Tiêu đề 8-modern-physics-.pdf ✅

Modern Physics 1. (B) When an electron moves in a circular path, then Radius, r = mv eB ⇒ r 2e 2B 2 2 = m2v 2 2 KEmax = (mv) 2 2m ⇒ r 2e 2B 2 2m = (KE)max Work function of the metal (W), i.e., W = hv − KEmax 1.89 − φ = r 2e 2B 2 2m 1 2 eV = r 2eB 2 2m eV [hv → 1.89eV, for the transition on from third to second orbit of H- atom ] = 100 × 10−6 × 16 × 10−19 × 9 × 10−8 2 × 9.1 × 10−31 φ = 1.89 − 1.6 × 9 2 × 9.1 = 1.89 − 0.79 = 1.1eV 2. (C) Let φ0 be the work function of the surface of the material. Then, According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons in the first case is Kmax1 = hc λ − φ0 and that in the second is Kmax2 = hc λ 2 − φ0 = 2hc λ − φ0 But Kmax2 = Kmax1 (Given) ∴ 2hc λ − φ0 = 3 ( hc λ − φ0) ; 3φ0 − φ0 = 3hc λ − 2hc λ 2hc λ − φ0 = 3hc λ − 3φ0 2φ0 = hc λ or φ0 = hc 2λ 3. (D) de-Broglie wavelength, λ = h p or λ ∝ 1 p , λp = constant. This represents a rectangular hyperbola 4. (D) 5. (C) Kmax = eV0 = h(v − v0 ) If v ′ = 2v ∴ Kmax ′ = eV0 ′ = h(2v − v0 ) = 2Kmax + hv0 ∴ Kmax ′ > 2Kmax and ⇒ v0 ′ > 2v0 6. (D) de-Broglie wavelength of neutrons in thermal equilibrium at temperature T is λ = h √2mkBT where m is the mass of the neutron kB is the Boltzmann constant h is the Plank's constant Here, m = 1.67 × 10−27 kg kB = 1.38 × 10−23JK−1 h = 6.63 × 10−34JS ∴ λ = 6.63 × 10−34 √2 × 1.67 × 10−27 × 1.38 × 10−23 × T = 3.08 × 10−34 × 1025 √T m = 30.8 × 10−10 √T = 30.8 √T Å 7. (B) The de-Broglie wavelength λ associated with the electrons is λ = 1.227 √V nm where V is the accelerating potential in volts. or λ ∝ 1 √V ∴ λ1 λ2 = √ V2 V1 = √ 100 × 103 25 × 103 = 2 or λ2 = λ1 2 8. (B) (KE)max = hc λ − φ ⇒ φ = hc λ − (KE)max ⇒ φ = 1240 400 − 1.68 ⇒ φ = 1.42eV 9. (D) When a beam of cathode rays (or electrons) are subjected to crossed electric (E) and magnetic (B) fields, the beam is not deflected, if Force on electron due to magnetic field = Force on electron due to electric field Bev = eE or v = E B If V is the potential difference between the anode and the cathode, then ∴ 1 2 mv 2 = eV e m = v 2 2V Substituting the value of u from the equation (i) in equation (ii), we get e m = E 2 2VB2 Specific charge of the cathode rays e m = E 2 2VB2 10. (C) We know λ = h mv and K = 1 2 mv 2 = (mv) 2 2m ⇒ mv = √2mK Thus, λ = h √2mK ⇒ λ ∝ 1 √K ⇒ λ2 λ1 = √K1 √K2 = √K1 √2K2 (∵ K2 = 2K1 ) or λ2 λ1 = 1 √2 or λ2 = λ1 √2 11. (A)
