Crackiitjee.in.Phy.ch27
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cr a
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iit je
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Table of Contents
Properties of Xrays ............................................................................................................................................. 3 Xray Spectrum ..................................................................................................................................................... 5 Continuous Xray Spectrum .............................................................................................................................................. 5 Characteristic Spectrum ..................................................................................................................................................... 7 Uses of Xrays ..................................................................................................................................................................... 8
Moseley’s Law ....................................................................................................................................................... 9 deBroglie Waves or Matter Waves ................................................................................................................. 13 Properties of Matter Waves ............................................................................................................................................. 14 Davisson – Germer Experiment ...................................................................................................................................... 15 Applications of deBroglie Wave Hypothesis ................................................................................................................. 16
in
Radiation Pressure/Force .................................................................................................................................. 19
e.
Laws of photoelectric effect (Experimental Observation) ............................................................................ 25 First Law ............................................................................................................................................................................ 25
iit je
Second Law ....................................................................................................................................................................... 25 Third Law ........................................................................................................................................................................... 25 Fourth Law......................................................................................................................................................................... 26
ck
Fifth Law ............................................................................................................................................................................ 26
Photoelectric Effect ............................................................................................................................................ 26
cr a
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
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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
cr a
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iit je
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Special Points .................................................................................................................................................................... 66
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Properties of Xrays
Xrays are invisible to human eyes Xrays affect photographic plate similar to visible radiation Xrays, under suitable condition cause photoelectric effect Xrays exhibit interference, diffraction and polarisation under suitable condition Xrays cause ionization of the gas through which they pass Xrays produce fluorescence and phosphorescence Xrays cause genetic mutation which leads to cancer Xrays are absorbed by the material through which they pass Due to interaction of Xray photons with electrons of atoms of material Io = Intensity of incident radiation x = Thickness penetrated I = Io ex
in
= Absorption coefficient
iit je
e.
N
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Intensity of Xrays
0
Depth of Penetration
(x)
Xray equipments are never operated continuously Xray equipments are always operated intermittently The process of Xray production can be looked upon as inverse photoelectric effect The intensity of Xrays 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.
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the quality of Xrays which is measured by their penetrating power is a function of potential difference
High Potential Difference High Penetrating Power Hard Xrays
Low Potential Difference Low Penetrating Power Soft Xrays
The efficiency of an Xray tube is given as Efficiency = 1.4 x 1019 Z V Z = Atomic no. of target
in
V = Potential difference
e.
For example:
V = 100 kV Efficiency 1 %
iit je
Z = 74 (Tungsten as target)
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Hence, 99% of the energy of striking electrons is converted into heat.
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Xray Spectrum
Xray Spectrum
Continuous Xray Spectrum
Characteristics Xray Spectrum
Continuous Xray Spectrum These continuous Xrays 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.
cr a
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iit je
e.
in
eV hfmax
min
hc min
hc eV
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I
Intensity of radiation
V = Voltage
min
(wavelength)
iit je
ck
The continuous Xradiations are also known as Bremsstrahlung Radiation, a german word meaning, literally breaking radiations. This word is describing the mechanism of production of Xrays. In practice, Xrays of wavelength min are not produced. There is wavelength (m) corresponding to which intensity of radiation is maximum
cr a
e.
in
0
Intensity of radiation
min
V = Voltage
max
(wavelength)
0 m V cons tant
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Characteristic Spectrum These Xrays are produced due to knockout of electrons of target metal atom When electron from Kshell 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 Kshell is knocked out, we get Kseries.
hc
ck
E hf
iit je
e.
in
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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 Kseries will have wavelength lesser than member of Lseries and Lseries members will have wavelength lesser than Mseries.
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iit je
ck
Characteristic Xray radiations are so called as these radiations are characteristic target used in Xray 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
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Uses of Xrays
Surgery Radiotherapy Industry Defective Departments Scientific Research Crystallography Metallography
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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 Xray production
f b (Z – a) a = screening constant b = A constant which depends on series
Depth of Penetration
cr a
0
ck
a
iit je
e.
in
Slope = b
For Kseries 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
f = RCZ2
1 1 2 2 n1 n2 1 1 2 2 n1 n2
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Z
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The value of b for different members of Kseries
K
Kβ
3 RC 2
Kγ
Kδ
15 RC 16
8 RC 9
24 RC 25
Example: A Xray 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.
in
Solution: We know that
e.
Q ne t t
iit je
I
Where, n is the number of electrons striking the target per second.
