DC Pandey Optics And Modern Physics

20 Optics & Modern Physics f 43. A freshly prepared sample of a e 45° g radioisotope of half-life 1386 s has activity 103 disintegrations per second. Given that m1 h ln 2 = 0.693, the fraction of the initial i number of nuclei (expressed in nearest m2 m3 integer percentage) that will decay in the first 80 s after preparation of the sample is (Single Integer Type, 2013) Match the paths in Column I with 44. The radius of the orbit of an electron in a conditions of refractive indices in Column II and select the correct answer using the Hydrogen-like atom is 4.5 a0 where a0 is codes given below the columns. the Bohr radius. Its orbital angular (Matching, Type 2013) momentum is 3h . It is given that h is 2p Column I Column II Planck constant and R is Rydberg P. e ® f Q. e ® g 1. m1 > 2 m 2 constant. The possible wavelength(s), R. e ® h 2. m1 >m1 and m 2 >m 3 S. e ® i 3. m1 = m 2 when the atom de-excites, is (are) 4. m 2 < m1 < 2 m 2 and (More than One Correct Option, 2013) m2 >m3 (a) 9 (b) 9 32R 16R 9 4 (c) 5R (d) 3R Codes RS P QR S Passage (Q. Nos. 45-46) PQ 14 (b) 1 2 4 3 23 (d) 2 3 4 1 The mass of a nucleus A X is less that the sum of (a) 2 3 Z (c) 4 1 the masses of (A - Z ) number of neutrons and Z 41. A point source S is placed at the bottom of number of protons in the nucleus. The energy a transparent block of height 10 mm and equivalent to the corresponding mass difference refractive index 2.72. It is immersed in a is known as the binding energy of the nucleus. A lower refractive index liquid as shown in heavy nucleus of mass M can break into two the figure. It is found that the light light nuclei of masses m1 and m2 only if (m1 + m2) < M. Also two light nuclei of masses m3 emerging from the Liquid and m4 can undergo complete fusion and form a block to the liquid heavy nucleus of mass M ¢ only if (m3 + m4 ) > M ¢. The masses of some neutral atoms are given in forms a circular Block bright spot of S the table below : (Passage Type, 2013) diameter 11.54 mm on the top of the block. The 11H 1.007825u 12H 2.014102u 13H 3.016050u refractive index of the liquid is 36Li 6.01513u 37Li 7.016004u 4 He 4.002603u 2 (Single Correct Option, 2013) (a) 1.21 (b) 1.30 (c) 1.36 (d) 1.42 16542Gd 151.919803u 82026Pb 205.974455u 70 Zn 69.925325u 30 42. A pulse of light of duration 100 ns is 82 Se 81.916709u 28309Bi 208.980388u 82041Po 205.974455u absorbed completely by a small object 34 initially at rest. Power of the pulse is 30 mV and the speed of light is 45. The correct statement is 3 ´ 108 ms-1. The final momentum of the (a) The nucleus 36Li can emit an alpha particle object is (Single Correct Option, 2013) (b) The nucleus 82140Po can emit a proton (c) Deuteron and alpha particle can undergo complete fusion (a) 0.3 ´ 10-17 kg-ms-1 (b) 1.0 ´ 10-17 kg-ms-1 Tchoemnpulectleeifu37s00iZonn and 82 (c) 3.0 ´ 10-17 kg -ms-1 (d) 9.0 ´ 10-17 kg-ms-1 (d) 34 Se can undergo

