Engg Physics

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Biyani's Think Tank

Concept based notes

Engg. Physics

(B.Tech)

Shamendra Sharma

Asst. Professor Deptt. of Engineering

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Think Tanks

Biyani Group of Colleges

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While every effort is taken to avoid errors or omissions in this Publication, any mistake or omission that may have crept in is not intentional. It may be taken note of that neither the publisher nor the author will be responsible for any damage or loss of any kind arising to anyone in any manner on account of such errors and omissions.

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Preface

I

am glad to present this book, especially designed to serve the needs of the students. The book has been written keeping in mind the general weakness in understanding the fundamental concepts of the topics. The book is self-explanatory and adopts the “Teach Yourself” style. It is based on question-answer pattern. The language of book is quite easy and understandable based on scientific approach.

Any further improvement in the contents of the book by making corrections, omission and inclusion is keen to be achieved based on suggestions from the readers for which the author shall be obliged.

I acknowledge special thanks to Mr. Rajeev Biyani, Chairman & Dr. Sanjay Biyani, Director (Acad.) Biyani Group of Colleges, who are the backbones and main concept provider and also have been constant source of motivation throughout this Endeavour. They played an active role in coordinating the various stages of this Endeavour and spearheaded the publishing work.

I look forward to receiving valuable suggestions from professors of various educational institutions, other faculty members and students for improvement of the quality of the book. The reader may feel free to send in their comments and suggestions to the under mentioned address.

Note: A feedback form is enclosed along with think tank. Kindly fill the feedback form and submit it at the time of submitting to books of library, else NOC from Library will not be given.

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Syllabus

Unit 1

Interference of light

Michelson’s Interferometer: Production of circular & straight line fringes,

Determination ofwavelength of light,Determination of wavelength separation of two nearby wavelengths.Newton’s rings and measurement of wavelength of

light.Optical technology: Elementary idea of anti-reflection coating and interference filters.

Unit 2

Polarization of light

Plane circular and elliptically polarized light on the basis of electric (light) vector, Malus law. Double Refraction: Qualitative description of double refraction phase retardation plates, quarter and half wave plates, construction, working and use of these in production and detection of circularly and elliptically polarized light. Optical Activity: Optical activity and laws of optical rotation, Specific rotation and its measurement using half -shade and bi- quartz devices.

Unit 3

Diffraction of light

Single slit diffraction: Quantitative description of single slit, position of maxima / minima and width of central maximum, intensity variation. Diffraction Grating: Construction and theory, Formation of spectrum by plane transmission grating, Determination of wavelength of light using plane transmission grating.

Resolving power: Geometrical & Spectral, Raleigh criterion, Resolving power of diffraction grating and telescope.

Unit 4

Elements of Material Science

Bonding in Solids: Covalent bonding and Metallic bonding. Classification of Solids as Insulators, Semiconductors and Conductors.

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Semiconductor. X-Ray diffractionand Bragg’s Law. Hall Effect: Theory, Hall Coefficient and applications.

Unit 5

Special Theory of Relativity

Postulates of special theory of relativity, Lorentz transformations, relativity of length, mass and time.

Relativistic velocity addition and mass - energy relation , Relativistic Energy and momentum.

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Chapter 1 Interference of light

In order for an interference pattern to be stable the waves must be emitted from coherent and monochromatic sources.

Most natural light sources are both non-coherent and polychromatic so interference is not widely observed in nature.

In order to create a stable interference pattern waves from different sources must maintain a constant phase relationship with each other.

It is much easier to arrange coherent sound sources than light sources. Because the frequency of light waves (around 1015 Hz) is so high coherence between separate sources is difficult to arrange

It is easy to introduce phase differences between the loudspeakers and constructive or destructive interference merely by either moving the loud speakers around or moving around in the field produced by the loudspeakers

Two Beam Interference

We‟ll add the electric fields of two light beams of identical frequency and wavelength that differ only by some initial phase difference

….(1) …(2)

where α is a constant that contains the phase difference between the waves.

…(3) We‟ll invoke the trig identity:

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Notice that we can plot each of these component waves as phasors as shown at right by plotting the magnitude and phase angle of each.

and it is easily shown that this yields

The resultant wave is another harmonic wave of the same frequency with an amplitude E0 and a phase α that is related to the original waves as shown in the phasor diagram above. The law of cosines may be applied to yield:

and the phase angle may determined by:

Now recall that irradiance (the time average value of the Poynting vector) of a light beam is:

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So the irradiance at a point P due to two beam interference is:

which may be recast in the form:

It is apparent that the first two quantities in the bracketed term are the individual contributions to the sum at P while the third term (the interference term) is the simultaneous contribution from both beams due to their interaction with each other.

The interference term is indicative of the wave nature of the beams of light.

• If light behaved like classical particles there would be no interference and the

irradiance would be: i.e. solely dependant on the

separate contributions of the individual waves. • The interference term varies between zero and 2 E1E2

depending on the orthogonality of the two beams. It may be shown that:

where δ is the phase difference between the two waves due to either a path length difference or an actual phase difference (for our purposes here they are functionally the same.)

When cosδ = +1constructive interference yields the maximum irradiance:

a condition that occurs whenever the phase difference is δ = 2mπ , where m = 0,±1,2,3,...

When cosδ = −1destructive interference yields the minimum irradiance

It is also apparent that complete destructive interference can occur when Ee1 = Ee2 = Ee3 and

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Eemax = 4E0 and Emin=0 Alternatively, it may be shown that the irradiance between two equal

interferingbeams may be written:

Young’s Double Slit:-

Consider a beam of light incident upon two slits, S1 and S2, as shown below. This arrangement insures that the beams that leave the slits are coherent. We‟ll choose an arbitrary point P on a distant screen and look at the pattern of interference which results from combining the two beams at this point. The path length difference between the two beams is S2 P – S1 P = Δ

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When S2 P − S 1P = mλ , constructive interference occurs

• When S2 P − S 1P=(m+1/2) λ destructive interference occurs

m = 0, ±1, ± 2,....

The conditions for interference are:

S2 P – S1 P =mλ ≈ asinθ (constructive) interference

m = 0, ±1, ± 2,.... ≈ asinθ (destructive interference )

Notice that a phase difference between the two beams, δ, is equivalent to a path length difference between the two beams, Δ so that

It then follows that the irradiance at the point of interference varies as a cosine function that also depends on λ and Δ

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As the cosine term varies from ±1 and 0 (usually as a function only of y), the intensity of the pattern on the screen varies from 4E0 to zero, i.e. constructive and destructive interference occur.