Also,
ck
I.t 5x103 x1 3.125x1016 / s Ans. e 1.6x1019
2eV m
cr a
n
= (2x1.6x1019x12x103/9.1x1031)1/2 = 6.49x107 m/s Ans.
Example: The wavelength of a certain line in the Xray 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
f1 f2
b(Z1 ) b(Z2 )
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c 1 (Z1 )2 c 2 (Z2 )2 2 = 1
(Z1 )2 (Z2 )2
= 200 x (741)2/(781)2 = 179.76 Å Ans.
Example: The mass absorption coefficient for aluminium for Xrays having λ = 0.32 Å is 0.6 cm2/g. If density of aluminium is 2.7 g/cm2, find (i) the linear absorption coefficient of aluminium, (ii) half value thickness, and
in
(iii) the thickness of the absorber needed to cut down the intensity of beam to 1/20 of initial value.
m = where is the density of material
iit je
e.
Sol. (i) The mass absorption coefficient m and linear absorption coefficient are related as
ck
= × m = 2.7 × 0.6
cr a
= 1.68 cm1 Ans.
(ii) Half value thickness =
=
= 0.428
(iii) We know that
I = I0 ex
According to problem or or
Ans.
= = e1.62x
or
20 = e1.62x
x = 1.85 cm Ans.
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Example: The wavelength of the characteristic Xray 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
= 0.27 Å
Ans.
iit je
e.
in
=
ck
or
=
cr a
∴
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deBroglie Waves or Matter Waves The waves associated with moving bodies are referred as deBroglie waves or matter waves. This concept of matter waves was given by deBroglie. DeBroglie had following two things in mind while giving this hypothesis.
Nature of Lone Symmetry
e.
iit je
E = hf
in
Dual Nature of Radiation
Planck’s Constant
Frequency of Radiation
ck
Energy
……….. (2)
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P= Momentum
Using above two relations, we get P=
……………. (1)
or
λ=
deBroglie relation λ = deBroglie wavelength P = momentum of moving body
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We know KE = P=
λ=
Properties of Matter Waves deBroglie waves are different from electromagnetic waves. Speed of matter wave is more than the speed of light. v = speed of wave = f λ =
=
=
∴
in
As speed of particle can never exceed speed of light.
e.
v>c
λ=
=
iit je
In ordinary situation, deBroglie wavelength is very small and wave nature of matter can be ignored. 4.8 × 1034 m
cr a
ck
The wave and particle aspects of moving bodies can never be observed at the same time i.e. the two natures are mutually exclusive.
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Davisson – Germer Experiment
in
(Proof of Matter Wave)
iit je
e.
In this experiment, high energy electron beam was made to impinge on a Nickel crystal and electron beam was found to be diffracted.
cr a
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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 × 1010 m = 65 n=1
∴ or
2 × 0.91 × 1010 sin 65 = 1 × λ 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= λ=
or we know that,
h = 6.626 × 1014 Js m = 9.1 × 1031 kg
or
λ=
Å
V = 54 V
Å = 1.67 Å
cr a
λ=
ck
For
e.
λ=
iit je
∴
in
e = 1.6 × 1019 C
Applications of deBroglie Wave Hypothesis 1. Electron Microscope 2. Quantization of Angular Momentum 2r=nλ=n
=
mvr =
=
=
de Broglie wavelength of a charged particle q = charge 16
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V = potential difference through which charged particle is accelerated.
KE = q v
charged particle
=
For electron, q = 1.6 × m = 9.1 ×
kg
=
K.E.gas molecule =
KT
iit je
e.
T= Absolute temperature K= Boltzmann’s constant
in
de Broglie wavelength of a gas molecule
gas molecule =
cr a
ck
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
thermal neutron
=
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Example: Find the ratio of deBroglie wavelength of proton and alpha particle which has been accelerated through same particle difference. Solution: We know that
= =
=
Alpha particle = doubly ionized helium particle
=2 =
=
=
Ans.
iit je
e.
in
=4
cr a
Solution:
ck
Example: Find the ratio of de Broglie wavelength of molecule of hydrogen and helium which are at 27°C and 127°C respectively.
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 =
Calculation of Radiation Pressure/Force
Particle nature of Photons
iit je
E = hf
e.
in
=
c = speed of light
I = Intensity of radiation
cr a
ck
Case (I) : Surface is perfectly reflecting
E = Energy received by surface per sec =IA
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.