Previous Years’ Questions (2018-13) 21 46. The kinetic energy (in keV) of the alpha excited state of the nucleus (126C*) at particle, when the nucleus 82140Po at rest 4.041 MeV above its ground state. If 12 B undergoes alpha decay, is 5 (a) 5316 (b) 5422 (c) 5707 (d) 5818 decays to 12 C *, the maximum kinetic 5 energy of the b-particle in units of MeV is (1 u = 931.5MeV / c2, where c is the speed 47. The isotopes 12 B having a mass 12.014 u 5 of light in vacuum) (Single Integer Type, 2013) undergoes b-decay to 12 C . 12 C has an 6 6 Answer with Explanations 1. (130.0 kW / m2) Þ IPQ = f PQ = èçæ f ÷øö çæ 2 ( AB)x ÷ö x x è fø I0 é PQ = 2( AB)x ù A0 ëêQ f ûú I'0 A'0 IPQ = 2 AB 2cm (Size of image is independent of x. So, final image will be of same height terminating at infinity) 20 cm 3. (a, c) 232 Th is converting into 82212Pb. 90 A0¢ æç 2 ÷ö 2 1 A0 Change in mass number ( A) = 20 A0 è 20 ø 100 100 = = Þ A0¢ = \\Number of a-particle emitted = 20 = 5 4 P = I0 A0 = I0¢ A0¢ Due to 5 a-particles, Z will change by 10 units. I0 A0 Þ I0¢ = A0 = 100I0 = 130 kW/m2 Since, given change is 8, therefore number of 100 b-particles emitted is 2. 2. (d) Image of point A 4. (24) \\ Power = nhf B (where, n = number of photons incident per second) Q Since, KE = 0, hf = work-funcition W 200 = nW = n [6.25 ´ 1.6 ´ 10-19] O f/2 Px C Þ n = 1.6 ´ 200 ´ 6.25 A 10-19 f/2 As photon is just above threshold frequency KEmax is zero and they are accelerated by potential difference of PQ = AB Þ PQ = 2 ( AB) x 500 V. x f /2 f \\ KEf = qDV P2 For A : 1 1 1 2m = qDV Þ P= 2 mq DV v (f /2 )] -f + [- = Þ v=f Since, efficiency is 100%, number of electrons emitted per second = number of photons incident per second. Þ IAB = - v = - f öø÷ Þ IAB = 2 AB As, photon is completely absorbed, force exerted AB u çèæ f - 2 = n(mV ) = nP = n 2 mqDV For height of PQ, = 200 ´ 2 (9 ´ 10-31) ´ 1.6 ´ 10-19 ´ 500 6.25 ´ 1.6 ´ 10-19 1+ 1 x)] = 1 v - [(f - -f = 24 Þ 1 = (f 1 x) - 1 Þ v = f(f - x) 5. (3) DE2-1 = 13.6 ´ Z2 éëê1 - 1ù = 13.6 ´ Z2 é3ù v - f x 4 úû êë 4ûú Þ IPQ = -v = f(f - x) = æç f ö÷ DE3- 2 = 13.6 ´ Z2 é1 - 1ù = 13.6 ´ Z2 é5ù PQ u x[(f - x)] è xø êë 4 9 ûú êë 36ûú

22 Optics & Modern Physics \\ DE2 = DE3-2 + 74.8 at P1 Dx = 0 13.6 ´ Z2 é3ù = 13.6 ´ Z2 é5ù + 74.8 at P2 Dx = 1. 8 mm = nl ëê 4úû ëê 36ûú Number of maximas will be 1. 8 mm Z2 é3 5ù n = Dx = 600 nm = 3000 êë 4 36 úû l 13.6 ´ - = 74.8 at P2 Dx = 3000l Z2 = 9 Hence, bright fringe will be formed. \\ Z=3 At P2, 3000 th maxima is formed. For (a) option 6. (a,b,c) The minimum deviation produced by a prism Dx = d sin q Þ dDx = d cosdq dm = 2i - A = A \\ i1 = i 2 = A and r1 = r2 = A /2 Þ r1 = i1 /2 Rl = d cos qRdq Þ Rdq = Rl Now, using Snell’s law d cos q sin A = m sin A /2 Þ m = 2 cos ( A /2 ) As we move from P1 to P2 q ­ cos q ¯ Rdq ­ A 9. (5) 1131 ¾T¾1/ 2¾= 8¾Da¾ys ® Xe131 + b i1 r1 i2 A 0 = 2.4 ´ 105 Bq = lN0 r2 Let the volume is V, t = 0 BC A0 = lN0 Þ t = 11. 