The pattern that results is an alternating series of bright and dark fringes.

The location of the bright fringes (intensity maxima): =Ym=mλS/a , m = 0,±1,±2.. and the

separation of the intensity maxima is:. ΔY=mλS/a

The Young‟s double slit experiment is very important historically because it is used to demonstrate the wave nature of light.

Young‟s double slit still has a lot of utility as an easy method of experimentally determining the wavelength of a beam of monochromatic light since:

ΔYa/s=λ

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We will begin by considering only the first two rays to emerge from a film of uniform thickness t and index of refraction n, as shown above (in reality internal reflection from the upper boundary produces many more refracted beams). We‟ll also assume near-normal incidence so that the path length of the refracted beam may be expressed in terms of the thickness of the film.

The wavelength, λn= in a medium whose refractive index is n is given by: λn=λo/n where λo is

the wavelength in free space

The path length difference between the reflected and refracted beams is: Δ = (AB + BC) = 2t or, in terms of the index of refraction in the film, nf(AB+ BC)=nf(2t)

It is apparent that when

or in terms of the wavelength in the film, 2t =λn/2 the two beams interfere destructively

based on the path length difference, Δ , since the path length difference is the equivalent of a phase shift of half a wavelength.

When 2nf t = λ0 f , or 2t = λn , the two beams interfere constructively since the path length

difference is the equivalent of a phase shift of a full wavelength.

There exists, however, a complication. Recall that waves often undergo a phase shift upon reflection. In beams of light: external reflections (nf>n0 ) f o n > n → 1800 (λ phase shift)

internal reflections (n0>nf ) n0 > nf → 00 (no phase shift)

If both reflections are internal or external – no relative phase shift between the beams occurs.

The reflected phase shifting causes beams that would interfere constructively due to path length difference alone to interfere destructively. Consider the thin film surrounded by air shown at right.

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Beam 1 is externally reflected with a phase change of 180° with respect to the incident wave. Beam 2 is internally reflected and undergoes no phase change upon reflection with respect to the incident wave. It is, however, 180 degrees out of phase with the Beam1

If only the reflections are considered the beams would interfere destructively but we must also consider the actual difference in path length 2t. Hence if

2t =λn/2 or t=λ/4

the waves recombine in phase which leads to the requirement for constructive interference for mixed reflections:

2t=(m+1/2) λn, where m=0,1,2,3….

2tnf=(m+1/2) λ0 where m=0,1,2,3….

If 2t is a multiple of λn the two waves combine out of phase and destructive

interference for mixed reflections results: 2t = λn

t = λn/2

=2t = mλn or 0 2tnf = mλ0 where m= 0, 1, 2, 3, …

These conditions apply only when the film is surrounded by a common medium (a glass slide in air, water between two glass slides, etc.) or when the index of refraction on both sides of the film is either lower or higher than that in the film resulting in a mixture of external and internal reflections.

If the film is located between two different media, one of lower index of refraction and one of higher index of refraction, the conditions for constructive and destructive interference are reversed by virtue of the fact that both reflections will be of the same type, either external or internal. In this case

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2t = mλn or 2tnfλ=m λ0where m= 0, 1, 2, 3, … (constructive interference)

2t=(m+1/2) λn or 2tnf=(m+1/2) λ0(destructive interference )

Example 1 Consider a non-reflecting film on a camera lens. Antireflection coatings on lenses are designed to exploit destructive interference to quench reflection from the lens in visible light. A commonly used coating is MgF2, n = 1.38.

Since the film is located between media of two different refractive indices, one lower and one higher than that of the lens, the reflections are matched (both external) and the path length difference is the sole determinant in the type of Interference that occurs . In this case the condition for destructive interference is:

2t=(m+1/2) λn or 2tn=(m+1/2) λ

If λf= λo/nf in the film and t= λf/4, the condition for destructive interference is

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For complete extinction to occur, the reflected and refracted waves should have the same amplitude. In practice the concept is to reduce reflection from 4 – 5% to less than 1%.

If we assume that 550nm If we assume that 550nm ( the minimum thickness (m = 0) of the coating is:

t= λo/4nf= / m =99.6nm

Notice that if the coating has an index of refraction greater than the glass of the lens then reflectivity is increased under the same conditions. This is how one way windows and reflecting sunglasses are made.

Example 2 Consider a layer of methylene iodide (n = 1.756), between two layers of glass (n = 1.5). What must the minimum thickness of the film be if light of 600 nm is to be strongly reflected?

The condition for maximum Constructive interference in a film surrounded by a common medium

2t

2

T=

So the minimum thickness is 85.4 nanometers.

What are the next three thicknesses of coating that will work?

Notice that Δt is about 171 nanometers. This represents a tight tolerance.

Example 3 Compute the minimum thickness for a nonreflecting coating for a solar cell.

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Assume: λ0 =550nm

Destructive interference with no relative phase change:

2

Example 4 In a thin film of gasoline on water the film appears yellow instead of white because blue light has been removed. What is the minimum thickness of the film?

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Notice that destructive interference occurs due to mixed reflections when t = 0 (m = 0) (what does this imply abut the edge of a film of gasoline on a wet surface?). The next minimum thickness that will work is:

2

Non–normal rays (arbitrary angle of incidence)

The phase difference along the plane defined by points C and D is due to path difference (Δ) between A D and A B C.

Unlike previous examples, the path in the film A B C is much greater than 2t for angles of incidence much greater than near-normal.

Δ =

Δ =

Δ[n (AE+ FC)- AD]+ (EB+ BF)

sin

AE= AG sin

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2AE=AC

AD=2 AE= (AE+ FC) Δ (EB+ BF)=2 EB

IF EB = t

Note that the path difference is expressed in terms of angle of refraction and that the angle of incidence may be determined by Snell‟s Law if needed.

For near-normal incidence Δ = 2 as before.

The phase difference between the waves is: = Δ= Δ

The net phase change must also take into account any phase changes occurring upon reflection (as before).

The net phase change must also take into account any phase changes occurring upon reflection (as before).

If is the path length difference and is the phase difference due to reflection then:

Δp + Δr = mλ constructive interference

Δp + Δr = destructive interference where m = 0, 1, 2, Fringes of Equal Thickness

If the thickness, t, of a film varies then the path length Δ= cosθ differs with out any variation of angle of incidence.

A bright or dark fringe for a given angle of incidence will be associated witha

particular thickness of the film as the Δ dependant conditions ofconstructive and destructive interference change with the change in thickness.

