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SCHEME OF EVALUATION

MANIPAL INSTITUTE OF TECHNOLOGY

MANIPAL UNIVERSITY, MANIPAL FIRST SEMESTER B.Tech. END-SEMESTER EXAMINATION - JANUARY 2014

SUBJECT: ENGINEERING PHYSICS (PHY101/102)

Time: 3 Hrs. Max. Marks: 50

Note: Answer any FIVE FULL questions. Each question carries 10 marks Answer all the sub questions of a main question in a continuous sequence. Write specific and precise answers. Any missing data may suitably be assumed. Write question number on the margin only. Draw neat sketches wherever necessary.

Physical Constants:

Speed of light in vacuum = 3.00 X 108 m/s Electron charge = 1.60 X 1019 C

Electron mass = 9.11 X 1031 kg atomic mass unit (u) = 1.66 X 1027 kg

Boltzmann constant = 1.38 X 1023 J/ K Planck’s constant = 6.63 X 1034 J.s

1A. Explain the terms with reference to lasers: (i) spontaneous emission

(ii) stimulated emission (iii) pumping. [5] 1B. Obtain an expression for the width of the central maximum in diffraction

pattern due to multiple slits. [3] 1C. Explain the terms: intrinsic semiconductor, extrinsic semiconductor. [2] 2A. Derive the Compton-shift equation. [5] 2B. Explain briefly hydrogen bonding. [3] 2C. Sketch the schematic graph of a travelling electromagnetic wave showing the

electric and magnetic vectors. Is this a polarized wave? If so, in which direction? [2]

3A. What is the necessary condition on the path length difference (and phase

difference) between two waves that interfere (i) constructively and (ii) destructively ? Draw a schematic plot of the intensity of light in a double-slit interference against phase-difference (and path-difference). [4]

3B. Sketch the energy level diagram of H-atom schematically, indicating the

energy value for each level and the transition lines for the Lyman series, Balmer series and Paschen series. [4]

3C. Explain the expression for the total potential energy of an ionic crystal.

Sketch schematically the plot of the same. [2]

MU - MIT I BTE C H – FIRST SEMESTER- END EXAMINATION JANUARY 2014 ENGINEERING PHYSICS SCHEME OF EVALUATION

2 ∕ 14

4A. A photon with wavelength λ is absorbed by an electron confined to a “box”. As a result, the electron moves from state n = 1 to n = 4. (i) Find the length (L) of the “box” in terms of λ. (ii) What is the wavelength (λ′ in terms of λ) of the photon emitted in the transition of that electron from the state n = 4 to the state n = 2 ? [5]

4B. Three electric fields from three coherent sources are described by E1 = Eo sin ωt,

E2 = Eo sin (ωt+φ) and E3 = Eo sin (ωt+2φ). Let the resultant field be represented by ER = Eθ sin (ωt+β). Use phasors to find Eθ and β when φ = 60°. Draw a schematic phasor diagram. [3]

4C. NaCl molecule contains Na+ and Cl− ions, each with a charge of magnitude

of that of an electron. The ionization energy of Na is 5.1 eV and the electron

affinity of Cl is 3.7 eV. What energy is needed to form Na+ and Cl− ions

from neutral atoms ? What energy is needed to break up a NaCl molecule

into Na+ and Cl− ions ? The dissociation energy of NaCl molecule is 4.2 eV. [2]

5A. The wave function for an electron in the 2p-state of H-atom is

( )

√ ⁄

⁄

. What is the most probable distance from the

nucleus to find an electron in the 2p-state [in terms of ao (= bohr radius)] ? [5] 5B. The current in a diode under forward bias of 100 mV is 200 mA at a

temperature of 300 K. What is the current in the diode if it is under reverse bias of 100 mV ? [3]

5C. The painting contains small dots (2 mm in diameter) of pure pigment, as indicated in figure. The illusion of colour mixing occurs because the pupils of the observer’s eyes diffract light entering them. Calculate the minimum distance an observer must stand from painting to achieve the desired blending of colour. (wavelength = 475 nm, diameter of pupil = 4.4mm) [2]

6A. The stopping potential for photoelectrons released from a metal is 1.48 V larger

compared to that in another metal. If the threshold frequency for the first metal is 40.0 % smaller than for the second metal, determine the work function for each metal. [5]