5 h, A = lN For this prism when the emergent ray at the second 115 = lçæè N ´ 2 . 5 ø÷ö 115 = l ´ 2 . 5 ´ (N0e -lt ) surface is tangential to the surface V V i 2 = p /2 Þ r2 = qc Þ r1 = A - qc 115 = (N0l) ´ (2 . 5) - 8lnd2ay(11.5 h) V ´e so, sin i1 = m sin( A - qc ) 115 = (2 . 4 ´ 105) ´ (2 . 5) ´ e -1/24 V so, i1 = sin-1éêsin A 4cos 2 A - 1 - cos ù ë 2 Aú 2 . 4 ´ 105 ´ 2 . 5 éëê1 - 1ù û = 115 24 ûú For minimum deviation through isosceles prism, the ray V inside the prism is parallel to the base of the prism if = 2 . 4 ´ 105 ´ 2 . 5éëê2243 ù ÐB = ÐC. 115 úû But it is not necessarily parallel to the base if, = 105 ´ 23 ´ 25 = 5 ´ 103 ml = 5 L ÐA = ÐB or ÐA = ÐC 115 ´ 102 7. (8) But this value of refractive index is not possible. 1. 6 sin q = (n - mDn)sin 90° 10. (d) According to photoelectric effect equation 1. 6 sin q = n - mDn KEmax = hc - f0 Þ p2 = hc - f0 [KE = p2 /2 m] l 2m l 1. 6 ´ 1 = 1. 6 - m(0.1) Þ 0. 8 = 1. 6 - m (0.1) 2 (h/ ld )2 2m = hc - f0 [ p = h/ l] m ´ 0.1 = 0.8 Þ m = 8 l 8. (c,d) Assuming small changes, differentiating both sides, P1 h2 ççèæ - 2dld öø÷÷ = - hc dl Þ dld µ l3d Dx= d sinq 2m l3d l2 dl l2 d sinq q P2 11. (a) According to question, S1 S2 d P q r1 N a r2 l = 600 nm M R Q

Previous Years’ Questions (2018-13) 23 Applying Snell's law at M, Radius of curvature of convex surface is 30 cm. n= sin a Þ 2 = sin 45° Faint image is erect and virtual. Focal length of lens is sin r1 sin r1 20 cm. Þ sin r1 = sin 45° = 1/ 2 = 1 r1 = 30° 13. (a,c,d) From Snell's law, n sin q = constant 2 2 2 \\ n1 sin qi = n2 sin qf sin qc = 1 = 1 Þ qc = 45° Further, l will depend on n1 and n( z). But it will be n 2 independent of n2. Let us take r2 = qc = 45° for just satisfying the condition 14. (d) l2 of TIR. In DPNM, 50 4–Ö3 4 5–0ÖÖ33 q + 90°+ r1 + 90°-r2 = 180° r or q = r2 - r1 = 45°- 30°= 15° Note If a > 45° (the given value). Then, r1 >30° (the a obtained value) \\ r2 > qc (as r2 - r1 = q O 30° 25 I1 50 30° or r2 = q + r1 ) 25/2 25Ö3 or TIR will take place. So, for taking TIR under all 2 conditions a should be greater than 45° or this is the minimum value of a. 12. (a,d) Case 1 For Lens 1 - 1 = 1 Þ v = uf vu f u+ f I Þ v = (-50)(30) = 75 cm - 50 + 30 n For Mirror 1 + 1 = 1 Þ v = uf vuf u- 30 cm 60 cm f Using lens formula, ççæè 25 3 ø÷÷ö ( 50) - 50 3 2 3 - 50 4- 3 1 + 1 = 1 Þ 1 = 1 + 2 Þ f1 = 20 cm Þ v = = cm 60 30 f1 f1 60 60 25 1 (n - 1)èçæ 1 1 øö÷ 2 f1 R ¥ Further, = - æç - 50 3 ÷ö Þ f = R = + 20 cm m= - v = h2 Þ h2 = - ç 4- 3 ÷ × 25 n-1 u h1 ç 25 3 ÷ 2 1 èçç ø÷÷ 2 Case 2 + 50 h2 = 4- 3 cm 10 cm nI The x-coordinate of the images 30 cm = 50 - v cos 30° + h2cos 60° » 25 cm The y-coordinate of the images Using mirror formula, = v sin 30° + h2sin 60° » 25 3 cm 1 - 1 = 1 Þ 3 - 1 = 1 = 2 15. (a,b) 10 30 f2 30 30 f2 30 f2 = 15 = R Þ R = 30 Þ R = 30 cm 2 d R 30 n-1 = +20 cm = n-1 O S1 S2 Þ 2 n - 2 = 3 Þ f1 = + 20 cm Path difference at point O = d = 0.6003 mm Refractive index of lens is 2.5. = 600300 nm.

