Title 11.DUAL NATURE OF RADIATION AND MATTER -Explanations.pdf ✅

1 (c) E = W0 + Kmax ⇒ hc λ1 = W0 + E1 and hc λ2 = W0 + E2 ⇒ hc = W0λ1 + E1λ1 and hc = W0λ2 + E2λ2 ⇒ W0λ1 + E1λ1 = W0λ2 + E2λ2 ⇒ W0 = E1λ1 − E2λ2 (λ2 − λ1) 2 (b) Vidicon is basically VID (eo) + ICON (oscope). It is a small televison camera tube that forms a charge density image on a photoconductive surface for subsequent electron-beam scanning 3 (b) λ = h √2mK ⇒ λ1 λ2 = √ K2 K1 = √ 16K K = 4 λ1 λ2 = 4 ⇒ λ2 = λ1 4 = 100 4 = 25 4λ = 100 − 25 = 75% 4 (a) The wave length of Lα line is given by 1 λ = R(z − 7.4) 2 ( 1 2 2 − 1 3 2 ) ⇒ λ ∝ 1 (Z − 7.4) 2 ⇒ λ1 λ2 = (z2 − 7.4) 2 (z1 − 7.4) 2 ⇒ 1.30 λ2 = (42 − 7.4) 2 (78 − 7.4) 2 ⇒ λ2 = 5.41 Å 5 (c) Here, E1 = E2 n1hv1 = n2hv2 So, n1 n2 = v2 v1 6 (a) 1 2 mv 2 = hv − φ0 = hv − hv0 For minimum kinetic energy of emitted photoelectron, v = v0 ∴ 1 2 mv 2 = 0 7 (d) Target should be of high atomic number and high melting point 10 (d) qE = mg ... (i) 6πη rv = mg 4 3 πr 3 ρg = mg ... (ii) ∴ r = ( 3mg 4πρg ) 1/3 ... (iii) Substituting the value of r in Eq. (ii), we get 6πηv ( 3mg 4πpg) 1/3 = mg or (6πηv) 3 ( 3mg 4πρg ) = (mg) 3 Again substituting mg = qE, we get (qE) 2 = ( 3 4πρg ) (6πηv) 3 Or qE = ( 3 4πρg ) 1/2 (6πηg) 3/2 ∴ q = 1 E ( 3 4πρg ) 1 2 (6πηv) 3/2 Substituting the values, we get q = 7 81π × 105 √ 3 4π × 900 × 9.8 × 216π 3 × √(1.8 × 10−5 × 2 × 10−3) 3 = 8.0 × 10−19 C 11 (b) Momentum of incident light per second p1 = E c = 60 3 × 108 = 2 × 10−7 Momentum of reflected light per second p2 = 60 100 × E c = 60 3 × 108 = 1.2 × 10−7 Force on the surface = change in momentum per second = p2 − (−p1 ) = p2 + p1 = (2 + 1.2) × 10−7 = 3.2 × 10−7N 12 (b) The graph between stopping potential and frequency is a straight line, so stopping potential and hence, maximum kinetic energy of photoelectrons depends linearly on the frequency. 15 (a) Refer to threshold frequency 16 (b) Energy E = hv = h c λ ∴ E1 E2 = λ2 λ1 = 5000 1 17 (c) λ = h mv = h √2meV = 6.6 × 10−34 √2 × (4 × 1.66 × 10−27) × (2 × 1.6 × 10−19) × V = 0.101 √V 18 (a) According to Einstein, the energy of photon is given by E = hv = hc λ Where h is Planck’s constant, c the speed of light and λ the wavelength. ∴ E1 E2 = λ2 λ1 Given, λ1 = 150 nm , λ = 300nm ∴ E1 E2 = 300 150 = 2 1 20 (d) Current i = ne t ⇒ n t = i e = 3.2 × 10−3 1.6 × 10−19 = 2 × 106 /s 21 (c)
E = hv/λ = hc eλ (in eV) = 6.6×10−34×3×108 1.6×10−19×0.21 = 5.9 × 10−6 eV 22 (d) Light consists of photons and cathode rays consists of electrons. However both effect the photographic plate 25 (c) v = √ 2eV m Since e, and m are constant, v1 v2 = √ V1 V2 or v2 = v1√ V1 V2 = 3.2 × 107√ 5824 2912 =4.525 × 107m/s 26 (a) Energy of photon E = hc λ = mc 2 ; momentum of photon = mc = h/λ 27 (b) E = hv ⇒ 100 × 1.6 × 10−19 = 6.6 × 10−34 × v ⇒ v = 2.42 × 1016Hz 28 (c) According to Mosley’s law v = a(Z − b) 2 and v ∝ 1 λ 29 (d) Stopping potential depends upon the energy of photon 30 (d) 1 2 mv 2 = hc λ − φ(in eV) = 6.6 × 10−34 × 3 × 108 2000 × 10−10 × 1.6 × 10−19 − 4.2 = 2eV = 2 × 1.6 × 10−19J v = √2 × 2 × 1.6 × 10−19/9.1 × 10−31 = √6.4/9.1 = 106ms −1 31 (b) λ0 = 12375 W0(eV) = 12375 2 = 6187.5Å ≃ 620 nm 33 (b) For φ = 900 , cos φ = 0 So, λ ′ = λ + h mec = 0.140× 10−9 + 6.63×10−34 (9.1×10−31)(3×108) = (0.140 × 10−9 + 2.4 × 10−12) m = 0.142 nm 34 (a) λmin = 12375 V Å ⇒ V = 12375 0.4125 = 30 kV 35 (a) Ek = hc λ − φ0(in eV) = 6.6 × 10−34 × 3 × 108 5000 × 10−10 × 1.6 × 10−19 − 1.9 = 2.48 − 1.9 = 0.58 eV 36 (d) Since the energy of incident electron, E = 80 keV. The minimum wavelength of X-rays produced is λ = hc E = 6.6 × 10−34 × 3 × 108 80 × 1000 × 1.6 × 10−19 = 1.55 × 10−10 m = 0.155 Å Since the energy of K-shell electron is −72.5 keV, so the incident electron of energy 80 keV will not only produce continuous spectrum of minimum wavelength 0.155Å but shell also knock electron of K shell out of atom, resulting emission of characteristics X-rays 38 (a) As q v B = mv 2 /r or m = qBr v = (2×1.6×10−19)×1×1 1.6×107 = 2 × 10−26kg = 2 × 1026 1.66 × 10−27 = 12 Therefore, particle must be c ++ 39 (d) From the symmetry of figure, the angle θ = 45°. The path of moving proton in a normal magnetic field is circular. If r is the radius of the circular path, then from the figure, AC = 2 r cos 45° = 2r × 1 √2 = √2r ...(i) As B q V = mv 2 r or r = mv Bq AC = √2 mv B q = √2 × 1.67 × 10−27 × 107 1 × 1.6 × 10−19 = 0.14 m 41 (a) When E, v and B are all along same direction, then magnetic force experienced by electron is zero while electric force is acting opposite to velocity of electron, so velocity of electron will decrease. 42 (b) Energy of a photon, E = hc λ λinfrared > λred > λBlue > λViolet Therefore, violet has the highest energy 43 (a) Number of photo electrons (N) ∝ Intensity ∝ 1 d 2 ⇒ N1 N2 = ( d2 d1 ) 2 ⇒ N1 N2 = ( 100 50 ) 2 = 4 1 ⇒ N2 = N1 4 44 (a) From conservation of energy the electron kinetic energy equals the maximum photon energy (we neglect the work function φ because it is normally so small compared to eV0). eV0 = hvmax or eV0 = hc λ min ∴ V0 = hc eλ min or V0 = 12400×10−10 10−11 =124 kv Hence, accelerating voltage for electrons in X–ray machine should be less than 124 kv.
48 (d) λ = h p ⇒ λ = h mv r = mv qB ⇒ mv = qrB ⇒ (2e)(0.83 × 10−2) ( 1 4 ) λ = 6.6 × 10−34 × 4 2 × 1.6 × 10−19 × 0.83 × 10−12 ⇒ λ = 0.01 Å 50 (a) The force on a particle is So, F = q(E + v × B) or F = Fe + Fm Fe = qE = −16× 10−18 × 104 (−k̂) =16× 10−14k̂ and Fm=−16 × 10−18(10î× Bĵ) =−16 × 10−17 × B(+k̂) =−16 × 10−17B × k̂ Since, particle will continue to move along + x-axis, so resultant force is equal to 0. Fe + Fm = 0 ∴ 16 × 10−14 = 16 × 10−17B ⇒ B = 16×10−14 16×10−17 = 103 B = 103Wb-m−2 52 (b) Let E and B be along X-axis. When a charged particle is released from rest, it will experience an electric force along the direction of electric field or opposite to the direction of electric field depending on the nature of charge. Due to this force, it acquires some velocity along X-axis. Due to this motion of charge, magnetic force can not have non-zero value because angle between v and B would be either 0 0 or 1800 . So, only electric force is acting on particle and hence, it will move along a straight line. 53 (c) Positive rays consist of positive ions 54 (d) According to Einstein’s photoelectric equation E = W0 + Kmax ⇒ V0 = hc e [ 1 λ − 1 λ0 ] Hence if λ decreases V0 increases 55 (b) The value of saturation current depends on intensity. It is independent of stopping potential 56 (b) Momentum of photon p = h λ = 6.6 × 10−34 10−10 = 6.6 × 10−24kg − m/s 57 (b) p = E c = hv c ⇒ v = pc h 59 (c) KE = hv − hv0 = 8 eV − ( 6×10−34×1.6×1015 1.6×10−19 eV) = 8-6= 2 eV 60 (c) Photoelectric current ∝ intensity of incident light. Therefore, the graph is a straight line having positive slope passing through origin 61 (c) From relation eVs = h(v − v0 ) or Vs = threshold or cut off voltage = h e (v − v0 ) = 6.6×10−34 1.6×10−19 (8.2 − 3.3) × 1014 = 6.6×4.9×10−1 1.6 = 2V 62 (a) The de-Broglie wavelength (λ) is given by λ = h p ...