to remember following energy levels.

1. Excitation Energy:

This is defined as energy difference between ground state and excited state. First excitation energy = E2 – E1

For Hydrogen atom = 10.2 eV Second excitation energy = E3 – E1

For Hydrogen atom = 12.1 eV

2. Excitation Potential:

This is defined as the potential difference which will be required to accelerate electron to acquire energy equal to given excitation energy.

Excitation potential =

For Hydrogen Atom, first excitation potential energy = 10.2 V

IInd _{Excitation potential =}

= = 12.1 eV

3. Ionization energy (IE):

 This is defined as the amount of energy required to ionize an atom.  Energy required to make electron move from ground state to infinity. 

 For hydrogen atom

IE = 0 - (- 13.6 eV)

= 13.6 eV 4. Ionization potential (I.P.):

o This is defined as the potential difference required to accelerate an electron so that it acquires energy to ionization energy.

o I.P. =

o For hydrogen atom

o I.P. =

= 13.6 eV

IE = E- E1

Hydrogen Spectrum

 It has only Lyman series. It has many series.

N = Number of spectral lines = n_C 2 = n = principal quantum of excited state

let n = 3 (2nd _{excited state) N =} 3_C 2 = 3 3 1 3 2 2 1  En = E2 - E1 = hf = _

Excited state Ground state R = Rydberg constant  Lyman Series: n1 = 1

Hydrogen Spectrum

Absorption spectrum Emission spectrum 

= R

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n2 = 2, 3, 4, ……….

 Balmer Series :

This is a series in which all the lines correspond to transition of electrons from higher state to the orbit having n = 2, n1 = 2,

n2 = 3, 4, 5,…………  Bracket Series : n1 = 3 n2 = 4, 5, 6, ………..  Paschen Series : n1 = 4  Pfund Series : n1 = 5

Special Points

(i) For any series, the wavelength of first member of series is longest and that of

limiting member is shortest.

(ii) Lyman series lies in ultraviolet region. Balmar series lies in visible region. Bracket series

Paschen series lies in infrared region. Pfund series

(iii) Absorption spectrum of H-atom consists of Lyman series only.

Properties of Electron in nth Orbit

(i) Energy of electron in nth orbit

En me

(ii) Bohr Radius

rn

En =

-

r

n

=

Å

(iii) Velocity of electron in nth orbit

Vn is independent of mass of electron

(iv) Current (In) In = e fn fn = (v) Time Period (Tn) Tn = fn =

(vi) Magnetic Dipole Moment (Mn)

Mn = In An = (e fn) ( rn2)

The value of magnetic moment in first Bohr orbit is called Bohr magneton (B). Its value is given by

B =

= 9.27 × 10

-24 _{A m}2

It is independent of atomic number. (vii) Magnetic Field (Bn)

The magnetic field at centre due to revolution of electron in nth orbit

is given by Bn = Vn = 2.18 × 106 m/s In



Tn



Mn

=

_

Mn



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Bn

Summary

 To calculate any property of atom, it is sufficient to remember dependency of ENERGY, RADIUS and VELOCITY on principal quantum number (n), atomic number (Z) and mass of electron (me).

 Dependency on mass of electron is required in situation where motion of nucleus is taken into consideration.

Limitations of Bohr’s Atomic Model

 It is applicable to one electron system only.

 Bohr could not explain why electrons are permitted to disobey law of electrodynamics while revolving in stable orbits.

 Bohr could not explain quantization of angular momentum.

 Bohr model could not account for splitting in magnetic and electric field.  Bohr atomic model could not explain hyperfine structure of spectral line.

Example: A gas of identical hydrogen like atom has some atoms in the lowest (ground) energy level A and some atoms in a particular upper (excited) energy level B and there are no atoms in any other energy level. The atom of the gas make transition to higher energy level by absorbing monochromatic light of photons have energy 2.7 eV. Some have more and some have less than 2.7 eV.

(i) Find the principal quantum number at initially excited level B. (ii) Find the ionization energy for the gas atoms.

(iii) Find the maximum and the minimum energies of the emitted photons. Solution:

Above figure shows the energy levels A, B of the hydrogen like atom. When light of photon energy 2.7 eV is absorbed, let the electrons go to an excited state C. Since subsequently the atom emits six different photons, state C should be such that six different transitions are possible. The possible transitions are shown in the above figure and it is obvious that energy level C must correspond to quantum number 4. The quantum number corresponding to state B must therefore be between 1 and 4. This means that it is either 2 or 3.

Also, EC – EB = 2.7 eV

If nB = 3, there will be no subsequent radiations with energy less than 2.7 eV. But we are

given that there are some subsequent radiations with energy less than 2.7 eV. This is possible only if there is some other energy state between B and C having a difference less

than 2.7 eV. Therefore, nB must be 2.