24 Optics & Modern Physics This path difference is equal to çæ1000 l + l ÷ö. èæç 1 øö÷ n è 2 ø 2 1 = 64 Þ Minima is formed at point O. Solving we get, n = 6 Line S1S 2 and screen are ^ to each other so fringe Now, t = n(t1/2) = 6(18 days) = 108 days pattern is circular (semi-circular because only half of screen is available) 20. (c) Electrostatic energy = Binding energy of 16. (b) hc - f = eV0 (f = work function) N – Binding energy of O l = {[7MH + 8Mn - MN ] - [8MH + 7Mn - MO ]} ´ C 2 = [- MH + Mn + MO -MN ] C 2 hc - f = 2e …(i) 0.3 ´ 10-6 = [- 1.007825 + 1.008665 + 15.003065 hc - f = 1e …(ii) - 15.000109] ´ 93.15 0.4 ´ 10-6 = + 3.5359 MeV Subtracting Eq. (ii) from Eq. (i) 3 ´ 1.44 ´8 ´7 3 ´ 1. 44 ´ 7 ´6 1 1 0.1 DE = 5 R - 5 R çæ 0.3 - 0.4 ö÷106 =e Þ æç 0.12 ´ 106 ö÷ =e hc è ø hc è ø = 3.5359 MeV h = 0.64 ´10-33 = 6.4 ´ 10-34 J-s R = 3 ´ 1.44 ´ 14 = 3.42 fm 5 ´ 3.5359 17. (a,b,d) As radius r µ n2 21. (3) z Þ Dr = çæ n + 1÷ö2 - çæ n ö÷2 = 2n + 1 » 2 µ 1 S1 x P r èz ø è zø n2 n n d æç n ÷ö2 è zø O z2 z2 d n2 + 1)2 S2 z2 DE - (n n2 E Energy E µ Þ = z2 m(S 2P) - S1P = ml Þ m d 2 + x2 - d 2 + x2 = ml (n + 1)2 = (n + 1)2 - n2 ×(n + 1)2 Þ DE = 2 n+ 1 ~- 2n µ 1 Þ (m - 1) d 2 + x2 = ml n2 × (n + 1)2 E n2 n2 n Þ èæç 4 - 1÷øö d2 + x2 = ml or d 2 + x2 = 3ml Angular momentum = nh 3 L 2p DL (n + 1)h - nh 1 1 Squaring this equation we get, L 2p 2p n n x2 = 9m2l2 - d 2 Þ p2 = 9 or p = 3 Þ = = µ nh 22. (7) Case I Reflection from mirror 2p 1= 1+ 1 Þ 1 = 1 + 1 18. (6) Energy of incident light (in eV) fvu -10 v -15 E = 12375 = 12.7 eV Þ v = - 30 970 air air After excitation, let the electron jumps to nth state, then O I2 - 13.6 = - 13.6 + 12.7 n2 15 Solving this equation, we get n = 4 30 I1 20 20 \\Total number of lines in emission spectrum, = n(n - 1) = 4(4 - 1) = 6 1= 1 -1 1 1- 1 2 2 f v u 10 v -20 n For lens Þ = Þ v = 20 19. (c) Using the relation, R = R 0 çèæ 1 öø÷ v1 v2 çæ 30 ÷ö æç 20 ö÷ 2 u1 u2 è 15 ø è 20 ø |M1| = = Here, R is activity of radioactive substance, R0 initial activity and n is number of half lives. =2 ´1=2 (in air)

