Revised Copy of Phy Pract Sem II

8/14/2019 Revised Copy of Phy Pract Sem II

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- 1 - A Journal of Applied Physic, SCOE, Pune 41.

Division: ___Roll No.: _______Date of Performance: __________------------------------------------------------------------------------------------------------

Expt. No.: 01: Energy Gapof a Semiconductor.

Aim: To determine the Forbidden Energy Gap of a semiconductor.

Apparatus: P-N junction diode, sand or oil bath, thermometer (0-1000C), 0-50 micro-ammeter, D.C. 1.5 V supply, heating coil, etc.

Circuit Diagram:

Formula: Eg = 2. k.F

Where, k = Boltzmans constant = 1.37 v 10-23 JK-1F = Slope of the graph ofln RT Vs 1/T.Eg = Energy gap.

Procedure:

1. Measure voltage of given cell or battery.2. Connect the circuit as shown in the diagram and get it

checked.

3. Start heating the sand or oil bath by using heater or burner.4. Measure current at 350C and then in steps of 50C each, up to

800C. OR

Heat the diode up to 800C, turn off the heater and measurethe current at 800C and in steps of 50C each, as diode coolsdown to 350C.

AQ

Thermometer

Beaker

Test TubeWater

Diode

Oil


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5. Calculate RT, ln RTand 1/T.6. Plot the graph ofln RTVs 1/T and find the slope of the graph

(F).

7. Substitute F in the given formula to find Eg in Joules andthen in eV.

8. Compare it with standard value of Eg for given diode.9. Also plot graph of R Vs T to verify that R falls non-linearly

with increase in temperature.

Observation Table:

D.C. Voltage (V) =________________Volt

Sr.No.

Temp.(t)

0C

I

(QA)

T=(t+273)

0K

RT =I

V

;

1/T

0K-1

ln RT

1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.


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Nature of Graph:

Calculations:

We have, Eg = 2. k.F (Slope of the graph ln RTVs 1/T)

= 2 v 1.37 v 10-23 v _______________Joules.

=______________________________ Joules.

= 19106.1 v eV.

=______________ eV.

Result:

The energy gap of given semiconductor diode is _____________ eV.

Signature of the Teacherwith date

Marks Total

ln RT

1/T

R

T


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Conduction band Conduction bandConduction band

Valance band Valance band Valance band

Eg = 0 eV Eg = 1.1 eVEg = 5 eV

For Conductors For Semiconductors For insulators

Theory:

Solids are classified according to energy band structure. Itconsists of two bands, namely valance band and conduction bandseparated by a gap known as forbidden energy gap (Eg). The band,which is occupied by valance electrons and has highest occupied band

energy at absolute zero temperature is called valance band. It may bepartially filled up depending upon the nature of the solid.

The lowest unfilled energy band at absolute zero temperature iscalled conduction band. The electrons in valance band can betransferred to conduction band by providing them the energy equal toEg.

On the basis of forbidden energy gap, solids are classified into

three groups, namely Conductors, Semiconductors, Insulators.Eg = 0 eV in conductors,0 00 K.

!

kT

EgT

2

exp0VV

where, VT is resistivity at temperature T, V0 is resistivity at absolutezero, Eg is forbidden energy gap of the semiconductor, and k isBoltzmanns constant.

Since resistance of given specimen is proportional to resistivity,we can write,


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!

kT

ERR

g

T2

exp0

Taking logarithm on both sides, we get,

kT

ERR

g

T

2lnln 0 !

This equation has the form cmxy ! , whereT

Ry ln! andT

x1

! .

Thus, the graph ofT

Rln VsT

1, is a straight line with slope equal to

k

Eg

2

and intercept equal to 0lnR . In this experiment, the above

considerations have been used to estimate the forbidden energy gap of

the given semiconductor material.

Questions:

1. Define (i) Valance band (ii) Conduction band (iii) energy gap2. How does the conductivity of a semi conducting material

depend on temperature?3. Define Fermi level and Fermi Dirac probability distribution

function4. In this experiment why is the diode connected in the

reverse bias?