6B. An HCl molecule (masses are 1 u and 35 u) is excited to its first rotational

energy level, corresponding to J = 1. If the distance between its nuclei is ro = 127.5 pm, what is the angular speed (ω) of the molecule about its centre of mass ? [3]

6C. A diffraction grating has 104 rulings uniformly spaced over 25.0 mm. It is

illuminated at normal incidence by a light of wavelength 589 nm. At what angle will the first order maximum occur? [2]

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -


3 ∕ 14

1A. Explain the terms with reference to lasers: (i) spontaneous emission (ii) stimulated emission (iii) pumping. [5]

(i) spontaneous emission: The average life time of the atomic system in the excited state is of the order of 10–8 s. After the life time of the atomic system in the excited state, it comes back to the state of lower energy on its own accord by emitting a photon of energy hf = E2– E1 (see figure). In an ordinary light source the radiation of light from different atoms is not coherent. The radiations are emitted in different directions in random manner. Such type of emission of radiation is called spontaneous emission. (ii) stimulated emission: When a photon (stimulating photon) of suitable frequency interacts with an excited atomic system, it comes down to ground state before its life time. Such an emission of radiation is called stimulated emission. In stimulated emission, both the stimulating photon and the stimulated photon are of same frequency, same phase and are in same state of polarization, they are emitted in the same direction. In other words, these two photons are coherent. Thus amplified radiation is got by stimulated emission. (see figure). (iii) pumping: Pumping is a process of exciting more number of atoms in the ground state to higher energy states, which is required for attaining the population inversion.

2 mark

2 mark

1 mark


4 ∕ 14

1B. Obtain an expression for the width of the central maximum in diffraction pattern due to multiple slits. [3]

The pattern contains central maximum with minima on either side. λ = wavelength of light. N = total number of slits.

θo = width of the central maximum.

ΔL = path difference between the light waves from the two adjacent slits separated by d at

the position θo.

Δ = phase difference between the light waves from the two adjacent slits separated by d

at the position θo.

At minima, the phase difference, Δ = 2 / N 1C. Explain the terms: intrinsic semiconductor, extrinsic semiconductor. [2] In an intrinsic semiconductor (pure semiconductor) there are equal number of

conduction electrons and holes. The doped semiconductors are called extrinsic semiconductors. In p-type extrinsic semiconductors the hole density is more than the conduction electron

density. (In p-type extrinsic semiconductors the dopant is a trivalent element). In n-type extrinsic semiconductors the conduction electron density is more than the hole density. (In n-type extrinsic semiconductors the dopant is a pentavalent element).

N2L

Nd

Ndsin

sindN

sindL

0

0

0

0

1½ mark

1 mark

1½ mark

½ mark

½ mark


5 ∕ 14

2A. Derive the Compton-shift

equation. [5]

Eo = hfo = hc/o = incident photon energy. Photon interacts elastically with free electron (mass = me) initially at rest as shown in figure. In the scattering process, the total energy and total linear momentum of the system must be conserved.

Applying the law of conservation of energy to the process gives

Where hc/ o = Eo is the energy of the incident photon,

hc/ ’ = E’ is the energy of the scattered photon, and Ke is the kinetic energy of the recoiling electron.

Substituting for Ke we get Applying law of conservation of momentum to this collision, both in x and y components of

momentum are conserved independently.

where h/o = po is the momentum of the incident photon

h/’ = p’ is the momentum of the scattered photon

mev = pe is the momentum of the scattered electron Rewriting the above equations as Squaring and adding the above equations give Rewriting equation (1) in terms of respective energy notations as Eo = E’ + Ee – mc2 i.e., Eo– E’ + mc2 = Ee ∙ ∙ ∙ (b) Square equation (b), substitute for and for from equation (a). Write the resulting equation in terms of respective

wavelengths, we get the Compton shift equation as

eo

K'

chch

)1(cm)1( 2e

o '

chch

coscos'

hhx vmeo

:COMPONENT

sinsin'

hy vm0 e:COMPONENT

coscosx eo:COMPONENT p'pp

sinsiny e:COMPONENT p'p

)a(p'p'pp2p 2e

2o

2o cos

4222e

2e cmcpE

2ep

)cos1(' cmhe

o

1 mark

1 mark

1 mark

1 mark

1 mark


6 ∕ 14

2B. Explain briefly hydrogen bonding. [3]

A hydrogen bond is a weak electrostatic chemical bond which forms between covalently bonded H-atoms and a strongly electronegative atom with a lone pair of electrons (eg: O, N, F). In the two covalent bonds in H2O, the electrons from the H-atoms are more likely to be found near the O-atom than near the H-atoms. Thus the protons (H-nuclei) are unshielded by electrons and can be attracted to the negative end of another polar molecule (eg: O of H2O). This way hydrogen bond is formed between two H2O molecules. This bond is strong enough to form a solid crystalline structure (ice). Similarly, DNA molecules are held together by N–H - - - N hydrogen bonds.