Previous Years’ Questions (2018-13) 25 Case II For mirror, there is no change. v = - 30 60° N medium 1.5 medium M O (7/6) (7/6) 60° rq 60°– r 30 I1 20 140 For lens, 1 = çæ 3 /2 - 1ö÷ èççæ 1 - 1 ÷ø÷ö Differentiating Eq. (i), n cos r dr + sin r = 0 fair è 1 ø R1 R2 dn or dr = - sin r = - tan r dn n cos r n 1 æç 3 / 2 - 1÷ö çæèç 1 1 ÷ø÷ö fmedium = è 7 / 6 ø R1 - R2 Þ cos q dq = - n cos (60°-r) æç - tan r ÷ö + sin (60°-r) dn è n ø with fair = 10 cm dq = 1 [cos(60°- r ) tan r + sin (60°-r)] We get 1 4 cm-1 1- 1 4 dn cos q fmedium = 70 Þ v -20 = 70 1 1 èæç 2 øö÷ çæè 2 ø÷ö 4 1 4 1 Form Eq. (i), r = 30° for n = 3 v 20 7 10 70 v 70 20 + = = Þ = - dq = 1 (cos 30 ´ tan 30 + sin 30) = 2 dn cos 60 v = 140, |M2| = v1 v2 = æç 30 ö÷ æç 140 ÷ö, 25. (a) 4 sin i = 45 sin(90 - q )= 45 cos qc u1 u2 è 15 ø è 20 ø 3 4 4 c = (2 ) æèç 140 ÷øö = 14 Þ ½½M 2½½ = 14 = 7 sin qc = n2 Þ cos qc = 1 - çæçè n2 ø÷ö÷ 2 20 ½M1½ 2 n1 n1 23. (b) R = 10 cm Þ 4 sin i = 45 3 ; sin i = 9 3 4 45 16 Applying m 2 - m1 = m 2 - m1 two times vu R In second case, sin qc = n2 = 7 n1 8 1 1.5 1 - 1.5 v - -50 = -10 15 16 8 8 3 15 5 1 1.5 0.5 Þ cos qc = Þ sin i = sin(90 - qc ) v 50 10 Þ + = 9 16 1 0.5 1.5 2.5 - 1.5 Simplifying we get, sin i = v 10 50 50 = - = (a) is correct. Þ v = 50 26. (d) sin i m = n1 sin (90 - qc ) MN = d, MI1 = 50 cm Þ NI1 = (d - 50) cm Þ sin i m = n1 cos qc Þ NA = n1 1 - sin2 qc +ve I1 = n1 1- n22 = n12 - n22 O MN n12 Again, 1.5 - 1 = 1.5 - 1 Substituting the values we get, ¥ (d - 10 - 50) NA1 = 3 and NA2 = 15 = 3 Þ NA2 < NA1 4 5 4 1 = 1 Þ d = 70 d - 50 20 Therefore, the numerical aperture of combined structure is equal to the lesser of the two numerical 24. (2) Applying Snell’s law at M and N, aperture, which is NA2. sin 60° = n sin r ...(i) 27. (3) Let initial power available from the plant is P0. After ...(ii) sin q = n sin (60 - r) çæ 1 ÷ö n è 2 ø Differentiating we get time t = nT or n half lives, this will become P0. cos q dq = - n cos (60 - r) dr + sin (60 - r) dn dn

26 Optics & Modern Physics Now, it is given that, Number of fringes in a given width æçè 1 ø÷ö n = Y or µ 1 Þ m2 < m1 as b 2 > b1 2 b b P0 = 12.5% of P0 = (0.125) P0 m m Solving this equation we get, n = 3 Distance of 3rd maximum of l2 from central maximum = 3l2D = 1800D 28. A-(p, q), B- (p, r), C-(p, s) D- (p, q, r) dd 29. (2) Let initial numbers are N1 and N2. Distance of 5th minimum of l1 from central maximum l1 = t2 = 2t = 2 = T2 (T = Half life) = 9l1D = 1800D l2 t1 t T1 2d d A = - dN = lN So, 3rd maximum of l2 will overlap with 5th minimum dt of l1 Angular separation (or angular fringe width) = Initial activity is same lµl \\ l1N1 = l2N2 …(i) d Activity at time t, Þ Angular separation for l1 