F= 19
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E = P.c
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Also, Radiation Pressure (P) =
=
Case (II) : Surface is perfectly absorbing 
=
F= ∴
=
e.
Case (III) : Surface is partially reflecting
+
are in same direction,
cr a
and
Due to absorbed photons
ck
Due to reflected photons
iit je
Reflection coefficient = 0.7
=
in
F=
F = F1 + F2 = 0.7
+ 0.3
F=
20
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Radiation is falling obliquely E = (I) (A cos )
For Perfectly Surface ∆ pone photon = 2
cos θ
F =
in
For perfectly absorbing surface
iit je
e.
∆p=
F =
cr a
ck
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 the direction of radiation)
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Projected Area
A=
= RH
×
F=
Projected Area A= = 2 RH
iit je
e.
in
F=
cr a
ck
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 =
=
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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.
in
Remark:
iit je
e.
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.
cr a
ck
Solution: Let us consider an elementary ring as shown in figure.
r = Radius of elementary ring = R sin Δ A = Area of elementary ring = (2r) (R d) = 2R2 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 )
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Δ N = Number of photon received per sec
=
=
Δ F = Force experienced = ΔN ΔPone photon
=
=
iit je
e.
in
The direction of this force is radial as shown in figure.
∴ Force experienced by elementary ring is given by
ck
D =
∴
cr a
The components of ΔF in the plane of elementary ring will cancel out. cos
dF =
=
=
=
ΔA
2 R2 sin d
The force on the entire sphere is given by = or
F=
cos3 d =
Ans.
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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
iit je
cr a
ck
IS = Saturation Photocurrent
e.
in
Saturation photocurrent is directly proportional to intensity of radiation
0
Second Law
I (Intensity)
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
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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
ck
iit je
e.
in
Photoelectric Effect
cr a
The phenomenon of ejection of electrons, when metallic surfaces are exposed to certain energetic radiations, is called photoelectric effect.
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Basic Terms
Photo Current
Photo electrons
Work Function
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
e.
in
The electrons which
called work function.
cr a
ck
field.
iit je
application of electric
metal bondage is
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
Work function () =
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Work Function 1.9 eV 2.2 eV 2.3 eV 2.5 eV 3.2 eV 4.5 eV 5.6 eV
Metal Cesium Potassium Sodium Lithium Copper Silver Platinum
Table: Work functions for Some Photosensitive Metals
Work Function can be overcome by many ways: By application of heat.
Field Emission:
By application of high electric field.
Secondary Emission:
By impinging a beam of high speed electrons.
iit je
Intensity of Radiation
e.
in
Thermionic Emission:
Point Source
cr a
Case I :
ck
This is defined as energy flowing per second per unit area normally.
P
PS = Power of Source
I
=
For a point source
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I
I
0
in e.
iit je
cr a
ck
Case II : Linear Source
r
l
r
P
I=
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.
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Two Theories of Radiation
Wave theory
Photon Theory
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
E = hf
continuously.
in
E = energy of photon
e.
h = Planck’s constant
* Energy from source is not emitted continuously.
cr a
ck
iit je
f = frequency of radiation
Photoelectric effect supports photon theory of radiation. A photon is a chargeless and massless particle. A photon is bound to move with speed of light.
P
=
P = momentum of photon c = speed of light Whenever photon interacts with mater, it transfers energy and momentum.
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Photocurrent (I)
iit je
e.
in
Experimental Study of Photoelectric Effect
cr a
ck
IS
VS
VP
Voltage (V)
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.
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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 108 cm. Hence, electrons are emitted out not only from surface but also from subsurface layer also.
e.
in
cr a
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.
ck
iit je
Photoelectric Effect and Wave Theory
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.
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Graphical Variation 1. Maximum Kinetic Energy Vs Frequency hf = + (K.E.)max or (K.E.)max = hf 
K.E.MAX
Metal 2 Metal 1
in
Metal 3
fo1
fo2
fo3
cr a
ck
0
iit je
e.
Slope = h

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f
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2. Stopping Potential Vs Frequency eVs = hf 
h Vs f e e
VS
in
Metal 2
e.
Metal 1
Metal 3
fo1
fo2
ck
0
iit je
Slope = h/e
fo3
cr a
or
/e
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f
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3. Stopping Potential Vs
1
hc eVs hc 1 Vs e e
in
VS
e.