Such fringe patterns are called fringes of equal thickness

Consider the arrangement of glass slides as shown at the right. The gap between the slides serves as

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If a beam of monochromatic light strikes this arrangement at near-normal incidence:

Δ= cos

and the conditions for constructive and destructive interference, respectively, are:

Δp + = +

Where Δr is either λ/2 or 0 depending on the whether or not a reflective phase shift exists. Note that as x decreases, t decreases and Δp → 0.

The pattern of equally spaced light and dark fringes seen in the film are known as Fizeau fringes.

A common application of this technique is measuring the smoothness of lenses or mirrors.

In this case the optical surface is not flat but has some radius of curvature and the air film beneath it will be in the form of a circular “wedge” producing Fizeau fringes in the form a series of concentric rings around the point of contact with the flat surface known as Newton‟s rings.

At the point of contact the thickness of the film, t, is zero and phase difference between the reflected rays is π (λ/2) due to the external and internal reflections.

The center of the fringe pattern is dark and is surrounded by a series of bright and dark concentric fringes.

If the lens surface is smooth the fringes are smooth.

This technique may also be used to measure the radius of curvature of the lens surface

If the flat surface is transparent a complementary fringe pattern is formed by the light transmitted through it.

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Chapter -2

Polarization of Light

Q.1 Define specific Rotation ? Describe the construction & working of Laurent’s half shade polarimeter How can it be used to determine specific rotation of glucose solution .

Ans. The arrangement of this type of polarimeter is similar to above fig except a Half shaped device is inserted b/w P & T.

Half Shade Device :-

It consists of two semicircular plates ; one being of simple glass & other of quartz . The quartz plate is a half wave plate whose optic axis is parallel to its face & perpendicular to vertical diameter . The HWP introduces a phase difference of or a path difference of /2 b/w o-ray & E-components . Thickness of the glass plate is so adjusted that it should absorb same amount of light as that by quartz plate . In this way the intensity of xmitted light from these semicircular plates remain same . This Laurenzts half shade plate is kept b in b/w p & T in such a way a that plane of vibration of incident plane polarized light remains perpendicular to the face of Laurent‟s plate.

Working :-

Suppose the plane of vibration of xmitted plane polarized light from polarizer is incident normally on the plate in the direction op makes an angle with the diameter A0B on passing through the glass, half of the vibrations will remain

along op, but on passing through the quartz (HWP) , half of these will be split op into E & O component along OA A phase difference of is introduced b/w these two vibrations . The o-vibrations will advance in phase by and will occur along OD instead of OC on emergence . The resultant vibration on emerging from the Quartz HWWP will there fore be along o Q such that - <poA = ,QPA = Q (Ref fig. below ) If the principal section A, OA2 of the analyser A is arlar to AOB as shown in fig

above (i) . then both half portions of the field of view will be seen equally bright . because the components of OP & OQ on A1 OA2 (UE1 = OE2) are equal . If the on A1

OA2 (OE1 = OE2) are equal . if the principal section A1 OA2 is rotated in the clock

wise direction . fig (i) then OE2 & OE, As a result of the left hand of the field of

view appears brighter than the right half again if A1 o A2 is rotated anticlockwise wise

fig (iii) . then oE2<OE, and right hand of the field of view appear brighter . than the

left half . In order to determine the angle of rotation , analyser A is rotated to such an extent . that both salves appear equally bright . Now the analyser is set in this position first by taking pure water & then by optically active soln in the polarimeter tube . The difference b/w the two readings will give the angle of rotation . Main defect of this

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polarimeter is that it can be used only for the wave length of light for which semicircular quartz plate acts as a half plate In general it is used with sodium light . Q.2 What is optical activity ? Mention the Law of optical rotation?

Ans. When plane polarized light is passed through water or glass . there is no change in the dirn of plane of polarisation of plane polarized light . But there are certain substance through which . if plane polarized light is passed . they rotate the plane of polarization. This phenomenon of rotation of plane of polarization of plane polarized light is called optical activity. Those substances which rotate the plane of polarization are called optically substances .

There are two types of optically active substance. Laws of optical Activity :-

→ Angle of Rotation ( ) of the plane of polarization of the polarized light of definite wave length is directly proportional to the distance covered by the polarize light in an optically active medium (solid, liquid, gas) Q r l

→ In the case of solutions the angle of rotation plane polarized light is directly proportional to the concentration of the solution .

→ The angle of rotation is inversely proportional to the square of the wave length . Thus „ ‟ will be east for red & greatest for violet i.e violet colour will be rotated maximum & red colour will be rotated minimum .

→ Thus when a white light is passed through an optically active substance, the xmitted rags dispersed into different colors . This phenomenon is called rotator dispersion .

→ The rotation produced by a number of optically . active substances is equal to the algebraic sum of the . Individual rotations , The rotation in clock wise dirn is taken as +ve, while the anti clock wise dirn is taken as –ve.

→ The angle of rotation also depends on the nature & temp of the substance . Q.3 A polarized examines two adjacent plane polarization beams A & B whose

planes of polarization are mutually perpendicular . In one position of analyser ,the beam B shows zero intensity from this position a rotation of 300 show that the two beams . have equal intensity what is the ratio of the Intensities of two beams .

Ans. From Males Lass I = Iv cos2

According to the question when rotated 300 from the direction beam B shows no intensity . The angle made by vibrating electric vector . with B & A are respectively 900-300 = 600 & 900-600 = 300

hence ,

IA cos2 300 = IB cos2 600

= cos260 = cos2300

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Q.4 A vo an long tube containing sugar soln rotates the plane of ploanisation . by 110. If specific rotation of sugar soln is 66 dm/gm/lit. Calculate strength of rotation . Ans. Given = 110 l = 2 dm S = 660 dm/gm/litre

S =

C = or C = = 0.083 gm/litre Q.5 Explain the working of polarimeter

Ans. In order ot find the angle of rotation fill the polarimeter to be with pure water & place the tube in its position inside the polarimeter . Now the analyser is rotated until the intensity of light as seen through the eye – piece becomes zero or the analyser is in a crossed position & the reading is noted on the circular late . Now the ploarimeter tube is filled with solution or liquid and placed inside the ploarimeter . This optically active liquid or solution rotates the plane of vibration of the ploarised light . As a result of the Intensily of light a reappears when seeing through the eye –piece . Now the analyser is rotated until the intensity of light as seen through the eye piece became zero . The position of analyser is again noted . The difference of the two position of the analyser gives the angle fo rotation the plane of polarization of the polarized light produced by an optically active slon . This ploarimeter is called simple polarimeter . It is less sensitive since at the crossed position of ploariser & analyser, even if we rotate . the analyzer through 40-50, the intensity of light is seen almost . hence the angle of rotation cannot be measured accurately . In order to the sensitivity of the ploarimeter; a particular derive is placed b/w the ploariser p & tube T .