will be lesser. A = lN = lN0 e - lt Þ A1 = l1N1 e - l1t 33. (a,c) 1 = (n1 - 1)æçè 1 - 1 ø÷ö Þ ffilm = ¥ (infinite) ffilm R R Þ = - dA1 = l21 e- l1 R1 dt N1 \\ There is no effect of presence of film. Similarly, R2 = l22 N2 e - l2t From Air to Glass Using the equation n2 - 1 = n2 - 1 After t = 2 t l1t = 1 (t ) = 1 (2 t ) = 2 vu R t1 t 1.5 1 1.5 - 1 v - ¥ = R Þ v = 3R Þ f1 = 3R 1 1 l2t = t2 (t ) = 1 = 2t (2 t ) = 1 From Glass to Air Again using the same equation RP = l12 N1 e -l1t Þ RP = l1 æèçç e -2 ÷øö÷ = 2 1- n2 = 1 - n2 Þ 1 - 1.5 = 1 - 1.5 RQ l22 N2 e -l2t RQ l1 e -1 e v u -R v ¥ -R 30. (2) Angular momentum = n çæè h ÷öø = 3æèç h ÷øö Þ v = 2 R Þ f2 = 2 R 2p 2p 34. (b) (P) 1= çæè 3 - 1÷øö çæè 1 + 1 ÷öø = 1 \\ n=3 f 2 r r r Now, rn µ n2 Þ r3 = (3)2 (a0 ) = 3a0 Þ f=r z 3 Þ 1 =1+ 1=2 Þ feq = r çæ h ÷ö 3æç h ö÷ feq f f r 2 Now, mv 3r3 = 3 è 2p ø Þ mv3(3a0) = è 2p ø or h = 2 pa0 Þ h = 2 pa0 (P = mv ) (Q) 1 = æç 3 - 1÷ö çæ 1ö÷ Þ f = 2r mv 3 P3 f è 2 ø èrø or l3 = 2 pa0 æç l = h ö÷ Þ 1+ 1=2 =1 Þ feq = r è Pø fffr \\ Answer is 2. (R) 1 = çèæ 3 - 1÷öø çæè - 1 öø÷ = - 1 f 2 r 2r 31. (d) From conservation laws of mass number and atomic number, we can say that x = n, y = n Þ f = -2r ( x = 10n, y = 10n) 1 = 1+ 1= 2 feq f f 2r \\ Only (a) and (d) options may be correct. Þ - Þ feq = -r From conservation of momentum,|Pxe| = |Psr| P2 From K = 2m Þ K µ 1 (S) Þ1 = 1+ 1 = 1 Þfeq = 2r m feq r -2 r 2r Ksr = mxe Þ K sr = 129 MeV Þ K xe = 86 MeV 35. (a) Energy corresponding to 248 nm wavelength K xe msr 32. (a, b, c) Fringe width b = lD or b µ l Þ l2 > l1 = 1240 eV = 5 eV 248 d So b 2 > b1

Previous Years’ Questions (2018-13) 27 Energy corresponding to 310 nm wavelength 40. (d) For e ® i = 1240 eV = 4 eV 4125°>>mmq12c Þ sin 45° > sin qc 310 Þ m1 > 2 m 2, (s-1) KE1 = u12 = 4 = 5 eV - W KE2 2 1 4 eV - W u 2 For e ® f Þ 16 - 4W = 5 - W Þ 11 = 3W angle of refraction is lesser than angle of incidence, so Þ W = 11 = 3.67 eV~= 3.7 eV m 2 > m1 and then m 2 > m 3 (P-2) 3 For e ® g, m1 = m 2 (Q-3) 36. (b) K a transition takes place from n1 = 2 to n2 = 1 for e ® h, m 2 < m1 < 2m 2 and m 2 > m 3 (R-4) \\ 1 = R (Z - b )2 é1 - 1ù 41. (c) At point Q angle of incidence is critical angle qC, l êë(1)2 (2 )2 ûú where For K-series, b = 1 Þ 1 µ (Z - 1)2 QP Rarer medium l qC Þ lCu = ( zMo - 1)2 = (42 - 1)2 qC h lMo ( zCu - 1)2 (29 - 1)2 = 41 ´ 41 = 1681 = 2.14 Denser medium 