5. Define specific resistance of a material.Answers:

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Division: ___Roll No.: _______Date of Performance: __________------------------------------------------------------------------------------------------------

Expt. No.: 02: Solar Cell.

Aim: To study the I-V characteristics of the given solar cell to

calculate the fill factor and value of resistance for maximum value

of workable power.

Apparatus: Solar cell, light source, variable load, mili-voltmeter, micro

ammeter, etc.

Circuit Diagram:

Formula:

Fill factor,

ocsc

VI

VIFF

v

v!

Where,

mI = Current corresponding to maximum power.

mV = Voltage corresponding to maximum power.

scI = Short circuit current. (Current recorded when load is zero)

ocV = Open circuit voltage. (Voltage recorded when load is max)

SolarCell

A

mV

Li ht

RL

+ -

-

+


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Procedure:

1. Connect the circuit, as shown in the circuit diagram and getit checked.

2. Start increasing the load (RL) and at selected voltage stepsmeasure current. Take maximum possible readings for graph.

Take the readings up to maximum load and record in the

observation table.

3. Calculate the values of power and resistance4. Plot the graph of, I Vs V. Draw a smooth curve from the

points plotted. Extend the curve upto x and y axes as per the

nature of the curve. We get the points Isc and Voc.

5. Draw perpendicular from Isc and Voc, we get a point (Isc,Voc).Draw a line joining origin (0,0) and the point (Isc,Voc). This line

cuts the graph at point (Im,Vm). Draw perpendiculars from

point (Im,Vm) on x and y axes. WE get the points Im and Vm.

6. Record Im, Vm, Isc, and Voc from the graph in the observations.7. Plot the graph ofP Vs R and record the value of resistance for

maximum workable power.

Graphs: 1. Graph of I Vs V

(Isc,Voc)

Im

Isc

Vm Voc(0,0)

V

I


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2. Graph of P Vs R

Observation Table:

Obs.No. Voltage(V)

Volt

Current(I

)

Ampere

Power(P)= VI

Watt

Resistance(R) =I

V

Ohm()

01

02

03

04

05

0607

08

09

10

11

12

13

14

15

16

17

18

19

20

P

RR at max workable P


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Observations from graph:

1. Open circuit voltage (Voc) =____________V.2. Short circuit current (Isc) =_____________A.3. Voltage corresponding to maximum power (Vm) =___________V.4. Current corresponding to maximum power (Im) =____________A.

Calculations:

2. Fill Factor,ocsc

VI

VIFF

v

v!

!@FF

!@FF ___________________

Results:

1. Fill factor of the solar cell =____________2. Resistance at maximum workable power (from graph) = _______.


Marks Total


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Theory:

The direct conversion of solar energy by means of thephotovoltaic effect, that is, the conversion of light (or otherelectromagnetic radiation) into electricity. The photovoltaic effect isdefined as the generation of an electromotive force as a result of the

absorption of ionizing radiation (later explained in detail). Energyconversion devices which are used to convert sunlight to electricity bythe use of the photovoltaic effectare called solar cells. A single

converter cell is called a solar cell or, more generally, a photovoltaiccell, and combination of such cells; designed to increase the electricpower output is called solar moduleor solar array.

The photovoltaic effect can be observed in nature in a variety of

materials, but the materials that have shown the best performance insunlight are the semiconductors. When photons from the sun areabsorbed in a semiconductor, they create free electrons with higherenergies than the electrons which provide the bonding in the basecrystal. Once these electrons are created, there must be an electricfield to induce these higher energy electrons to flow out of thesemiconductor to do useful work. The electric field in most solar cells

is provided by a junction of materials which have different electricalproperties.