Q.6 Using the accept of electric vector of electromagnetic wave discuss circularly & elliptically polarized light .

Ans. It the electric vector rotates with constant amplitude in an plane perpendicular to the direction of proprgaton ie . the head of electric field vector rotates in a circular path then the light is called circularly ploarised . Such type of ploarised light is produced by the combination of two perpendicular periodic vibrations of equal amplitude & having a phase difference of .

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Chapter 3 Diffraction of Light

Q.1 Show that the Intensity of light diffracted from a plane x mission grating is given by

I = I ( ) = ( ) =

where symbols have their usual meanings find the positions of maximum & minima & calculate

(a) width of n the principal maximum. (b) Resolving power of grating .

Ans. Plane x mission grating is a spectral . arrangement of frauhofer . diffraction due to N slits . let a plane wave front be incident normally on N slits . Then according to Huygens principle all points in each s lit . become a source of secondary wavelets . By the theory of diffraction at a single s lit discussed . the resultant amplitude in the direction q is given by AQ = Ao sin2 (1)

2 where r is the phase of the resultant have , and it is given by

L = e sin o ---(2)

The resultant wave in each slit is supposed to be coming out from the middle point of the slit . Now suppose S1, S2, S3, ---SN producing interference

According to the fig. 3.8 the path difference b/w the successive wavelets diffracted at angle Q from each slit is same & is equal to (e+b) sin Q

The equivalent phase difference will be (e+b) sin let 2 = (e+b) sin

S1

e+b S2

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So, = (e+b) sin ---(3)

The resultant of all the wavelets at the point p1 is the vector sum of all the amplitude s

of the coherent wavelets obtained from N identical sources The amplitude of each wavelets is Ao & its phase by an amount 2 in successive wavelet .

Now applying vector addition method the resultant amplitude will be mpn (ref fig 3.9) so,

resultant amplitude

R = M PN = 2MP = 2r sin N ---(4)

& MP1 = r sin ---(5)

here MP1 = AQ = Ao sin2 amplitude of first vector & r = radius of curvature divide

eqn (5) by (4) we get . = = MPN = A from eqn. (1) MPN = A0 So resultant amplitude R = A0 ---(6)

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Now intensity at any point on screen due to plane transmission grating I R2

I = KR2

I = KA02

I = I0

here

(i) Width of principal maxima

The condition of nth order principal maxima of the wavelength l in the direction of angle „on‟ is given by (e+b) sin Q n = n ---(1)

let the first minimum on either side of principal maxima can be obtained in the direction of Qn – d o n & Q n + do n . Then 2donis called the annular width of principal maximum . S G Gth maxima d n 2 n ( n-d n) n

For the position of minimum --- (e+b) sin (Qn + d Qn) = ---(2)

where m is an integer excluding the values of O , N , 2N--- so m = nN will give the position of principal maxima thus for the first minima . which is adjacent nt to the n th order principal maxima .

m=nN 1

(e+b) sin( n + d n) = (nN+1) = n + ---(3) (e+b) sin( n + d n) = (nN-1) = n + ---(4) Substituting eqn. (4) from (3) we get ,

(e+b) [ sin( n + d n)-sin( n + d n)] = ---(5) I0 = kA02

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since d n is small sin d n d n 2(e+b) cos n d n =

2d n =

Thus the angler width of principal maxima is inversely proportional to the total no of lines ruled on the grating and the order of principal maxima and directly proportional to tanθn . since the in tan q n Since the in tan n since the in tan n is less than the

in n there fore the angler width of principal maxima for higher values of no. (b) Resolving power of a plane diffraction grating

Q +d f n+d n

n

Grating

In the position of just resolution , the resolving power of the grating R =

let a parallel beam of light of have length & + d be incident normally on the diffraction grating if n th principal maximum of & + d are formed in the direction „ n‟ & n + d n respectively (sec above fig) then ,

For the principal maximum of wave long the at p (e+b) sin n = n --- (1)

& for a principal maximum of wave long the ( +d ) at Q (e+b) sin ( n + d n) = n ( + d ) --- (2)

where n = 1,2,3---& (e+b) is the grating element The nth principal maximum of eave length . is formed at p . If first minimum of this wave length is formed in the direction ( n+d n) ie at the point . then according to Rayleigh criteria the spectral

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line of wave length & +d will appear just resolved for minimum of wave length formed in the direction ( n+d n) ie at Q (e+b) sin ( n+d n) ie at

(e+b) sin ( n + d n) = m --- (3) N

The first minimum adjacent to the nth principal maximum in the direction ( n+d )

will be obtained for m= nN 1 ; where N is the total no of rulings on the grating. (e+b) sin ( n+d n) = ---(4)

Comparing eqn., S (1) & (2) we get , n + = n( + d )

nd =

Resolution limit of grating = = & Resolving power of grating R = = Nn --- (5)

Thus the resolving power of grating is equal to the product of the order of the spectrum & the total no of rulings on the grating.

It is directly proportional to the order of the spectrum & the total no of lines on the grating surface .(N). It is independent of the grating elements (e+b) of the lines .

n = [ substitute the value of n in eqn (5) we get,

Resolving power = Nn = =

where W = N (e+b) = Total width of ruled surface

Resolving power of the grating would not be affected , if the no of lines N in a given width of ruled space is changed .

if Q = 900 then maximum resolving power R = = =

Q.2 Explain, what is meant by Resolving power of a diffraction grating . Define an expression for the same.

Ans. Resolving power of a grating is the ability to separate. the lines very close to each other expression will do in the last expression

Q.3 Give the theory of plane xmission grating & show how would you use it to find the wave length of light. How are spectral lines affected if the ruling are made closer ?

Ans. Already do in the Question No. 1

Q.4 Resolving power of diffraction grating ? Ans. Already do in the Question

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Q.5 Derive an expression for the intensity of diffracted light in the fraunhofer ‘s diffraction due to single slit ?