28 ´ 28 784 S QP = r 37. (c) m = lair = 1 = 3 sin qC = ml lmedium (2/3) 2 m block ml Further, |m| = 1 = v In DPQS,sin qC = r Þ m block = r 3 u r2 + h2 r2 + h2 \\ |v| = |u| Þ u = - 24 m (Real object) Þ ml = r h2 ´ 2.72 = 5.77 ´ 2.72 = 1.36 3 r2 + 11.54 \\ v = + 8m (Real image) Final momentum of object = Power ´ time Speed of light Now, 1 1 1 = (m 1)æçè 1 1 ÷øö 42. (b) v u f +R ¥ - = - - 1 1 æç 3 1ö÷ æç 1 ÷ö = 30 ´ 10-3 ´ 100 ´ 10-9 = 1.0 ´ 10-17 kg-m/s -1 8 24 è 2 øè R ø 3 ´ 108 \\ + = - Þ R = 3m 38. (a) Component along the plane = 1 43. (4) Number of nuclei decayed in time t , 2 3 Nd = N0(1 - e -lt ) 2 and component perpendicular to the plane = \\ % decayed = èççæ Nd ÷÷öø ´ 100 = (1 - e -lt d ) ´ 100 …(i) N0 æèç 1 ö÷ø \\ tan i = 2 = 1 Here, l = 0.693 = 5 ´ 10-4 s-1 3 1386 èççæ 3 ÷÷øö 2 \\% decayed » (lt ) ´ 100 = (5 ´ 10-4 ) (80) (100) = 4 \\ i = 30° = angle of incidence 44. (a, c) L = 3 çæ h ö÷ è 2p ø f 39. (b) I = Imax cos 2 2 …(i) l1 l2 n=3 …(ii) l3 n=2 Given, = Imax I 2 \\From Eqs. (i) and (ii), we have n=1 f = p , 3p , 5p \\ n = 3, as L = n æç h ÷ö Þ rn µ n2 2 2 2 è 2p ø z Or path difference, Dx = æèç l ÷öø × f r3 = 4.5a0 \\ z = 2 2p 1 çæè 1 1 ÷öø çæè 1 1 ÷øö \\ Dx = l , 3l , 5l K æç 2 n+ 1÷ö l l1 = Rz2 22 - 32 = 4R 4 - 9 4 4 4 è 4 ø

28 Optics & Modern Physics \\ l1 = 9 46. (a) 84Po 210 ¾® 2He 4 + 82Pb206 5R Mass defect Dm = (mPo - MHe - mPb ) = 0.005818 u Þ 1 = Rz2 æç 1 - 1 ÷ö = 4R çæ1 - 1 ö÷ \\ Q = (Dm) (931.48) MeV l2 è 12 32 ø è 9 ø = 5.4193 MeV = 5419 keV Þ l2 = 9 a Pb 32 R Þ 1 = Rz2 æç 1 - 1 ÷ö = 4R çæ1 - 1 ÷ö From conservation of linear momentum, pPb = pa l3 è 12 22 ø è 4 ø \\ 2 mPb kPb = 2 ma ka Þ l3 = 1 Þ ka = mPb = 206 3R kPb ma 4 45. (a) 3Li7 ® 2He 4 + 1H3 \\ ka = ççèæ 206 4 öø÷÷ (ktotal ) 206 + Dm = [MLi - MHe - MH3 ] = [6.01513 - 4.002603 - 3.016050] = æç 206 ÷ö (5419) = 5316 keV = - 1.003523 u è 210 ø Dm is negative so reaction is not possible. 47. (9) 12 B ¾®162 C + -01e +n (b) 84Po 210 ® 83Bi209 + 1P1 5 Dm is negative so reaction is not possible. Mass of 162C = 12.000 u (by definition of 1 a.m.u.) (c) 1H2 ® 2He 4 + 3Li6 Q-value of reaction, Q = (MB - MC ) ´ c 2 Dm is positive so reaction is possible. (d) 30Zn70 + 34Se 82 ® 64Gd152 = (12.014 - 12.000) ´ 931.5 = 13.041 MeV 4.041 MeV of energy is taken by 162C * Dm is positive so reaction is not possible. Þ Maximum KE of b-particle is (13.041- 4.041) = 9 MeV