Metal 2
iit je
Metal 1
Metal 3
ck
Slope = h/e
0
cr a
or
fo1
fo2
fo3
/e
35
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f
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4. Maximum speed of photoelectron Vs frequency (K.E.)max = hf 
1 m2max = hf  2
f0
cr a
ck
0
iit je
e.
in
VS
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f
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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
hc
(K.E.)max = hf  =
hc K.E.max
Also, K.E.max = eVs = 0.36eV Ans.
(6.62x1034 )(3x108 ) 0.36x1.6x1019 10 5893x10 = 1.746 eV Ans.
iit je
2.794x1019 4.22x1014 hertz 34 h 6.62x10
ck
fo
e.
The threshold frequency is given by:
in
cr a
Example : A beam of light has three wavelengths 4144 Å, 4972 Å and 6261 Å with a total intensity of 3.6 × 103 Wm2 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 × 107 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 × 107 watt
37
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Energy incident on surface for each wavelength in two second E = (1.2 × 107) × (2) = 2.4 × 107 Joule Number of photons n1 due to wavelength 4144 Å = 4144 × 1010 m = 0.5 × 1012
n1 = Number of photons n2 due to wavelength 4972 Å
= 0.575 × 1012
n2 =
N = n1 + n2 = 0.5 × 1012 + 0.575 × 1012 Ans.
cr a
ck
iit je
e.
in
= 1.075 × 1012
38
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Atomic Excitation Ways of Atomic Excitation Two ways of Atomic Excitation
By Photon
By Collision
Absorption Whenever an atom is
in
An atom will not
excited by collision,
e.
absorb photon of any
For a photon to be
ck
absorbed by an atom, it
iit je
arbitrary energy.
inelastic. If loss of K.E. as permitted by conservation of linear
cr a
must have energy equal
collision must be
to difference of energy
momentum is less than
of ground state and any
the minimum
excited state
excitation energy, then collision will be elastic.
39
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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’
e.
mv2 
m(v/2)2 =
ck
=
iit je
Loss of K.E. =
in
v’ = v/2
= 50 % of initial kinetic energy KEmin = (10.2 eV) × 2 = 20.4 eV
Ans.
cr a
∴
Example: A neutron of kinetic energy 65 eV collides inelastically with a singly ionized helium atom at rest. It is scattered at an angle of 90 with respect to its original direction. (i) (ii)
Find the allowed values of the energy of neutron and that of the atom after the collision. If the atoms get deexcited 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 m2 cos m1 = 42 sin Eliminating ‘’, we get
40
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2 + 12 = 16 22 [sin2 + cos2 ] = 16 22 22 =
2 12 16
If ΔE represents the loss of energy in the collision, then COE gives
1 65 eV = 1 m 12 + (4m) 22 + ΔE 2
2
= 1 m 12 + 1 (4m) + 2 2 =
u2 12 16
+ ΔE
1 m 2 + 1 u2 + 1 m 2 + ΔE 1 1 2 8 8
e.
cr a
ck
iit je
= 5 m 12 + 1 (65 eV) + ΔE 4 8
in
= 1 m 12 + 1 1 mu2 + 1 m 12 + ΔE 2 4 2 8
or
5 3 65 4 eV = 8
We know that, En =
m 12 + ΔE
13.6Z 2 n2
Here,
Z=4
∴
E1 =  54.4 eV
E2 =  13.6 eV 41
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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 m 2 = 65 × 3  ΔE = 1 4 8
3 65 4 40.8 eV
1 m 2 + 4 × 7.95 eV = 4 × 1.59 eV = 6.36 eV 1 2 5
iit je
=
e.
∴ Kinetic energy of scattered neutron is
in
= (48.75 – 40.8) eV = 7.95 eV
Hence 1st allowed values of the energy of neutron
ck
after collision = 6.36
cr a
2 + 12 = 16 22 or
1 m 2 + 1 m 2 = 1 m × 16 2 2 1 2 2 2
or
1 m 2 + 1 m 2 = 4 × 1 (4m) 2 1 2 2 2 2
or
65 eV + 6.36 eV = 4 × K.E. of atom
∴ K.E. of atom =
71.36 eV = 17.84 eV 4
Ans.
Second Possibility: Let ΔE = 54.4 – 6.04 = 48.36 eV In this case,
5 m 2 = 65 × 3  ΔE = 1 4 8
3 eV 65 4 48.36
= (48.75 – 48.36) eV = 0.39 eV 42
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4 ∴ 1 m 12 = × 0.39 eV = 2.56 eV = 0.312 eV 2 5 5 Hence 2nd allowed value of the energy of neutron after collision = 0.312 eV 2 + 12 = 16 22 or
1 m 2 + 1 m 2 = 4 × 1 (4m) 2 1 2 2 2 2
or
65 eV + 0.312 eV = 4 × K.E. of atom
∴ K.E. of atom =
65.312 eV = 16.328 eV 4
Ans.