To obtain a useful power output from photon interaction in asemi-conductor three processes are required.1. The photons have to be absorbed in the active part of the material

and result in electrons being excited to a higher energy potential.2. The electron hole charge carriers created by the absorption must be

physically separated and moved to the edge of the cell.3. The charge carriers must be removed from the cell and delivered to

a useful load before they loose their extra potential.For completing the above processes, a solar cell consists of:

(a) Semi-conductor in which electron hole pairs are created byabsorption of incident solar radiation.

(b) Region containing a drift field for charge separation, and(c) Charge collecting front and back electrodes.

The photo-voltaic effect can be described easily for p-n junction in

semi-conductor. In this junction after the photons are absorbed, thefree electrons of the n-side will tend to flow to the p-side, and the holesof p-side will tend to flow to the n-region to compensate for theirrespective deficiencies. This diffusion will create an electric field EFfrom the n-region to the p-region. This field will increase until itreaches equilibrium for Ve, the sum of the diffusion potentials for holesand electrons.


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If electrical contacts are made with the two semiconductormaterials and the contacts are connected through an external

electrical conductor, the free electron will flow from the n-type materialthrough the conductor to the p-type material as shown in the figure.Here the free electrons will enter the holes and become boundelectrons; thus both free electrons and holes will be removed. The flowof electrons through the external conductor constitutes an electriccurrent which will continue as long as more free electrons and holesare being formed by the solar radiation. This is the basis ofphotovoltaic conversion, that is, the conversion of solar energy intoelectrical energy. The combination ofn-type andp-type semiconductors

thus constitutes a photovoltaic (PV) cell or solar cell. All such cells

generate direct current which can be converted into alternatingcurrent if desired.

Questions:

1. What is solar cell?2. What is photovoltaic effect?3. What is short circuit current?4. What is open circuit voltage?5. What are merits and demerits of solar cell?

Answers:

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P-REGIONBASE MATERIAL

N- REGION

LOAD

DIFFUSED LAYERMETAL CONDUCTOR

NGATIVE CONTACTS

SUN LIGHTCURRENT COLLECTION

GRID (METAL FINGERS

0.2 m

300 m


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Division: _____Roll No.: _________Date of Performance: __________------------------------------------------------------------------------------------------------

Expt. No.: 03: Characteristics of a Photocell.

Aim: To study I-D and I-V characteristics of a given photocell.

Apparatus: Photocell, power supply, micro ammeter, voltmeter, lamp,meter-scale, etc.

Circuit Diagram:

1. For I-D Characteristics:

Procedure:

A

V BRhPhoto cell

Light

Graph of I Vs 1/D2

I V1

V2

1/D2


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D1

D2

1. Keep certain constant voltage (V1) across the photocell.2. Now by varying the distance between photocell and the lamp,

note down the photocurrent at different positions of the lamp.

3. Record observations for another constant voltage (V2).4.Plot the graph of 21

DVsI for both V1and V2.

Observation Table:

ObsNo.

V1 = _________Volts V2 = _________Volts

D(cm)

1/D2(cm-2)

I

(QA)

Dcm)

1/D2(cm-2)

I

(QA)

1.2.3.4.5.6.7.8.9.10.

2. For I-V Characteristics:

Procedure:

Graph of I Vs V

I

I

V


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1. Keep certain constant distance (D1) between lamp and photocell.2. Vary voltage across photocell using the rheostat.3. Note down the photocurrent in micro ammeter at every step.4. Repeat the procedure for another distance (D2).5. Plot the graph of I Vs V for both D1 and D2.

Observation Table:

Obs

No.

D1 = _______cm. D2 = _______cm.

V(Volt)

I

(QA)

V(Volt)

I

(QA)

1.2.3.4.5.6.7.8.9.10.11.12.

Result:

1. __________________________________________________________________________________________________________________________

2. __________________________________________________________________________________________________________________________


Marks Total


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I

Vo V

Intensity 1

Intensity 2

Theory:

Photo diode works on the principle of photoelectric effect. Whenlight is made incident on the surface of certain metals, electrons are

emitted by the surface. This is called photoelectric effect. A photo diodeconsists of an evacuated tube with two electrodes.