(a) and (b) represent radiations of the same frequency because their kinetic energies are the same. But satuation photocurrents are different. Therefore intensities are different. 12. (D) Power of monochromic light beam is P = Nhu where N is the number of photons emitted per second. Power P = 2 × 10−3 W Energy of one photon E = hν = 6.63 × 10−34 × 6 × 1014 J Number of photons emitted per second, N = P/E = 2 × 10−3 6.63 × 10−34 × 6 × 1014 = 0.05 × 1017 = 5 × 1015 13. (C) K.E. = hu − W i.e. 1 2 mvmax 2 = hv − W ⇒ 1 2 m × (4 × 106 ) 2 = 2hv0 − hv0 or 1 2 m × (4 × 106 ) 2 = 2hv0 Another case, 2hv0 → 5hv0 1 2 mvmax 2 = 4hv0 ⇒ 1 2 mv 2 = 4 × 1 2 × m × (4 × 106 ) 2 ⇒ vmax 2 = 64 × 1012 ⇒ vmax = 8 × 106 m/s 14. (D) As λ is increased, there will be value of λ and which photoelectron will be cease to come out, so photocurrent will becomes zero. 15. (A) 16. (D) According to Einstein's photoelectric equation, 1 2 mv 2 = hc W − W0 or hc λ = 1 2 mv 2 + W0 and 1 2 mv1 2 = hc 3λ/4 − W0 = 4 3 ( 1 2 mv 2 + W0) − W0 So, v1 is greater than v(4/3) 1/2 . 17. (D) Einstein's photoelectric equation is KEmax = hv − φ The equation of line is y = mx + c Comparing above two equations m = h, c = −φ Hence, slope of graph is equal to Planck's constant (non-variable) and does not depend on intensity of radiation. 18. (C) Kinetic energy (E) = 100eV; Mass of electron (m) = 9.1 × 10−31 kg; 1eV = 1.6 × 10−19 J and Plank's constant (h) = 6.6 × 10−34 J − s. Energy of an electron (E) = 100 × (1.6 × 10−19)J or λ = h √2mE = 6.6×10−34 √2×9.1×10−31×100×1.6×10−19 = 1.2 × 10−10 m = 1.2Å 19. (A) The kinetic energy of an electron 1 2 × mv 2 = eV or final velocity of electron (v) = √ 2eV m . 20. (D) Momentum of electrons (pe ) = √2meV and momentum for proton (pp) = √2MeV Therefore, λp λe = h/pp h/pe = √ 2meV 2MeV = √( m M ) Therefore λp = λ√( m M ). 21. (B) The photoelectric effect is an instantaneous phenomenon (experimentally proved). It takes approximate time of the order of 10−10 S. 22. (B) Here, hc λ = 103 eV and hv = 106 eV. Hence, v = 103c λ = 103×3×108 1.24×10−9 = 2.4 × 1020 Hz 23. (C) hc λ0 = φ ⇒ λmax = hc φ = 6.6×10−34×3×108 4×1.6×10−19 24. (A) Momentum of the photon = hv c ⇒ c v = h p = λ v = c λ = cp h = 3 × 108 × 3.3 × 10−29 6.6 × 10−34 = 1.5 × 1013 Hz Where, v = frequency of radiation
25. (C) For hydrogen atom, we get : 1 λ = RZ 2 ( 1 1 2 − 1 2 2 ) ⇒ 1 λ1 = R(1) 2 ( 3 4 ) ⇒ 1 λ2 = R(1) 2 ( 3 4 ) ⇒ 1 λ3 = R(2) 2 ( 3 4 ) ⇒ 1 λ4 = R(3) 2 ( 3 4 ) ∴ 1 λ1 = 1 4λ3 = 1 9λ4 = 1 λ2 26. (A) 27. (B) 28. (B) I2 I1 = (r1 ) 2 (r2 ) 2 ⇒ I2 I1 = (1) 2 ( 1 2 ) 2 ⇒ I2 = 4I1 Now, since number of electrons emitted per second is directly proportional to intensity, so number of electrons emitted by photocathode would increase by a factor of 4 . 