2C. Sketch the schematic graph of a travelling electromagnetic wave showing the

electric and magnetic vectors. Is this a polarized wave? If so, in which direction? [2]

= electric vector

= magnetic induction vector x = direction of propagation of the

wave y = direction of polarization of the

wave = direction of vector YES, PLANE POLARIZED ELECTROMAGNETIC WAVE

1 mark

½ mark

1½ mark

1 mark

1 mark


7 ∕ 14

3A. What is the necessary condition on the path length difference (and phase difference) between two waves that interfere (i) constructively and (ii) destructively ? Draw a schematic plot of the intensity of light in a double-slit interference against phase-difference (and path-difference). [4]

Maximal constructive interference of two waves occurs when their path length

difference is integral multiple of wavelength [ = m λ] or phase difference is zero or

integral multiple of 2 , [ = 2 m ] . Complete destructive interference of two waves occur when their path length

difference is odd integral multiple of half-wavelength [ = (m+½) λ] or phase

difference is integral multiple of , [ = m ] .

1 mark

1 mark

½ mark for the proper curve ½ mark for the proper values on y-axis and mentioning

that Io is the intensity due to one source

½ mark for the proper values on x-axis

½ mark for the proper values on x-axis


8 ∕ 14

3B. Sketch the energy level

diagram of H-atom schematically, indicating the energy value for each level and the transition lines for the Lyman series, Balmer series and Paschen series. [4]

3C. Explain the expression for the total potential energy of an ionic crystal.

Sketch schematically the plot of the same. [2]

Total potential energy of an ionic crystal: r = nearest-neighbour distance α = Madelung constant ke = electrical constant m = small integer B = a constant Ionic cohesive energy of the crystal: ro = equilibrium separation distance

EN

ERG

Y (

eV)

→

−0.85

−1.5

−3.4

−13.6

1 mark for proper energy levels 1 mark for proper transition lines 1 mark for proper energy values

1 mark for proper labels of n-values and series names

m

2

eTOTALr

B

r

ekU

ATTRACTIVE POTENTIAL

REPULSIVE POTENTIAL

m

11

r

ekU

o

2

eo

1 mark 1 mark


9 ∕ 14

4A. A photon with wavelength λ is absorbed by an electron confined to a “box”. As a result, the electron moves from state n = 1 to n = 4. (i) Find the length (L) of the “box” in terms of λ. (ii) What is the wavelength (λ′ in terms of λ) of the photon emitted in the transition of that electron from the state n = 4 to the state n = 2 ? [5] (i) (ii)

( )

( )

( )

√

(

)

4B. Three electric fields from three coherent sources are described by E1 = Eo sin ωt,

E2 = Eo sin (ωt+φ) and E3 = Eo sin (ωt+2φ). Let the resultant field be represented by ER = Eθ sin (ωt+β). Use phasors to find Eθ and β when φ = 60°. Draw a schematic phasor diagram. [3]

At t = 0, E1 = Eo sin ωt = 0 E2 = Eo sin (ωt+φ) = Eo sin 60°

E3 = Eo sin (ωt+2φ) = Eo sin 120° Phasors are shown in the diagram. E1X = Eo E1Y = 0 E2X = Eo cos 60° = 0.5 Eo E2Y = Eo sin 60° = 0.866 Eo E3X = Eo cos 120° = −0.5 Eo E3Y = Eo sin 120° = 0.866 Eo EX = E1X + E2X + E3X = Eo EY = E1Y + E2Y + E3Y = 1.73 Eo E = √(EX

