MANIPAL INSTITUTE OF TECHNOLOGY

MANIPAL INSTITUT OF TCHNOLOGY MANIPAL UNIVRSITY, MANIPAL SCOND SMSTR B.Tech. ND-SMSTR XAMINATION - JULY 01 SUBJCT: NGINRING PHYSICS (PHY101/10) Time: Hrs. Max. Marks: 50 Note: Answer any FIV FULL questions. ach 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 =.00 X 10 8 m/s lectron charge = 1.60 X 10 19 C lectron mass = 9.11 X 10 1 kg atomic mass unit (u) = 1.66 X 10 7 kg Boltzmann constant = 1.8 X 10 J/ K Planck s constant = 6.6 X 10 4 J.s 1A. Obtain an expression for the intensity of light in double-slit interference using phasor-diagram. [5] 1B. Sketch schematically the graph of wavelength vs. intensity of radiation from a blackbody. xplain Planck s radiation law. [] 1C. The ionization energy of Na-atom is 5.1 ev. The electron affinity of Cl-atom is.7 ev. The dissociation energy of NaCl-molecule into atoms is 4. ev. Sketch schematically a graph of total energy of NaCl-molecule vs. distance between the Na + and Cl ions, indicating the relevant energy values and the bond-length (r o ). [] A. Sketch the potential-well diagram of finite height U and length L, obtain the general solution of the schrodinger equation for a particle of mass m in it. [5] B. xplain Fermi-level, donor-levels and acceptor-levels; indicate their positions in the energy band diagram of a semiconductor. [] C. xplain with diagram, the polarization of reflected light, incident at Brewster s angle. [] A. Write a note on a semiconductor junction diode. [5] B. Discuss qualitatively the diffraction due to multiple slits (eg, 5 slits), with diagrams. [] C. xplain metastable state with reference to lasers. [] 1 / 1

MU - MIT I BT C H SCOND SMSTR- ND XAMINATION JULY 01 NGINRING PHYSICS 4A. Monochromatic light with wavelength 58 nm falls on a slit with width 5. m. The distance from the slit to a screen is.48 m. Consider a point on the screen 1.1 cm from the central maximum. Calculate (i) the diffraction angle (ii) the phase angle (iii) the ratio of the intensity at this point to the intensity at the central maximum. [5] 4B. An electron is accelerated through a potential difference of 54 volt. Calculate its momentum and de Broglie wavelength. [] 4C. A thin rod of superconducting material.50 cm long is placed into a 0.540-T magnetic field with its cylindrical axis along the magnetic field lines. Find the magnitude of the surface current on the curved surface of the rod. Magnetic constant = 4 10 7 Tm/A. [] 5A. The wavelengths of the first two lines in Lyman series of the hydrogen spectrum are 11.5 nm and 10.5 nm respectively. From these values calculate the wavelength of the first line in Balmer series. [5] 5B. Show that the average kinetic energy [ ] of a conduction electron in a metal at zero K is where F = Fermi-energy; the density of conduction electrons, ; the density of states, ; m = electron mass. [] 5C. Two polarizing sheets have their polarizing directions parallel so that the intensity of the transmitted light is a maximum. Through what angle must either sheet must be turned if the intensity is to drop by one-half? [] 6A. The J = 0 to J = 1 rotational transition of the CO molecule occurs at a frequency of 1.15 x 10 11 Hz. (i) Use this information to calculate the moment of inertia of the molecule. (ii) Calculate the bond-length of the molecule. (iii) What happens if another photon of frequency 1.15 x 10 11 Hz is incident on the CO molecule while it is in the J = 1 state? [5] 6B. A thin film of acetone (index of refraction = 1.5) is coating a thick glass plate (index of refraction = 1.50). Plane light waves of variable wavelengths are incident normal to the film. When one views the reflected wave, it is noted that complete destructive interference occurs at 600 nm and constructive interference at 700 nm. Calculate the thickness of the acetone film? [] 6C. Use Moseley s relation to calculate the wavelength of K X-rays from copper (Z=9) target. The value of the constant in Moseley s relation is 49.6 10 6 Hz ½. [] - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 1

MU - MIT I BT C H SCOND SMSTR- ND XAMINATION JULY 01 NGINRING PHYSICS 1A. Obtain an expression for the intensity of light in double-slit interference using phasor-diagram. [5] lectric field components at P due to S 1 and S are, 1 = o sin ωt & = o sin (ωt + φ) respectively. Phasor Rotating vector. Let two vectors be, 1 = o sin ωt & = o sin (ωt + φ) Resultant field = 1 + From phasor diagram, = 1 + = sin(t + ) = o cos sin(t + ) But = φ/. So above eqn can be written as, = o cos(/) sin(t+/) So intensity at an arbitrary point P on the screen due to interference of two sources having phase difference of : 1 ωt φ Where is the intensity due to single slit. Since = d sin θ / λ, = d sin θ / λ θ = angular position of the point P on the screen d = slit separation, = wavelength 1