Proceeding as above, if ΔE = 54.4 – 3.4 = 51 eV
in
Kinetic energy of neutron is
iit je
e.
1 m 2 = 4 × (48.75  51) =  4 × 2.25 eV 1 2 5 5 =  4 × 0.45 eV =  1.80 eV
ck
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 Heatom are 17.84 eV and 16.328 eV.
cr a
(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 = Z2 R 1 1 = 4 × 1.097 × 8 × 107 12 32 9 1 f1 =
c = (1.097 × 107) × 4 × 8 × (3 × 108) 1 9 = 11.7 × 1015 Hz
(b)
Ans.
1 = Z2 R 1 1 = 4 × 1.097 × 107 × 5 22 32 2 36 f2 =
c = (1.097 × 107) × 4 × 5 × (3 × 108) 2 36 = 1.827 × 1015 Hz
Ans. 43
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(c) 1 = Z2 R 1 1 = 4 × 1.097 × 107 × 3 12 22 4 3 = (1.097 × 107) × 4 ×
Ans.
iit je
e.
in
= 9.85 × 1015 Hz
3 × (3 × 108) 4
ck
c 3
cr a
f3 =
44
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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)
iit je
e.
in
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.
ck
e
e
e
cr a
e
Positively Charged Matter
e
e
e
e
e e
e
Electron s
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.
45
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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 .
in
Ni = Total number of alpha particles that reach the screen
iit je
Z = Atomic number of the foil atoms
e.
n = Number of atoms per unit volume in the foil
r = Distance of the screen from the foil
t = Foil thickness.
ck
KE = Kinetic energy of the alpha particles
cr a
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. 46
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N
Scattering angle (θ) = 180°
r
e.
in
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
iit je
= Constant
cr a
ck
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. 106 m. When passing through the metal foil, the  particles get scattered through different angles. 47
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(iii)
(iv)
iit je
(ii)
The fact that most of the  particles passed undeviated led to the conclusion that an atom has a lot of empty space in it.  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. 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. The difference in deflection of various particles can be explained as follows:
ck
(i)
cr a
Conclusion
e.
in
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.
 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.
48
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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.
(iv)
(v)
e.
iit je
(iii)
ck
(ii)
An atom consists of equal amounts of positive and negative charge so that atom, as a whole, is electrically neutral. 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. The negative charge, which is contained in the atom of electrons, is distributed all around the nucleus, but separated from it. 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. The nuclear diameter is of the order of 1014 m.
cr a
(i)
in
Rutherford’s Atomic Model:
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
=
49
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In one of the experiments,  particles of velocity 2 × 107 ms1 was bomdarded upon gold foil. Here,
Z = 79 e = 1.59 × 1019 C m = 4 × 1.67 × 1027 kg v = 2 × 107 ms1 r0 = 4 × 9 × 109 × = 2.69 × 1014 m
This gives the radius of the nucleus.
e.
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
iit je
(i)
in
Failure of Rutherford’s Atomic Model
centripetal acceleration
. So, it must radiate energy continuously. As a
cr a
ck
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}
(ii)
Thus, the whole atomic structure should collapse. This is contrary to the actual fact that atom is very stable. 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.
50
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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
=
∴
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. Law of quantization of angular momentum: When electron revolves around nucleus in stationary orbit, its angular momentum is quantized and is given by
(v)
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. 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.
cr a
(iv)
ck
mvr =
iit je
(iii)
e.
in
(ii)
51
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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 ∴
mv2 
En =
We can write
En = 
e.
∴
in
mv2 =
ck
r=
iit je
Now, we get
This is the equation for the radius of the permitted orbits.
cr a
Now, we get En =
On substituting the value of different constants, we get En =
eV
52
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Special Points (i) a) b) (ii)
In the derivation of above expression for the energy of the electron in the nth orbit, following assumptions are taken: Gravitational potential energy has been neglected and for writing electrostatic potential energy, infinity is taken as zero potential position. Nucleus is assumed to be infinitely massive i.e. nucleus is at rest. The quantity
is known as Rydberg constant ‘R’ and its value is
1.09 × 107 m1. Rhc = 13.6 eV
Also,
=  KE
(iii)
Total energy, (TE) =
(iv)
For Hydrogen atom, Z = 1 ∴ En =
cr a
ck
iit je
e.
in
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
53
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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
in
3. Ionization energy (IE): This is defined as the amount of energy required to ionize an atom.
iit je
e.