M L1

Fraunhofer Diffraction due to single slit S = monochromatic Source ( wave length ) S1 = screen

L1 , L2 = Converging (convex ) lens

Wf = plane wave front

AB = s lit with width „e‟ (e ) = diffracted angle

AFP1 + BEP1 = diffracted waves Bk = path difference b/w the wave p = central (bright) maxima at screen

p1 = Any arbitrary point on screen (to calculate Instead ) MN = spectral wave front

A slit . rectangular aperture of karge length . compared to its breadth is placed perpendicular to the plane of paper let the slit . AB. be illuminated . by a parallel beam of monochromatic light of wave length from a source which is placed at the principal focus of the lens . L1 let the width of the narrow slit be „e‟ light is diffracted

by the slit and is focused by another lens L2 on the screen . S1 Instead of a sharp

image of the slit , a contral bright band & alternate dose & bright bands of decreasing intensity . symmetrical on both sides are obtained . This pattern is called diffraction pattern of a single slit

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Now a icular xG is drawn from the centre k of an are on the line AB, which will bisect . the line AB in two equal parts . (BG & AG)

< AKG = < BKG = = & sin =

So , AB = 2AG = 2r sin [in ∆ AKG] [ AK =r]

Substitute the value of r from eqn. (1) we get AB = sin = = A (Resultant Amplitude )

Where is given by eqn. (1) = e sin ---(2)

The resultant in intensity at point p1 will be I a2

I = k a2 = kf2 a2 sin2 [k is proportionality const] 2

or I = I0 sin2 ---(3)

2

where I0 = kA0 z = kp2a2 is the maximum in intensity at the point where all the waves

reach in the same phase eqn. (3) represent intensity at d any point p1 on screen.

Q.6 What do you understand by resolution ? Explain what is mean by resolving power of a grating ?

Ans. Where tow objects ( or their image are very near to each other , they may appear as one and it may be impossible for the nailed eye to see them separately .

The method of seeing two objects or images which are very dose to each other as separate using some optical instrument is called resolution

The resolving power of an grating is the ability to separate the spectral lines very close to each other . when two spectral lines in the spectrum . produced by grating are just resolved then in this position . the ratio of the wave length difference d and them mean . wave length of the spectral . lines is called reciprocal is called . resolving power of the grating .

Q.7 A grating has 9600 lines uniformly spaced over a width of 3.0 cm and is illuminated by light from mercury vapor Lamp find .

(a) dispersion in the third order in the vicinity of green line of wave length jH 600 A0

(b) Resolving power of a grating in 5th order Ans. Grating element (e+b) = = 3 10-2 = 3.126 10-6m

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9600 form grating eqn (e+b) sin = n

= sin-1 = sin-1 3 5460 10-10 3.126 10-6 = sin-1 (0.524) = 31.6 dispersion is given by = = 3 (3.126 10-6)cos (31.60) = 1.13 10-3 rad/m = 0.06460 n/m

(b) Resolving power of grating in 5th order RP = nN = 5 9600 = 4.8 104

Q.8 What should be the minimum no of lines in a grating which will just resolve in the second order , the lines whose wave lengths are 5890 A0 & 5896 A0

soln Given that n =z

1 = 5890 A0 = 5890 10-8cm 2 = 5896 A0 5896 10-8 cm d = 2 - 1 = 6A0 = 6 108cm Resolving power = = nN 5893A0 = 2 N 6A0 N = = 491

Q.9 How many lines per cm are there in a grating . which gives an angle of diffraction of 30 in first order spectrum of light of wave long the 6 10-5 cm? Ans. The grating eqn. is

(e+b) sin = n

(e+b) = = 1 6 10-5 = 12 10-5 cm sin300

No of lines per cm = = 105 = 8333.3/ cm 12

Q.10 A plane transmission grating has 6000 lines/cm Calculate . the bighest order of spectrum which can be seen with light of wave longth 4000 A0.

Ans. We know that (e+b) sin = n

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n

max = =

= 2 = 100 = 4.16 4 24 10-2 24

Q.11 Light composed of two spectral lines with wave length 6000A0 & 6000 s A0 . It falls normally on a diffraction grating 10mm wide . At a certain diffraction angle ‘ ’ these lines are close to be being resolved (according to lay leih crilerion ) Ans. Given that = 6000A0 & + d = 6000.5 A0

d = 0.5 A0

Total width of grating = w = N (e+b) = 10mm =1cm The resolving power of

grating = nN ---(1)

& the position of principal maxima is given by (e+b) sin = n sin = = = [using (i)] sin = 2 [ w = N (e+b)] wd sin = (6000 10-8)2 = 0.72 1cm 0.5 10-8cm = sin-1 (0.72) = 460

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Chapter-4

Elements of Material Science

We will encounter different cases such as ionic, covalent, or metallic bonding. It has to be kept in mind that these are just idealized limiting cases. Often mixed bonding types are found, for example, a combination of metallic and covalent bonding in the transition metals. As in conventional chemistry, only a restricted number of all the electrons participate in the bonding. These so-called valence electrons are the electrons in the outermost shell(s) of an atom. The electrons in the inner shells, or core electrons, are bound so tightly to the nucleus that they do not feel the presence of other atoms in their neighborhood.

Attractive

Two different forces must be present to establish bonding in a solid or in a molecule. Anattractive force is necessary for any bonding. Different types of attractive forces are discussed below. A repulsive force, on the other hand, is required in order to keep the atoms from getting too close to each other. An expression for an interatomic potential can be written as

………(1)

where n >m, that is, the repulsive part has to prevail for short distances (sometimes this is achieved by assuming an exponential repulsion potential). Such a potential and the resulting force are shown in Figure 1.1. The reason for the strong repulsion at short distances is the Pauli exclusion principle. For a strong overlap of the electron clouds of two atoms, the wave functions have to change in order to become orthogonal to each other, because the Pauli principle forbids having more than two electrons in the same quantum state. The orthogonalization costs much energy, hence the strong repulsion.

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Inter atomic distance, r (equilibrium distance)

Figure 1.1 (a) Typical interatomic potential for bonding in solids according to (1.1) with n¼6 and m¼1. (b) Resulting force,that is, _gradf(r). Ionic Bonding:

Ionic bonding involves the transfer of electrons from an electropositive atom to an Electro negative atom. The bonding force is the Coulomb attraction between the two resulting ions. Ionizing both atoms usually costs some energy. In the case of NaCl, the ionization energy of Na is 5.1 eV but the electron affinity of Cl is only 3.6 eV. The net energy cost for creating a pair of ions is thus 5.1_3.6¼1.5 eV. The energy gain is given by the Coulomb potential. For just one Na and one Cl ion separated by a distance a¼0.28 nm, this is _e2/4pe0a, which amounts to 5.1eV.