MU - MIT I BT C H SCOND SMSTR- ND XAMINATION JULY 01 NGINRING PHYSICS 1B. Sketch schematically the graph of wavelength vs. intensity of radiation from a blackbody. xplain Planck s radiation law. [] Planck s radiation law: INTNSITY λ MAX WAVLNGTH I(, T) hc 5 e 1 I (,T) dλ is the intensity emitted in the wavelength interval dλ, at the wavelength λ from a blackbody, at temperature T. h = Planck s constant k = Boltzmann constant c = speed of light in vacuum 1 hc λkt 1C. The ionization energy of Na-atom is 5.1 ev. The electron affinity of Cl-atom is.7 ev. The dissociation energy of NaCl-molecule into atoms is 4. ev. Sketch schematically a graph of total energy of NaCl-molecule vs. distance between the Na + and Cl ions, indicating the relevant energy values and the bond-length (r o ). [] r BOND LNGTH Curve with proper shape and position - Proper labelling of axes - Proper labeling of the dissociation energy 4. ev and bond-length - 4 1

MU - MIT I BT C H SCOND SMSTR- ND XAMINATION JULY 01 NGINRING PHYSICS A. Sketch the potential-well diagram of finite height U and length L, obtain the general solution of the schrodinger equation for a particle of mass m in it. [5] [ for each main point. Max marks: 5 ] I II III U(x) = 0, 0 < x < L, U(x) = U, x < 0, x > L Particle energy = < U ; classically the particle is U permanently bound in the potential well. However, according to quantum mechanics, a finite probability exists that the particle can be found outside the well even if < U. That is, the wave function is generally 0 L nonzero in regions I and III. This is possible due to X the uncertainty principle. In regions II, where U = 0, the allowed wave functions are again sinusoidal. But the boundary conditions no longer require that the wave function must be zero at the ends of the well. SCHRÖDINGR QUATION OUTSID TH FINIT WLL IN RGIONS I & III: d m (U ) d dx C m, C (U ) dx x C x GNRAL SOLUTION OF TH ABOV QUATION: (x) A e C B A must be 0 in Region III and B must be zero in Region I, otherwise, the probabilities would be infinite in those regions. Solution should be finite. ie., the wave functions outside the finite potential well are e I = A e C x for x < 0 III = B e C x for x > L SCHRODINGR QUATION INSID TH SQUAR WLL POTNTIAL IN RGION II, WHR U = 0 or where GNRAL SOLUTION OF TH ABOV QUATION ψ F II m sin x G k m cos x k Results show that the wave function outside the potential well decay exponentially with distance. The boundary conditions require that I = II AT x = 0 AND II = III AT x = L So the wave function is smooth where the regions meet. To determine the constants A, B, F, G, & the allowed values of energy, apply the four boundary conditions and the normalization condition ψ ( 0) ψ ( 0) ψ ( L) ψ ( L) dψi dx I x0 II dψ dx II x0 dψ dx II II xl III dψ dx III xl dx 1 5 1

MU - MIT I BT C H SCOND SMSTR- ND XAMINATION JULY 01 NGINRING PHYSICS B. xplain Fermi-level, donor-levels and acceptor-levels; indicate their positions in the energy band diagram of a semiconductor. [] G CONDUCTION BAND VALNC BAND SMICONDUCTOR 1 D F A Fermi-level ( F ) is the energy level in the energy band diagram of a semiconductor where the probability of electron-occupation is half, at temperatures above zero kelvin. It lies in the energy band gap. Donor levels ( D ) are the localized energy levels (near the donor atoms) in the energy band diagram of an N-type semiconductor, which lie in the energy band gap, just below the conduction band. Acceptor levels ( A ) are the localized energy levels (near the acceptor atoms) in the energy band diagram of a P-type semiconductor, which lie in the energy gap, just above the valence band. C. xplain with diagram, the polarization of reflected light, incident at Brewster s angle. [] UNPOLARISD LIGHT DILCTRIC MATRIAL θ P θ R θ P POLARISD LIGHT AIR PARTIALLY POLARISD LIGHT When light is incident on the surface of a dielectric material at polarizing angle of incidence, (θ p = Brewster s angle) the reflected light is completely plane polarized and the transmitted ray is perpendicular to reflected ray. 6 1

MU - MIT I BT C H SCOND SMSTR- ND XAMINATION JULY 01 NGINRING PHYSICS A. Write a note on a semiconductor junction diode. [5] [ for each main point. Max marks: 5 ] pn-junction is formed when a p-type semiconductor is joined to an n-type semiconductor. A pn-junction has three distinct regions: a p-region, an n-region, and a depletion region at the junction [Figure (a)]. The depletion region has no movable charges (conduction electrons and holes) because the conduction electrons on the n-side have crossed the junction due to diffusion and neutralized the holes on the p-side. Because of the fixed ion cores in the depletion region, an electric field exists [Figure (b)], due to which there is a potential difference V o across the junction [Figure (c)]. In the forward bias, the p-side of the junction is made positive with respect to the n-side, by application of an external voltage ΔV. Then the internal potential difference ΔV o across the junction decreases. This gives rise to a current I which increases exponentially with the increase in forward bias ΔV. Δ In the reverse bias, the p-side of the junction is made negative with respect to the n-side, by application of a negative voltage ΔV. Then the internal potential difference ΔV o across the junction increases. This gives rise to a reverse current that quickly reaches a saturation value I o (reverse saturation current). FORWARD BIAS CHARACTRISTIC CURV OF A pn-junction RVRS BIAS The current in an ideal diode under the biasing voltage ΔV at temperature T is e V I Io exp 1 k T 7 1