Energy required to make electron move from ground state to infinity. IE = E E1
ck
For hydrogen atom IE = 0  ( 13.6 eV)
cr a
= 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
54
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Hydrogen Spectrum Hydrogen Spectrum
Emission spectrum
Absorption spectrum
It has many series. N = Number of spectral lines = nC2
in
It has only Lyman series.
ck
iit je
e.
=
n = principal quantum of excited state let n = 3 (2nd excited state) N = 3 C2 = 3
cr a
3
1
3
En = E2  E1 = hf =
Excited state
Ground state
=R
R = Rydberg constant Lyman Series: n1 = 1 55
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2
2
1
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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
e.
(ii)
For any series, the wavelength of first member of series is longest and that of limiting member is shortest. Lyman series lies in ultraviolet region. Balmar series lies in visible region.
in
(i)
ck
Absorption spectrum of Hatom consists of Lyman series only.
cr a
(iii)
iit je
Bracket series Paschen series lies in infrared region. Pfund series
Properties of Electron in nth Orbit (i)
Energy of electron in nth orbit
En =

En me (ii)
Bohr Radius
rn =
Å
rn
56
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(iii)
Velocity of electron in nth orbit
Vn = 2.18 × 106
m/s
Vn is independent of mass of electron (iv)
Current (In) In = e fn fn =
In (v)
Time Period (Tn)
e.
iit je
Tn
in
Tn =
fn =
ck
Magnetic Dipole Moment (Mn) Mn = In An = (e fn) ( rn2)
cr a
(vi)
Mn
=
The value of magnetic moment in first Bohr orbit is called Bohr magneton (B). Its value is given by = 9.27 × 1024 A m2
B =
Mn
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 = 57
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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
cr a
ck
iit je
e.
in
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.
58
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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.
cr a
ck
iit je
e.
in
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 = EB = E 2 = 
Z2 Z2 =  3.4 Z2
59
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Z2 =  0.852 Z2 – ( 3.4 Z2) = 2.55 Z2
EC = E 4 = 
and
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
and
Emin = E4 – E3 = 15.1 – 0.85
Ans.
in
= 0.66 eV Ans.
e.
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.
cr a
ck
iit je
Solution: Consider the situation as shown in figure
mv = mv’ + or
m(v – v’) =
………………… (1)
mv2 + En =
mv’ 2 + Em + hf
(v2 – v’ 2)+ En – Em = hf …………. (2) or
(v + v’) (v – v’) + hf0 = hf
[∵ En – Em = hf0]
Using equation (1), we get
60
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(2v)
+ hf0 = hf
f = f0
or As
< 1, expanding binomially and neglecting higher powers, we get Ans.
f = f0
Example: Suppose the potential energy between electron and proton at a distance r is . Use Bohr’s theory to obtain energy levels of such a hypothetical atom.
It is given that
U=F=
Hence,

=
iit je

in
=F
Solution: We know that
e.
given by

=
……………… (1)
=
cr a
∴
mvr =
Also, Hence, v =
ck
According to Bohr’s theory, this force provides the necessary centripetal force for orbital motion.
………………………. (2)
…………………………. (3)
Substituting this value in equation (1), we get =
or
r=
Substituting this value of r in equation (2), we get v=
61
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Total Energy (E) = K.E. + P.E. =
mv2 
=
E=
k=
Ans.
ck
iit je
e.
in
,
cr a
or

62
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Nuclear Motion
r = rN + re From the property of mass
ck
mN rN = me re
iit je
e.
in
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.
cr a
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 =
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where me = mass of electron mN = mass of nucleus Now for Hatom, 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.
in
For hydrogen atom with stationary nucleus, Rydberg constant is given by
e.
R=
iit je
Hence, Rydberg constant considering motion of nucleus is given as
R’ =
=
0.055 %
cr a
=
ck
Therefore, relative difference in two values of Rydberg constant is given as Ans.
I = Ie + IN
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 = 64
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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 =
in
is less than one, hence it is called reduced mass.
cr a
ck
iit je
e.
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 =
65
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Special Points (i) (ii)
The concept of nuclear motion led to the discovery of deuterium isotopes. 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
cr a
ck
iit je
e.
in
In all situation of motion of nucleus, replace mass of electron by its reduced mass.
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