Potential energies for more complicated structures are discussed below. It is important to distinguish between the different energy contributions involved: The ionization energy is the energy needed to turn atoms into ions, the lattice energy is the electrostatic energy gain by assembling a lattice, and the cohesive energy is the lattice energy minus the ionization energy, that is, it represents the total energy balance to form the ionic solid.

The fact that the total potential energy curve in Figure 1.1 contains a repulsive part in addition to the Coulomb potential means that the actual potential minimum for a given inter atomic distance a is a bit shallower than expected from the pure Coulomb potential (10% or so). Therefore, the lattice energy can be calculated rather accurately using classical physics if the inter atomic distance a is known. Predicting this distance, however, requires the inclusion of quantum effects. In any event, ionic bonding is very strong. The cohesive energy per atom is in the order of several electron volts.

Covalent Bonding

Covalent bonding is based on the true sharing of electrons between different atoms. The simplest case is that of the hydrogen molecule that we will discuss quantitatively below. In solids, covalent bonding is often found for elements with a roughly half filled

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outer shell. A prominent example is carbon that forms solids as diamond or graphite as well as complex molecules such as Buckminster Fullerene C60 or carbon nanotubes. The covalent bonds in diamond are constructed from a linear combination of the 2s orbital and three 2p orbital‟s. This results in four so-called sp3 orbital‟s that stick out in a tetrahedral configuration from the carbon atoms. In graphite, the 2s orbital is combined with only two 2p orbital‟s, giving three sp2

orbital‟s, all in one plane and separated by an angle of 1200, and one p orbital oriented perpendicular to this plane. This linear combination of orbitals already reveals an important characteristic for the covalent bonding: it is highly directional. In addition to this, it is also very stable and the bonding energies are several electron volts. A very instructive example for covalent bonding is the hydrogen molecule H2 for which we

will sketch a solution here. We go into some detail, as much of this will be useful in later chapters. As a starting point, take two hydrogen atoms with their nuclei at RA and RB and we

call |RB_RA|=R. We do, of course, know the solution of the Schrödinger equation for each of

the atoms. Let these ground-state wave functions be ψA and ψB, respectively. The Hamilton

operator for the hydrogen molecule can be written as

……(2)

where r1 and r2 are the coordinates of the electrons belonging to the A and the B nucleus,

respectively. The first two terms refer to the kinetic energy of the two electrons. The operatorsΔ12and Δ22 andr22 act only on the coordinates r1 and r2, respectively.

The electrostatic term contains the repulsion between the two nuclei and the repulsion between the two electrons as well as the attraction of each electron to each nucleus.

The solution of this problem is not simple. It would be greatly simplified by removing the electrostatic interaction between the two electrons because then the Hamiltonian could be written as the sum of two parts, one for each electron. This could then be solved by a product of the two wave functions that are solutions to the two individual Hamiltonians, that is, the two-particle wave function would look like Ψ(r1, r2)= ΨA (r1) ΨB (r2). Actually, this is not

quite right because such a wave function is not Ψ in accordance with the Pauli principle. Since the electrons are fermions, the total wave function must be ant symmetric with respect to particle exchange and the simple product wave function does not fulfill this requirement The total wave function consists of a spatial part and a spin part and therefore there are two possibilities for forming an ant symmetric wave function. We can either choose a symmetric spatial part and an ant symmetric spin part or vice versa. This is achieved by constructing the special wave function of the form

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Ψ (r1, r2)α ΨA (r1) ΨB (r2)+ ΨA (r2) ΨB (r1)………..(1)

Ψ (r1, r2)α ΨA (r1) ΨB (r2)-ΨA (r2) ΨB (r1)………..(2)

The plus sign in (1.3) returns a symmetric spatial wave function that we can take for an ant symmetric spin wave function with the total spin equal to zero (the so-called singlet state); the minus in (1.4) results in an ant symmetric spatial wave function for a symmetric spin wave function with the total spin equal to 1 (the so-called triplet state). The ant symmetric wave function (1.4) vanishes if r1¼r2, that is, the two electrons cannot be at the same place simultaneously. This leads to a depletion of the electron density between the nuclei and hence to an ant bonding state. For the symmetric case, on the other hand, the electrons have opposite spins and can be at the same place, which leads to a charge accumulation between the nuclei and hence to a bonding state (see Figure 1.2). An approximate way to calculate the eigenvalues of (1.2) was suggested by W. Heitler and F. London in 1927. The idea is to use the known single-particle 1s wave functions for atomic hydrogen for ΨA and ΨB to form a

two-electron wave function

Figure 1.2 The energy changes Δ E and Δ E for the formation of the hydrogen molecule. The dashed lines represent the approximation for long distances. The two insets show gray scale images of the corresponding electron probability density

The energy changes Δ E and ΔE for the formation of the hydrogen molecule. The dashed lines represent the approximation for long distances. The two insets show gray scale images of the corresponding electron probability density. Ψ (r1, r2), which is given by either (1.3) or

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The calculation is quite lengthy and shall not be given here. The resulting ground state energies for the singlet and triplet states can be written as

………(1) …………..(2)

E0 is the ground-state energy for one hydrogen atom that appears here twice because we start

with two atoms. The energies ΔE1 and ΔE2 are also shown in Figure 1.2. ΔE

is always larger than zero and does not lead to any chemical bonding. DE"#, however, shows a minimum below zero at approximately 1.5 times the Bohr radius.

This is the binding state.

For long distances between the nuclei, (1.6) and (1.7) can further be simplified to

Give E=2E0+C ……… that is, the energy change upon bonding has two parts, one that

does depend on the relative spin orientations of the electrons (_X) and one that does not (C). The energy difference between the two states is then given by 2X, where X is called the exchange energy. In the case of the hydrogen molecule, the exchange energy is always negative.

We will encounter similar concepts in the chapter about magnetism where the underlying principle for magnetic ordering is very similar to what we have here. The total energy of a system of electrons depends on their relative spin directions through the exchange energy, and therefore a particular ordered spin configuration is favored. For two electrons, the .magnetic. character is purely given by the sign of X. For a negative X, the coupling with two opposite spins is favorable (the anti ferro magnetic case) whereas a positive X would lead to a situation where two parallel spins give the lowest energy (the ferromagnetic case).