When light is incident on plate E, electrons are emitted from its

surface. The positive potential at cathode C accelerates these electronstowards C and current flows through the external circuit. It is observedthat current flows even when the applied voltage is zero and a negativepotential V0 has to be applied to reduce the current to zero. V0 iscalled the stopping potential. As the intensity of incident radiationincreases, the current increases even for constant voltage however Voremains the same. These observations cannot be explained using the

classical concepts of light as an electromagnetic wave. Einstein gavethe explanation of photoelectric effect based on the concept of photonand thereby established the concept of wave-particle duality of

radiation.

According to Einsteins explanation, the electron is ejected out fromthe metal surface as a result of the energy transfer between photonand electron in an elastic collision. If the energy of the incident photon

(E = hR, by Plancks theory) exceeds the binding energy of the valanceelectron, the electron is knocked out. If the intensity of light isincreased, the number of incident photons increases, leading to more

collisions and emission of more electrons. However, the energy of theelectrons remains the same. Therefore current increases with intensitybut V0 remain constant. Increasing the voltage does not lead to an

increase in current, because the current is dependent on the numberof incident photons, rather than the K.E. of the electrons.


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Questions:

1. What is photoelectric effect?2. On which factors does the photocurrent depend?3. What is stopping potential?4. Why does the current go to saturation after threshold

voltage if the intensity of incident light is constant?5. Explain the working of photocell in short.

Answers:

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Mili-voltmeter

Constant Current

Power Supply


Expt. No.: 04: Hall-effect.

Aim:To study the Hall-effect in semiconductor to determine,

i) Whether the semiconductor is P-type or N-type.ii) Hall coefficient of a semiconductor.iii) Concentration of majority carriers.iv) Mobility of majority carriers.

Apparatus: Hall-effect setup, Hall probe, electromagnet, constant

current power supply, digital gauss meter etc.

Diagram:

I E

B

N S

Formula:

1.Hall coefficient =BI

tVR HH v

v!

2.The semiconductor type can be identified from the polarity ofHall Voltage.

3.The concentration of majority charge carriers,eR

noreR

pHH

11 !!

4. Mobility of charge carriers = HRWQ ! Where, VH= Hall voltage.

t = Thickness of probe (semiconductor wafer).I = Current flowing through the specimen.

e = Charge of electron. = Conductivity of material.


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Procedure:

1. Connect the width-wire contacts of Hall probe to theterminals marked voltage.

2. Without connecting the current terminals, the setup isswitched on and zero adjustment is made. The unit is then

switched off.

3. The length-wire contacts of Hall probe are connected to thecurrent terminals of the setup.

4. Now the Hall probe is placed in the air gap of theelectromagnet. Power supply for the electromagnet is

switched on and magnetic field is adjusted to the suitable

value. Rotate the Hall probe till it is perpendicular to

magnetic field. Hall voltage will be maximum for this

position.

5. The current through the semiconductor is adjusted to asuitable value. Hall voltage and polarity are noted down.

6. Measure the Hall voltage as a function of current keeping themagnetic field as constant. Take ten observations.

7. Plot the graph of VH against I at constant magnetic field.Calculate slope of the graph.

Observations:

1. Thickness of semiconductor wafer = t = ________ mm.= ________ m.

2. Resistivity = V = ______________ ohm m.


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Observation Table:

Magnetic field = B = _____________ k gauss = _____________ Wb/m2

Sr.

No.

Current (I)

mA

Hall Voltage (VH)

mV

1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.

Graph 1:

VH

I


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Calculations:

1. Calculation of Hall coefficient:1Graphofslope

B

t

I

V

B

tR

H

Hv!

!

=

= _______________m3/C

2. Calculation of concentration of charge carriers:

HRe

pRn.

1! =

= _______________ / m3

3. Calculation of mobility of charge carriers:

!!!HH

RRV

WQ1

.

=______________m2/V.sec.