(i) Where h is Planck’s constant, p the momentum. Also, p = mv = m0 √1− v2 c2 ...(ii) Where m0 is rest mass, v the velocity and c the speed of light. From Eqs. (i) and (ii), we get λ = h√1− v2 c2 m0v Given, v = c ∴ λ = 0 63 (c) λ = h mv ⇒ λ ∝ 1 m 64 (d) According to Einstein’s photoelectric equation Kmax = hv − φ0 eVs = hc λ − φ0 ⇒ Vs = hc λe − φ0 e Where, λ = Wavelength of incident light φ0 = Work function Vs = Stopping potential According to given problem V E Y Z X B B E Y Z X
V1 = hc λe − φ0 e V2 = hc ( λ 2 ) e − φ0 e V2 = 2hc λe − φ0 e = 2hc λe − 2φ0 e + 2φ0 e − φ0 e = 2 ( hc λe − φ0 e ) + φ0 e V2 = 2V1 + φ0 e [Using (i)] ∴ V2 > 2V1 65 (c) λmin = 12375 50 × 103 Å = 0.247 = 0.25Å 66 (b) v = E B = 20 0.5 = 40 m/sec 67 (c) The wavelength of X-ray lines is given by Rydberg Formula 1 λ = RZ 2 ( 1 n1 2 − 1 n2 2 ) For Kα line, n1 = 1 and n2 = 2 ∴ 1 λ = RZ 2 ( 3 4 ) ⇒ Z = ( 4 3Rλ) 1/2 = [ 4 3(1.097 × 107m−1)(0.76 × 10−10m) ] 1/2 = 39.99 ≈ 40 68 (b) 1 2 mv 2 = eV ⇒ e m = v 2 2V = (8.4 × 106) 2 2 × 200 = 1.76 × 1011 C kg 69 (c) Work function is the intercept on K. E. axis i. e. 2eV 70 (c) E = hc λ ⇒ E1 E2 = λ2 λ1 ⇒ 3.32 × 10−19 E2 = 4000 6000 ⇒ E2 = 4.98 × 10−19J = 3.1 eV 71 (b) For an electron Mass, me = 9.11 × 10−31kg Kinetic energy, K = 10eV = 10 × 1.6 × 10−19J de Broglie wavelength, λe = h √2meK ...(i) For the person Mass, m = 66kg Speed, v = 100kmhr −1 = 100 × 5 18 ms −1 de Broglie wavelength, λ = h mv ...(ii) Diving (i) by (ii), we get λe λ = h √2meK × mv h = mv √2meK = 66 × 100 × 5 18 √2 × 9.11 × 10−31 × 10 × 1.6 × 10−19 = 1.07 × 1027 72 (c) A charged particle is deflected by electric and magnetic fields. If the cathode rays is deflected by electric and magnetic fields then this is the strong argument for the particle nature of cathode rays. 73 (d) In photoelectric effect particle nature of electron is shown. While in electron microscope, beam of electron is considered as electron wave 74 (c) Potential difference V = hc eλ = 6.6×10−34×3×108 1.6×10−19×2×10−10 = 6200 V 75 (b) Let K and K′ be the maximum kinetic energy of photoelectrons for incident light of frequency v and 2v respectively. According to Einstein’s photoelectric equation, K = hv − E0 ...(i) and K ′ = h(2v) − E0 ...(ii) = 2hv − E0 = hv + hv − E0 K ′ = hv + K [using (i)] 76 (b) From the given graph it is clear that if we extend the given graph for A and B, intercept of the line A on V axis will be smaller as compared to line B means work function of A is smaller than that of B 77 (a) λ = hc eV = 12375 59000 = 0.20Å [∵ hc e = 12375] 78 (d) hv − W0 = 1 2 mvmax 2 ⇒ hc λ − hc λ0 = 1 2 mvmax 2 ⇒ hc ( λ0 − λ λλ0 ) = 1 2 mvmax 2 ⇒ vmax = √ 2hc m ( λ0 − λ λλ0 ) When wavelength is λ and velocity is v, then v = √ 2hc m ( λ0 − λ λλ0 ) ... (i) When wavelength is 3λ 4 and velocity is v′ then v ′ = √ 2hc m [ λ0 − (3λ/4) (3λ/4) × λ0 ] ... (ii) Divide equation (ii) by (i), we get v ′ v = √ [λ0 − (3λ/4)] 3 4 λλ0 × λλ0 λ0 − λ v ′ = v ( 4 3 ) 1/2 √ [λ0 − (3λ/4)] λ0 − λ i. e. v ′ > v ( 4 3 ) 1/2 79 (b)