2 + EY2) = 2 Eo tan β = EY / EX = 1.73 β = 60°

ER = Eθ sin (ωt+β), Eθ = 2 Eo & β = 60°

½ mark

½ mark

½ mark for phasor diagram & ½ mark for labelling phasor diagram

½ mark

½ mark

E

o

E

o

E

o

φ

2φ

β

Eθ

½ mark ½ mark

2 mark 2 mark


10 ∕ 14

4C. NaCl molecule contains Na+ and Cl− ions, each with a charge of magnitude

of that of an electron. The ionization energy of Na is 5.1 eV and the electron

affinity of Cl is 3.7 eV. What energy is needed to form Na+ and Cl− ions

from neutral atoms ? What energy is needed to break up a NaCl molecule

into Na+ and Cl− ions ? The dissociation energy of NaCl molecule is 4.2 eV. [2]

Energy needed to form Na+ and

Cl− ions from neutral atoms = 5.1 eV − 3.7 eV = 1.4 eV

Energy needed to break up a NaCl molecule into Na+ and Cl− ions

= (Energy needed to form Na+ and Cl− ions from neutral atoms) + (dissociation energy)

= 1.4 eV + 4.2 eV

= 5.6 eV

1 mark

1 mark


11 ∕ 14

5A. The wave function for an electron in the 2p-state of H-atom is

( )

√ ⁄

⁄

. What is the most probable distance ro from

the nucleus to find an electron in the 2p-state [in terms of ao (= bohr radius)] ? [5]

( ) | | ⁄

⁄

⁄ (

) ⁄

, at the most probable value of distance r = ro .

[(

) ⁄ ]

(

)

5B. The current in a diode under forward bias of 100 mV is 200 mA at a

temperature of 300 K. What is the current in the diode if it is under reverse bias of 100 mV ? [3]

I1 = 200 mA , ΔV1 = 100 mV

I2 = ? ΔV2 = –100 mV

I2 = – 4.2 mA

1 mark

1 mark

1 mark

1 mark

2 mark

1

Tk

VeexpII 1

o1

1

Tk

VeexpII 2

o2

1Tk

Veexp

1Tk

Veexp

I

I

1

2

1

21 mark

1 mark


12 ∕ 14

5C. The painting contains small dots (y 2 mm in diameter) of pure pigment, as indicated in figure. The illusion of colour mixing occurs because the pupils of the observer’s eyes diffract light entering them. Calculate the minimum distance an observer must stand from painting to achieve the desired blending of colour. (wavelength λ = 475 nm, diameter of pupil d = 4.4mm). [2]

= 1.32 x 10–4

RAD For the desired blending of colour = 15 m

1 mark d

22.1R

RD

y

R

yD

1 mark


13 ∕ 14

6A. The stopping potential for photoelectrons released from a metal is 1.48 V larger compared to that in another metal. If the threshold frequency for the first metal is 40.0 % smaller than for the second metal, determine the work function for each metal. [5]

ΔVS1 − ΔVS2 = 1.48 V, (fC2 – fC1) / fC2 = 40%

e ΔVS1 = KMAX1 = hf – φ1 SUBTRACT

e (ΔVS1– ΔVS2) = φ2– φ1 e ΔVS2 = KMAX2 = hf – φ2

φ2– φ1 = 1.48 eV (fC2 – fC1) / fC2 = 0.400 (h fC2 – h fC1) / (h fC2) = 0.400 (φ2 – φ1) / φ2 = 0.400

φ2 = 3.70 eV

φ1 = 2.22 eV 6B. An HCl molecule (masses are 1 u and 35 u) is excited to its first rotational

energy level, corresponding to J = 1. If the distance between its nuclei is ro = 127.5 pm, what is the angular speed (ω) of the molecule about its centre of mass ? [3]

h = 6.63 x 10 –34 J.s , m1 = 1 u , m2 = 35 u , ro = 127.5 pm = 0.9726 u = 1.6145 x 10–27 kg = = 2.62 x 10–47 kg.m2 = 5.68 x 1012 rad/s

1 mark

1 mark

2 mark

2 mark

21

21

mm

mm

2orI

IE

2

1

2

121 I

I21

)1J(JI2

E2

ROT

1 mark

1 mark


14 ∕ 14

6C. A diffraction grating has 104 rulings uniformly spaced over 25.0 mm. It is illuminated at normal incidence by a light of wavelength 589 nm. At what angle will the first order maximum occur? [2]

N = 104 , d = 25.0 mm , λ = 589 nm , m = 1

d sin θ = m λ , , θ = 13.6° - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

nm2500N

wd

d

msin

1 mark

1 mark

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