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Results:

1. The polarity of Hall voltage is ______________. Therefore thegiven semiconductor is of ______ type.

2. The value of Hall coefficient is _______________m3/C.3. The concentration of majority carriers is found to be

_____________ m2/V.sec.

4. The mobility of charge carriers is found to be__________________ m2/V.sec.


Precautions:

1. The probe is properly centered and oriented in magnetic fieldsuch that maximum value of Hall voltage is generated.

2. The potentiometer control of the electromagnet power supplyis kept at its minimum position while switching on and off the

power supply.

3. The potentiometer control of the current flowing through theprobe is brought to zero position before switching on the

current source.

4. The magnetic field is varied gradually in steps to avoid damageto the electromagnetic coils.

Marks Total


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Theory:

Hall observed that, when a strip of metal carrying current wasplaced in a transverse magnetic field, a potential difference wasdeveloped across the strip perpendicular to both direction of currentand the magnetic field.

It was later shown that, semiconductors also exhibits Hall effect,and the direction of electric field developed because of inducedpotential depends on whether the current is due to electrons or holes.

In order to understand Hall Effect consider a rectangular strip ofn-type semi-conductor material. Under the application of somepotential difference, let current I flow through it along positive X-direction. Let the magnetic field B act along the positive Z-direction as

shown in figure.

Y

Side 2

d

w

I XSide 1

B

Z

As the magnetic field is applied, the magnetic force acts alongnegative Y- direction. As the semiconductor is n-type, the charge

carriers are electrons.The magnitude of force is given by,

veBF ..1 ! ---------------------- (1)

Where, !v drift velocity.

e= charge on electron.

Thus due to this force the electrons are forced to move innegative Y-direction and soon obstructed by the walls of specimen andget accumulated there. Because of charge accumulation towards side1, it becomes negatively charged w. r. t. side 2. The electric field isdeveloped between side 1 and side 2 directed along negative Y-direction. This electric field opposes further movement of electronstowards side 1. In equilibrium condition the force due to electric field

is balanced to the force due to magnetic field and the current flows inthe positive X-direction only. At this stage a steady potential differenceis produced between side 1 and side 2 called as Hall Voltage, VH.


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Let E be the electric field intensity due to Hall Voltage then,

get,we(2)and(1)equationsFrom

)2(2

! eEF

.2&1tan,)4(,

)3(.

sidesidebetweencedisdwheredVEBut

vBE

BeveE

H !!

!@

!

.sec

)7(..

.

1

..

..

,)5(

)6(..

...

,

)5(..

.

,)4()3(

striptheofareationcrossA

ionconcentratcarriernwhere

A

dIB

enAen

dIBV

becomesEquation

AenIv

AvenI

thatknowWe

dvBV

orvBd

V

getweandequationsFrom

H

H

H

!

!

!!

@

!@

!

!

!

wA

d

wdmaterialtheofareationcrossA

andvolumeunitperechordensityechenwhere

dJBenA

dIB

enV

densitycurrentJA

I

Now

H

1

.sec

.argarg.,

)8(...

1..

.

1

,

!@

!!

!

!!@

!!

)9(..1

,)7(

!

@

w

IB

enV

beco es

quation

H

Here, VH is positive at side 2 w. r. t. side 1 as we considered the

conduction is due to electrons. (n-type semiconductor). If the side 2 isnegative w. r. t. side 1 then the charge carriers are holes i.e. thesemiconductor isp-type.

Thus measurement of Hall Voltage helps us to identify whetherthe given semiconductor isp-type or n-type.


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In equation (9) B, I, w, and VH can be measured and then thecharge density n.e can be determined and from that the carrierconcentration (n) can be determined.

In equation (9) the quantityen.

1is the property of the material of

the specimen and is called as Hall coefficient RH

)12(.

1.

.,

.

..

)3(.,

)11(.

.

.

,)8(

)10(.

1

!@

!@

!!

!@

!

!@

!