Metallic Bonding : In metals, the outer valence electrons are removed from the ion cores, but in contrast to ionic solids, there are no electronegative ions to bind them. Therefore, they are free to migrate between the remaining ion cores. These delocalized valence electrons are involved in the conduction of electricity and are therefore often called conduction electrons. One can expect metals to form from elements for which the energy cost of removing outer electrons is not too big. Nevertheless, this removal always costs some energy that has to be more than compensated by the bonding. Explaining the energy gain from the bonding in an intuitive picture is difficult

But we can at least try to make it plausible. The ultimate reason must be some sort of energy lowering.

One energy contribution that is lowered is the kinetic energy of the conduction electrons. Consider the kinetic energy contribution in a Hamiltonian, T=-ħ2 2/2m. A matrix element

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(Ψ | T| Ψ) measures the kinetic energy of a particle. TY is proportional to the second spatial derivative of the wave function, that is,

the curvature. For an electron that is localized to an atom, the curvature of its wave function is much higher than that for a nearly free electron in a metal and this is where the energy gain comes from. The other contribution to the electron energy is the potential energy. One should think that the average electrostatic potential of any single electron in a solid is almost zero because there are (almost) as many other electrons as there are ions with the same amount of charge. But this turns out to be wrong. In fact, the electrons see an attractive potential. The reason is again partly due to the Pauli principle that, loosely speaking, does not allow two electrons with the same spin direction to be at the same place and therefore the electrons go .out of each others way.. In addition to this, there is also a direct Coulomb interaction between the electrons, which makes them avoid each other. We will discuss this in more detail when dealing with magnetism. Typically, metallic bonding is not as strong as covalent or ionic bonding but it amounts to a few electron volts per atom. Stronger bonding is found in transition metals, that is, metals with both s and p conduction electrons and a partially filled d shell. The explanation for this is that we have a mixed bonding. The s and p electrons turn into delocalized metallic conduction electrons, whereas the d electrons create much more localized, covalent-type bonds.

1.Why is a typical inter atomic potential, such as in Figure 1.1, so asymmetric? 2. Which elements are likely to form crystals through ionic bonding?

3. What kind of forces are important for ionic bonding?

4. How does the lattice energy in an ionic crystal depend on the inter atomic distance?

5. Explain the difference between cohesive energy and lattice energy. 6. Which elements are likely to form metals?

7. Where does the energy gain in metallic bonding come from?

8. What is the difference between a simple metal and a transition metal (definition and typical physical properties)?

Conductors

Conductors are generally substances which have the property to pass different types of energy. In the following, the conductivity of electricity is the value of interest.

Metals: The conductivity of metals is based on the free electrons (so-called Fermi gas) due to the metal bonding. Already with low energy electrons become sufficiently detached from the atoms and a conductivity is achieved.

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Metallic bonding: fixed ions and free valence electrons (Fermi gas)

The conductivity depends, inter alia, on the temperature. If the temperature rises, the metal atoms swing ever stronger, so that the electrons are constrained in their movements. Consequence, the resistance increases. The best conductors, gold and silver, are used relatively rare because of the high costs (gold e.g. for the contacting of the finished chips). The alternatives in the semiconductor technology for the wiring of the individual components of microchips are aluminum and copper.

Salts: In addition to metals, salts can also conduct electricity. There are no free electrons, so the conductivity depends on ions which can be solved when a salt is melting or dissolving, so that the ions are free to move (see chapter chemical bonds for details).

Insulators

Insulators possess no free charge carriers and thus are non-conductive.

The atomic bond: The atomic bond is based on shared electron pairs of nonmetals. The elements which behave like nonmetals have the desire to catch electrons, thus there are no free electrons which might serve as charge carriers.

The ionic bond: In the solid state, ions are arranged in a grid network. By electrical forces, the particles are held together. There are no free charge carriers to enable a current flow. Thus substances composed of ions can be both conductor and insulator.

Semiconductors

Semiconductors are solids whose conductivity lies between the conductivity of conductors and insulators. Due to exchange of electrons - to achieve the noble gas configuration - semiconductors</a> arrange as lattice structure. Unlike metals, the conductivity increases with increasing temperature.

Increasing temperatures leads to broken bonds and free electrons are generated. At the location at which the electron was placed, a so-called defect electron ("hole") remains.

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Cut-out of a silicon lattice

The electron flow is based on the conductivity properties of semiconductors</a>. The electronic band structure illustrates why semiconductors</a> behave like this.

The band model

The electronic band structure is an energy schema to describe the conductivity of conductors, insulators, and semiconductors</a>. The schema consists of two energy bands (valence and conduction band) and the band gap. The valence electrons - which serve as charge carriers - are located in the valence band, in the ground state the conduction band is occupied with no electrons. Between the two energy bands there is the band gap, its width affects the conductivity of materials.

The energy bands If we consider a single atom, there are according to the Bohr model of atoms sharply distinct energy levels, which may be occupied by electrons. If there are multiple atoms side by side they are interdependent, the discrete energy levels are fanned out. In a silicon crystal, there are approximately 1023 atoms per cubic centimeter, so that the individual energy levels are no longer distinguishable from each other and thus form broad energy ranges.

Energy levels of atoms which are in interdependency with other atoms

The width of the energy bands depends on how strongly the electrons are bound to the atom. The valence electrons in the highest energy level interact strongly with those of neighboring atoms and can be solved relatively easily from an atom; with a very large number of atoms, a single electron can no longer be assigned to one single atom. As a result, the energy bands of the individual atoms merge to a continuous band, the valence band.

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Energy bands of atoms which are in interdependency with other atoms

The band model of conductors: In conductors, the valence band is either not fully occupied with electrons, or the filled valence band overlaps with the empty conduction band. In general, both states occure at the same time, the electrons can therefore move inside the partially filled valence band or inside the two overlapping bands. In conductors there is no band gap between the valence band and conduction band.

The band model of insulators: In insulators the valence band is fully occupied with electrons due to the covalent bonds. The electrons can not move because they're "locked up" between the atoms. To achieve a conductivity, electrons from the valence band have to move into the conduction band. This prevents the band gap, which lies in-between the valence band and conduction band.

Only with considerable energy expenditure (if at all possible) the band gap can be overcome; thus leading to a negligible conductivity.

The band model of semiconductors</a>: Even in semiconductors</a>, there is a band gap, but compared to insulators it is so small that even at room temperature electrons from the valence band can be lifted into the conduction band. The electrons can move freely and act as charge carriers. In addition, each electron also leaves a hole in the valence band behind, which can be filled by other electrons in the valence band. Thus one gets wandering holes in the valence band, which can be viewed as positive charge carriers.