(i) En = - Z2 EB = E2 = - Z2 _{= - 3.4 Z}2

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and EC = E4 = -

Z2 _{= - 0.852 Z}2 _{– (- 3.4 Z}2_{) = 2.55 Z}2

∴ EC – EB = 2.7 eV = - 0.852 Z2 = - 13.6 eV

∴ Z = 1 Ans

(ii) The ionization energy = E1 = - 13.6 Z2 = - 13.6 eV Ans.

(iii) The maximum energy of the emitted radiation Emax corresponds to a transition from n =

4 to n =1

∴ Emax = - 0.852 – (- 13.6) = 12.748 eV Ans.

and Emin = E4 – E3 = 15.1 – 0.85

= 0.66 eV Ans.

Example: Consider an excited hydrogen atom in state n moving with a velocity v (v < c). It emits a photon in the direction of its motion and changes its state to a lower state m. Find

the frequency of emitted radiation in terms of frequency f0 emitted if the atom were at rest.

Solution: Consider the situation as shown in figure

mv = mv’ + or m(v – v’) = ……… (1) mv2 _{+ E} n = mv’ 2 + Em + hf (v2 _{– v’} 2_{)+ E} n – Em = hf …………. (2) or (v + v’) (v – v’) + hf0 = hf [∵ En – Em = hf0]

Using equation (1), we get

(2v)

+ hf0 = hf

or f = f0

As < 1, expanding binomially and neglecting higher powers, we get

f = f0

Ans.

Example: Suppose the potential energy between electron and proton at a distance r is

given by

. Use Bohr’s theory to obtain energy levels of such a hypothetical atom.

Solution: We know that - = F It is given that U = - Hence, F =

-

=

-

=

According to Bohr’s theory, this force provides the necessary centripetal force for orbital motion. ∴ = ……… (1) Also, mvr =  ………. (2) Hence, v =  ………. (3)

Substituting this value in equation (1), we get

 =

or r = 

Substituting this value of r in equation (2), we get v =



Total Energy (E) = K.E. + P.E. = mv2 _- =



-

 or E = , k = Ans.

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Nuclear Motion

When motion of nucleus is taken into consideration, it is dealt with the concept of reduced mass.

In situation, where nucleus as well as electrons is moving, it is better to say that both revolve around common centre of mass.

The motion of nucleus is taken into account when the physical properties of the system (atom) are affected significantly compared to the situation when nucleus is assumed to be stationary. When motion of nucleus is accounted, it is more appropriate to say that both electron and nucleus revolve around their common centre of mass.

r = rN + re

From the property of mass mN rN = me re

On solving above equations, we get rN =

re =

Example: Taking into account the motion of the nucleus of a hydrogen atom, find the expressions for the electron’s energy in the ground state and for the Rydberg constant. How much (in percent) do the binding energy and the Rydberg constant, obtained without taking into account the motion of the nucleus, differ from the more accurate corresponding values of these quantities ?

Solution: We know that when motion of nucleus is taken into consideration, mass of electron is replaced by reduced mass of system which is given by

 =

where me = mass of electron

mN = mass of nucleus

Now for H-atom, binding energy is given by Eb = - E1 =

∴ E’b =



Hence, relative difference in binding energy of the electron in the two cases is given by

=

=

 0.055 % Ans.

For hydrogen atom with stationary nucleus, Rydberg constant is given by R =

Hence, Rydberg constant considering motion of nucleus is given as

R’ = 

Therefore, relative difference in two values of Rydberg constant is given as

=

=

 0.055 % Ans.

where I = moment inertial of system (atom)

Ie = moment of enertia of electron

IN = moment of enertia of nucleus

I = me re2 + mN rN2 = me + mN =  r2  = reduced mass = I = Ie + IN

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is angular momentum of system is angular momentum of electron is angular momentum of nucleus

L = me ve re + mN vN rN

= me w re2 + mN vN rN2

=  w r2

 =

is less than one, hence it is called reduced mass.

To calculate any property of such system, we can look upon as system of

particle having mass equal to reduced mass () revolving around stationary

nucleus at a separation r.

Law of Circular Orbit:

= =

Law of Quantisation of Angular Momentum:

me ve re + mN vN rN =

= +

Special Points

(i) The concept of nuclear motion led to the discovery of deuterium isotopes.

(ii) When motion of nucleus is considered in hydrogen atom, all the energy levels are changed by the fraction

= 0.99945

This represents an increase of 0.55 percent of energy.

Summary

In all situation of motion of nucleus, replace mass of electron by its reduced mass.