@

!@

JB

R

JI

wd

wd

I

A

IJalso

IB

wdER

equationfrodEVhere

IB

wVR

w

IBRV

beco esequationen

R

H

H

H

H

H

HH

H

Thus, Hall coefficient RH of a semiconductor material may bedefined as electric field per unit surface current density due to unittransverse magnetic field.

)13(.

.1

.

1..

,

!!@

!

!@!

WQ

QW

QW

H

H

H

RMobility

R

enRasen

thatknowWe

Thus measurement of conductivity and Hall coefficientdetermines mobility of charge carriers.

Thus Hall Effect can be used for,

y Determination of type of semiconductory Determination of carrier concentration.y Determination of mobility of carriers.


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Questions:

1. What is Hall Effect?2. What is Hall probe?3. Does strength of the magnetic field produced depend upon the

distance between the pole pieces?4. What is Hall field and Hall voltage? Why are they produced in

the material?

5. Why does Hall Effect assume importance in semiconductors?Answers:

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Expt. No.: 05: He-Ne Laser.

Aim:To determine the diameter of a given wire by using the diffraction

pattern of He-Ne Laser.

Apparatus: He-Ne Laser, Thin wire, Screen, Scale etc.

Diagram:

Formula:

The diameter of a given wire can be found by using the formula,

-

! ''

mmxx

mm

fd P

Where,d =diameter of a given wire.

= wavelength of He-Ne Laser.f= distance between the source and the screen.

', mm = orders of diffraction.

',mm

xx = distance of orders from center.

He-Ne laser gun

WireScreen

Diffraction attern on the Screen

f


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Procedure:

8. Fix the wire, whose diameter is to be determined in the pathof a laser beam and obtain the diffraction pattern on the

screen in front of the laser gun.

9. Measure the distance f between the source (laser gun) andthe screen.

10. Measure the distance of secondary maximum of differentorders from the center on either side.

11. Find the mean distance for each order.12. Find the value of )'(

mmxx by using half table method and

then find mean value of )'(mm

xx .

Observations:

1. Wavelength of He-Ne Laser == ________________cm.2. Distance between the source and the screen =f=________cm

Observation Table:

Obs.No.

Orderm

Distance of orderfrom center

xm (cm.)

Meandistancexm (cm.)

'mm

xx (cm.)

By half table

method

Mean

'mm

xx (cm.)Right Left

1.2.3.4.5.6.7.8.9.10.11.12.


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Calculations:

-

!

'

'

mmxx

mm

fd P

=

=______________________cm.

Result:

The diameter of given wire is _________________cm.


Theory:

LASER stands for Light Amplification by Stimulated Emissionof Radiation. Laser is a quantum electronic device which produces

very intense, unidirectional, monochromatic and coherent beam oflight. Laser beam of visible light is highly parallel. For this reason, itcan travel many miles without appreciable divergence.

Laser action:

When an atom absorbs energy it goes to higher energy states andit emits light during its transition from this excited higher energy state

to the lower energy state.Consider two energy levels E1 and E2 ground and excited statesrespectively in an atom as shown in figure (a). If the atom is in lowerstate E1 and when a photon of energy (E2-E1) strikes the atom, it goesin the excited state E2 by absorbing the energy of photon.

Marks Total


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The atom can stay in the ground state for unlimited time. On theother hand atom can remain in a excited state for limited time, knownas Life Time. Life time of excited H2 atom is of the order of 10-8

seconds. However some of the excited states have longer life time of theorder of 10-3 seconds. These states are called as Meta-stable States.

These states play very important role in Laser operation.Now in figure (b) the atom is in excited state E2 and there is no

radiation nearby.

After the life time of this state if the transition of an atom is

allowed on its own accord spontaneously from level E2 to level E1. Itemits a photon of frequency E2-E1/h. This is called as SpontaneousEmission. Spontaneous emission depends on the type of the atom andtype of transition but is independent of outside circumstances. It is

random in character. The radiation is random mixture of quantahaving various wavelengths. The waves coincide nether neither inwavelength nor in phase. Thus radiation is in-coherent and has broadspectrum.