There are always pairs of electrons and holes, so that there are as many negative as positive charges, the semiconductor crystal as a whole is neutral. A pure undoped semiconductor is known as intrinsic semiconductor. Per cubic centimeter there are about 1010 free electrons and holes (at room temperature).

Since the electrons always assume the energetically lowest state, they fall back into the valence band and recombine with the holes if there is no energy supply. At a certain temperature an equilibrium is arranged between the electrons elevated to the conduction band and the electrons falling back. With increasing temperature the number of electrons that can leap the band gap is increased, and thus increasing the conductivity of semiconductors</a>.

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The band model

Since the width of the band gap represents a certain energy corresponding to a particular wavelength, one tries to alter the width selective in order to obtain certain colors of light emitting diodes (LED). This may be achieved by combining different materials. Gallium arsenide (GaAs) has a band gap of 1.4 eV (electron volts, at room temperature) and thus emits red light.

The intrinsic conductivity of silicon is of no interest for the functioning of components, since it depends, inter alia, on the supplied energy. Which means that it changes with the temperature; in addition a conductivity comparable to metals is only possible at very high temperatures (several hundred degrees Celsius). In order to deliberately influence the conductivity of semiconductors</a>, impurity atoms can be introduced into the regular silicon lattice to alter the number of free electrons and holes.

Conductivity of Semiconductors

Now let‟s consider the conductivity of semiconductors.

Consider silicon which, like carbon, has the diamond cubic cyrstal structure. The valance electrons are all covalently bonded in sp3 orbitals. These orbitals are completely filled. However, in this case, the next available energy level (in the Conduction Band) is 1.1 eV above the highest occupied level.

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So the picture is exactly the same as that for the insulating materials, except that the the size of the energy gap is smaller. Hence for Si, with an energy gap of 1.1 eV, at room temperature some electrons will be promoted into the conduction band which accounts for its intermediate value of conductivity. Furthermore, for every electron promoted into the conduction band, a "hole" is left in the valance band!

These holes are also considered to be charge carriers. So there are two types of carriers for electrical conduction: electrons and holes. Electrons are called n-type carriers (for negative) and holes are called p-type carriers (for positive).

For intrinsic semiconductors (no impurities), the number of electrons will be equal to the number of holes.

ne = nh

Thus, the conductivity for an intrinsic semiconductor can be calculated: sigma = ne q me + nh q mh

And since ne = nh the conductivity for an intrisic semiconductor is given by this equation:

sigma = ni q (me + mh)

where i stands for intrinsic carrier concentration.

Overall Conclusion on the Conductivity of Semiconductors:

Semiconductors are semi-good electrical conductors because although their valence band is completely filled, the energy gap between the valance band and the conduction band is not too large. Hence some electrons can bridge it to become charge carriers. The difference between a semiconductors and an insulator is the magnitude of the energy gap. For semiconductors Eg < 2eV and for Insulators Eg > 2eV.

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Temperature Affect the Conductivity of a Conductor?

Let‟s Consider the effect of increasing the temperature on the conductivity of conductors.

Let's look at the factors that go into conductivity and consider how each of these are affected: sigma = n q m

First consider what will happen to n as temperature increases. The electrons that are charge carriers in a conductor will gain energy and go into higher energy levels. However, these energy levels are all still in the valance band. So the number of charge carriers will not change for a conductor with an increase in temperature. Now consider q. As temperature increases, the charge on each carrier will not change.

Finally, what happens to the mobility? Recall that mobility is the drift velocity divided by the electric field strength. Temperature won't affect the electric field strength. But it will decrease the drift velocity because as the temperature increases, the atomic vibrations will increase, which will cause more collisions of the electrons with the crystal lattice. Hence the drift velocity will decrease.

Conclusion:

The electrical conductivity of a conductor will decrease with an increase in temperature!

The relationship is not linear, however, if we consider the resistivity, which is the reciprocal of conductivity, we do get a linear relationship:

rho = rhoroomTemp [1 + alpha(T - Troom)]

where rhoroomTemp is the room temperature resisitvity and alpha is the temperature

coefficient of resistivity.

Let‟s Consider the effect of increasing the temperature on the conductivity of semiconductors.

Let's look at the factors that go into conductivity of a semiconductor and consider how each of these are affected:

sigma = ni q (me + mh)

First let's consider q. As with conductors, as temperature increases, the charge on each carrier will not change.

Now consider mobility. The effect of an increase in temperature on mobility is the same as it was for conductors. With the same reasoning, we see that the drift velocity will decrease causing the mobility to decrease.

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Lastly, let's consider what will happen to ni for semiconductors as temperature

increases. The electrons in the valance band will gain energy and go into the higher energy levels in the conduction band where they become charge carriers! So this term will increase. Not only will it increase, but it will increase exponentially! (Promoting electrons from the valance band into the conduction band is a thermally activated process.)

o ni = C e – (E – Eave)/kT

o ni = C e – Eg/2kT

So even though mobility decreases, the exponential increase in the number of charge carriers will dominate.

Conclusion:

The electrical conductivity of a semiconductor will increase exponentially with an increase in temperature!

sigma = C e – Eg/2kT

We can graph this equation on log vs. 1/T axes to get a linear plot (as with all Arrhenius type equations):

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Now let's consider adding impurities to a semiconductor.

When we add impurities to semiconductors we call them dopants and the process is called doping. The result is a dilute (100 -1000 ppm) substitutional solid solution.

There are two kinds of dopants: one will give negative charge carriers (to make an n-type semiconductor) and the other will give positive charge carriers (to make a p-type semiconductor).

N-type Semiconductor

N-type semiconcdutors have dopants from the VA group, such as P+5. These donor impurity atoms are in substitutional solid solution. The extra valance electron not needed for the sp3 tetrahedral bonding is only loosely bound to the P atom in a donor energy level, Ed. The

energy of this donor energy level is close to the lowest energy level of the conduction band (in Si it is 0.4 eV) and so it is easy to promote an electron from the donor level to the conduction band. These promoted electrons become charge carriers that contribute to the material's conductivity. Since they are negative, the result is called an n-type semiconductor.

As temperature increases, more and more of these donor electrons will be promoted into the conduction band. Eventually, a temperature will be reached such that there will be none left. The donor electrons will be "exhausted". During this process the relationship of conductivity to temperature will look like this:

sigma = sigma0 e – (Ec-Ed) /kT