MU - MIT I BT C H SCOND SMSTR- ND XAMINATION JULY 01 NGINRING PHYSICS B. Discuss qualitatively the diffraction due to multiple slits (eg, 5 slits), with diagrams. [] [ for each main point. Max marks: ] The diffraction of multiple slit arrangement will be the interference pattern multiplied by the single slit diffraction envelope. Condition for principal maxima, at diffraction angle θ is d sin θ = m where d is the separation between adjacent slits, is the wavelength of light, m is the order of the principal maximum. Location of principal maxima is independent of number of slits. IF ALL TH LABLS AR SHOWN IN TH DIAGRAM IF SCONDARY MAXIMA AR SHOWN BTWN TWO PRINCIPAL MAXIMA C. xplain metastable state with reference to lasers. [] A metastable state is an excited energy state of an atomic system from which spontaneous transitions to lower states is forbidden (not allowed by quantum mechanical selection rules). The average life time of the atomic system in the metastable state is of the order of 10 s which is much longer than that in an ordinary excited state. Stimulated transitions from the metastable state are allowed. An excited atomic system goes to metastable state (usually a lower energy state) due to transfer of its extra energy by collision with another atomic system. 8 1

MU - MIT I BT C H SCOND SMSTR- ND XAMINATION JULY 01 NGINRING PHYSICS 5A. The wavelengths of the first two lines in Lyman series of the hydrogen spectrum are 11.5 nm and 10.5 nm respectively. From these values calculate the wavelength of the first line in Balmer series. [5] H-atom energy level diagram ma h λ From the figure h λ h λ OR 1 1 λ 1 = 11.5 nm, h = 6.6 x 10 4 J.s, λ 1 = 10.5 nm c =.00 x 10 8 m/s = 1.94 x 10 18 J = 1.67 x 10 18 J =.0 x 10 19 J OR = 6.56 x 10 7 m = 656 nm = 656 nm [ Half the marks may be given for the correct answer obtained by other methods ] 10 1

MU - MIT I BT C H SCOND SMSTR- ND XAMINATION JULY 01 NGINRING PHYSICS 5B. Show that the average kinetic energy [ ] of a conduction electron in a metal at zero K is where F = Fermi-energy; the density of conduction electrons, ; the density of states, ; m = electron mass. [] For < F N() N() m h exp k T 1 F 8 m 8 m AV d OR AV n h 5 n h e 0 8 8 m h 1 e 1 F 5 F 1 n e 8 m h F AV 5 F 5C. Two polarizing sheets have their polarizing directions parallel so that the intensity of the transmitted light is a maximum. Through what angle must either sheet be turned if the intensity is to drop by one-half? [] or θ = 45 OR θ = 15 11 1

MU - MIT I BT C H SCOND SMSTR- ND XAMINATION JULY 01 NGINRING PHYSICS 6A. The J = 0 to J = 1 rotational transition of the CO molecule occurs at a frequency of 1.15 x 10 11 Hz. (i) Use this information to calculate the moment of inertia of the molecule. (ii) Calculate the bond-length of the molecule. (iii) What happens if another photon of frequency 1.15 x 10 11 Hz is incident on the CO molecule while it is in the J = 1 state? [5] (i) (ii) f 1 = 1.15 x 10 11 Hz, h = 6.6 x 10 4 J.s, m 1 = 1 u, m = 16 u u = 1.66 X 10 7 kg, J = 1,,,... h for J = 1 = 1.46 x 10 46 kg.m μ = 1.18 x 10 6 kg (iii) No excitation, photon simply passes by OR stimulated emission occurs. 6B. A thin film of acetone (index of refraction = 1.5) is coating a thick glass plate (index of refraction = 1.50). Plane light waves of variable wavelengths are incident normal to the film. When one views the reflected wave, it is noted that complete destructive interference occurs at 600 nm and constructive interference at 700 nm. Calculate the thickness of the acetone film? [] n = 1.5, 1 = 600 nm, = 700 nm Destructive interference: n d = (m ½) 1 Constructive interference: n d = (m 1) => m = 4 = 840 nm 6C. Use Moseley s relation to calculate the wavelength of K X-rays from copper (Z=9) target. The value of the constant in Moseley s relation is 49.6 10 6 Hz ½. [] C = 49.6 10 6 Hz ½, c =.00 x 10 8 m/s, Z = 9, Moseley s relation: f = C (Z 1) = 1.55 x 10 10 m 1 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 1 1