In addition to spontaneous emission there is another possibilityas shown in figure (c).

In excited state spontaneous transition spontaneous emissionFig. (b)

Photon

before transition during transition after transitionFig. (a)

E1

Photon

E2

In excited state stimulated transition stimulated emissionFig. (c)

TwoPhotons

Photon


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Figure shows the atom again in excited state E2. If a photoncomes along of exactly the frequency (E2 - E1)/h, it can force the atomto fall to the lower level and emit a photon of the same frequency. Thismeans that under the effect of another external photon an atom canpass from an excited state to lower state emitting photon not onlyspontaneously but also forced to it. When the atoms are in the excitedenergy level, emission is stimulated before spontaneous emission

occurs. The absorption of incident photon causes two photons to beemitted, one by spontaneous and other by stimulated emission. Boththe photons travel in the same direction have the same frequency andphase i.e. they are coherent. The emission of two photons with aninput of only one photon implies amplification.

The occurrence of spontaneous emission is directly proportionalto the number of atoms in the specified energy level, whereas instimulated emission the rate of occurrence is proportional not only tothe number of atoms in the excited state but also number of incidentphotons.

Population Inversion:

The process of getting large number of atoms in excited statethan lower level is called as population inversion. If large number of

atoms can be excited to upper energy levels then the probability ofstimulated emission i.e. light amplification increases. The system inwhich population of higher energy state is more than population oflower energy state is called as Negative Temperature State. Negativetemperature state is not a physical quantity; it just indicates that the

system is non-equilibrium.When the atomic system interacts with photon it undergoes with

the process of stimulated absorption, spontaneous emission as well asstimulated emission.

If, N1 = Number of atoms in lower energy state E1.N2 = Number of atoms in higher energy state E2.

Normally,N1 > N2

If the photons of energy 12 EEh !R are incident on the atom,

photons get absorbed and some of the atoms get excited to state E2.

This process of stimulated absorption depopulates level E1.The rate at which process occurs is expressed as,R12 = Pa.N1 -------- (1)

Where Pa = probability of stimulated absorption.Similarly, the stimulated emission depopulates energy level E2

resulting in the emission of photons.The rate at which process occurs is,R21 = Pe.N2 -------- (2)

Where Pe = probability of stimulated emission.At equilibrium, Pa = Pe.


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Then comparing two rates in equation (1) and (2), it is observedthat more energy is absorbed than emitted.

i.e. Pa.N1 > Pe.N2 21 NN "3

To produce more emission it is essential to have N2 > N1. This iscalled as Population Inversion. The system in which populationinversion is achieved is called as active system. The method of risingatoms from lower energy level to higher energy level is called as Optical

Pumping.

Optical Pumping:

To achieve steady state population inversion a three level systemis used as shown in figure below.

The three level systems consist of energy levels E1, E2 and E3. Let

N1, N2 and N3 be the population densities of these levels respectively.

In equilibrium,N1 > N3

The pump lifts atoms from level E1 to level E3 from which theydecay rapidly to level E2 through some non-radiation process. If thisnon-radiation process is very fast between the levels E3 and E2, theelectrons from level E3 get transferred to level E2 rather than return tolevel E1. Level E2 is a meta-stable state with a longer life time. As aresult the rate at which the atoms are pumped to level E3 exceeds. Thesituation is raised where the number of atoms reaching and remainingin level E2 exceeds than the number of atoms in level E1. Thus

population inversion is achieved. When electrons make transition fromE2 to E1 radiations of frequency (E2-E1)/h are given out.

Pumping

Rh

E3

E2

E1

N3

N2

N1


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Questions:

1. What are the properties of laser?2. Explain spontaneous emission and stimulated emission.3. Define optical pumping and active system.4. What is population inversion?5. Why laser beam is used in this experiment?

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Documents

Revised Copy of Phy Pract Sem II