allen physics part 2

ALLENÒ CBSE 149 SOLUTION SECTION - A 1. The loss of electrons from the n-region and the gain of electron by the p-region causes a difference of potential across the junction of the two regions. The polarity of this potential is such as to oppose further flow of carriers so that a condition of equilibrium exists. Figure shows the p-n junction at equilibrium and the potential across the junction. The n-material has lost electrons, and p-material has acquired electrons. The n-material is thus positive relative to the p material. Since this potential tends to prevent the movement of electron from the n region into the p region, it is often called a barrier potential. 2 (a). node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper www.notesdrive.com (b). The photocurrent is directly proportional to the intensity of light this can be used for measuring the intensity of incident light. 3(a). (i) The below table gives a few differences between isotopes and isobars. Isotopes Isobars The mass number are different The mass number are same Atomic numbers are the same Atomic numbers are different (ii) No, because the different mass numbers A1 and A2 may be due to different atomic numbers or different neutron numbers. Only in the case where atomic number is the same would the two nuclei be isotopes of the same element. OR E

150 Physics ALLENÒ 3(b). (i) Metal should be photo- sensitive. (ii) For a given photosensitive material, threshold frequency is the minimum frequency of radiation that is required for photoelectric emission from the material. SECTION – B 4.(a) Energy E =hn n= = . × = 0.98 × 10 = 9.8 × 10 Hz .× (b) Work function f0 = 2.14 eV Energy of photon in eV E= . × = 4.06 .× KEmax = E - f0 KEmax. = 4.06 – 2.14 = 1.92 eV www.notesdrive.com 5. Given: l = 600 nm , nW = 4 = 1.33 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper3 Incident ray Reflected ray f, l, c f, l1, n 1 1 ii r Refracted ray f, l2, n 2 2 Frequency of incident ray C = fl Þ f = c = 3 ´ 108 l 600 ´10-9 f = 5´1014 Hz Reflected ray Refracted ray Wavelength (l1) = 600 nm Wavelength l2 = n1l1 = 1´ 600 = 450 nm n2 4 3 Frequency (f1) = 5 × 1014Hz Frequency Speed (v1) = 3 × 108 m/s (f2) = 5 × 1014Hz (Constant) Speed: V2 = C nw V2 = 3 ´ 108 4/3 V2 = 2.25 × 108 m/s E

ALLENÒ CBSE 151 6. (a) (i) In radar system Þ microwaves In water purifies Þ UVrays In remote switches in TV Þ Infrared rays (ii) Microwaves are produced by special vacuum tubes (Klystrons, Magnetrons and Gunn diodes) (iii) UV radiation in produced in welding arc and the sun is an important source of ultraviolet light. (iv) Infrared waves are produced by hot bodies and molecules. OR (b) (i) Two sources are said to be coherent if- node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper(1) These produces waves of same frequency with www.notesdrive.com(2) Constant or zero phase difference (ii) Interference is the result of superposition of secondary wavelets from two different slits while diffraction results due to superposition of secondary wavelets from different parts of the same source slit. In interference pattern all maxima are equally bright while in case of diffraction maxima are of decreasing intensity. 7. Bohr's postulate: Electron revolves around the nucleus in those orbits for which the angular momentum is an integral multiple of h/2p. mnr = nh 2p Speed of e– in nth orbit For hydrogen radius of nth orbit is given by rn = e0n2h2 vn pme2 From Bohr's Postulate Å rn e– mn n rn = nh 2p mnn æ e0n2h2 ö = nh ç pme2 ÷ 2p è ø e2 nn = 2e0hn nn µ 1 n E

152 Physics Semiconductors ALLENÒ Insulators 8. Energy band diagrams Conductors Conductors C.B C.B E C.B Over Eg = EC – EV Eg > 3eV lapping V.B V.B. E V.B For Conductor Eg = 0 For Conductor Eg < 3eV For Insulator Eg > 3eV www.notesdrive.comConduction band determines the electrical conductivity of solid. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question PaperOn increasing the temperature of semiconductor electrical conductivity increases. Because when we increase temperature, some of the covalent bond got broken and e-hole pairs are generated. 9. Energy of proton = 4.1 MeV, Z = 82 (i) Speed of a proton 1 mP n2p = 4.1 × 106 × 1.6 × 10–19 2 n2p = 2 ´ 4.1´1.6 ´10-13 1.67 ´10-27 = 7.85 × 1014 n 2 = 2.8´107 m /s p (ii) Distance of closest approach K.E = ke(ze) r0 r0 = KZe2 = 9´109 ´82 ´ (1.6 ´10-19 )2 = 288 × 10–16 r0 4.1´106 ´1.6 ´10-19 r0 = 2.9 ´10-14 m 10. Spacing between fringes in double slit experiment is given by b = lD d Where, l Þ Wavelength of light; d Þ Slit separation; D Þ Distance between slit and screen (i) On increasing 'd' fringe width decreases (ii) lred > lblue , So 'b' decreases (iii) When whole apparatus is immersed in oil of refractive index 1.2, then wavelength decreases. l ' = l 1.2 b ' = l'D Þ b ' = b d 1.2 It means fringe width decreases. E

ALLENÒ CBSE 153 11. (a) (i) Light ray should travel from denser to rarer medium. (ii) Angle of incidence should be greater than critical angle (i > ic). (b) A 45° 45° B C 60° 30° 60° 60° 60° node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper OR www.notesdrive.com(a) (i) For negative magnification object is placed beyond the focus: m= v A u B' using sign convention BF m = - v u 0 F u focus onptical between ceAn'tre. (ii) For positive magnification object is placed and m= v A' u A Using sign convention m = –v B' F B u O –u n F v m= u A (b) B DI v u' O 3r r I' C 9r Refraction through spherical surface ABC n2 - n1 = n2 - n1 ìRe fractive index of glass = 3/2 v' u R íîu = –3r 3 - 1 = æ 3 - 1 ö÷ø (-3r) çè 2 r 2 v' 3 = 1 - 1 Þ v' = 9r 2v' 2r 3r E

154 Physics ALLENÒ Refraction through spherical surface ADC For refracting surface ADC, image I' acts as a virtual object and I is final image n2 - n1 = n2 - n1 v u' R 1 - 1.5 = 1 - 1.5 (u' = 9r – 2r = 7r) v u' -r 1 - 1.5 = 0.5 v 7r r 1 = 1 + 1.5 v 2r 7r 1 = 1.5 + 0.5 = 1.5 + 3.5 Þ v = 7r v 7r r 7r 5 Distance of final image from 'B' www.notesdrive.com = 7r + 2r = 17r = 3.4r node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\04_ 05_CBSE Sample Paper & Question Paper55 SECTION – C 12. (I) B (II) A (III) D (IV) D Hint : M = M0 × Me = 10 × 20 = 200 (V) C Sol. f0 = 1.2 cm, fe = 3.0 cm u0 = –1.25 cm, V0 = ? Using lens formula 1 = 1 - 1 Þ 1 = 1 + 1 f0 v0 u0 1.2 v0 1.25 5 = 1 + 4 Þ 1 = 5 - 4 6 V0 5 V0 6 5 1 = 25 - 24 = 1 Þ V0 = 30cm V0 30 30 When final image is formed at infinity then magnifying power M = V0 æ Dö u0 çè fe ÷ø = 30 æ 25 ö Þ M = 200 1.25 èç 3 ø÷ E