29. (A) de-Broglie wavelength, λ = h p = h mv 30. (B) The wavelength of a spectral line in the Lyman series is 1 λL = R ( 1 1 2 − 1 n2 ) , n = 2,3,4, ..... and that in the Balmer series is 1 λB = R ( 1 2 2 − 1 n2 ) , n = 3,4,5, ... ... For the longest wavelength in the Lyman series, n = 2 ∴ 1 λL = R ( 1 1 2 − 1 2 2 ) = R ( 1 1 − 1 4 ) = R ( 4 − 1 4 ) = 3R 4 or 1 λL = 4 3R For the longest wavelength in Balmer series, n = 3 ∴ 1 λB = R ( 1 2 2 − 1 3 2 ) = R ( 1 4 − 1 9 ) = R ( 9−4 36 ) = 5R 36 or λB = 36 5R Thus, λL λB = 4 3R × 5R 36 = 5 27 31. (D) Radius of the nucleus R = R0A 1/3 ∴ RAl RTe = ( AAl ATe ) 1/3 Here, AAl = 27, ATe = 125,RTe =? RAl RTe = ( 27 125) 1/3 = 3 5 ⇒ RTe = 5 3 RAl 32. (D) Energy of electron in He+3 rd orbit E3 = −13.6 × Z 2 n2 eV = −13.6 × 4 9 eV = −13.6 × 4 9 × 1.6 × 10−19 J ≃ 9.7 × 10−19 J As per Bohr's model, Kinetic energy of electron in the 3 rd orbit = −E3 energy of electron in the 3 rd orbit = −E3 ∴ 9.7 × 10−19 = 1 2 mev 2 v = √ 2 × 9.7 × 10−19 9.1 × 10−31 = 1.46 × 106 ms−1 33. (C) Work function, W = hc λ [Here they are interested in asking threshold wavelength] where, h = Planck's constant, c = velocity of light Therefore, WNa WCu = λCu λNa or λNa λNa = WCu WNa = 4.5 2.3 = 2 (nearly) 34. (D) Binding energy of 3 7Li nucleus = 7 × 5.60MeV = 39.2MeV Binding energy of 2 4He nucleus = 4 × 7.06MeV = 28.24MeV The reaction is 3 7Li + 1 1H → 2( 2 4He) + Q ∴ Q = 2(BE of 4 2He) − (BE of 3 7Li) = 2 × 28.24MeV − 39.2MeV = 56.48MeV − 39.2MeV = 17.28MeV 35. (D) ΔE = hv where, ΔE is energy of radiation, h is Plank's constant and v is frequency. ⇒ v = ΔE h = k [ 1 (n − 1) 2 − 1 n 2 ] = k2n n 2(n − 1) 2 = 2k n 3 ∝ 1 n 3 36. (C) Nuclear radius, R = R0A 1/3 where R0 is a constant and A is the mass number ∴ RAl RCu = (27) 1/3 (64) 1/3 = 3 4 or RCu = 4 3 × RAl = 4 3 × 3.6 fermi = 4.8 fermi 37. (A) In steady state, electric force on drop = weight of drop i.e., qE = mg ⇒ q = mg E = 9.9 × 10−15 × 10 3 × 104 = 3.3 × 10−18C
38. (B) According to radioactive decay law N = N0e −λt where N0 = Number of radioactive nuclei at time t = 0 N = Number of radioactive nuclei left undecayed at any time t λ = decay constant At time t2, 2 3 of the sample had decayed ∴ N = 1 3 N0 ∴ 1 3 N0 = N0e −λt2 At time t1, 1 3 of the sample had decayed, ∴ N = 2 3 N0 ∴ 2 3 N0 = N0e −λt1 Divide (i) by (ii), we get 1 2 = e −λt2 e −λt1 1 2 = e −λ(t2−t1 ) λ(t2 − t1 ) = ln 2 t2 − t1 = ln 2 λ = ln 2 ( ln 2 T1/2 ) (∵ λ = ln 2 T1/2 ) = T1/2 = 50 days 39. (C) P Q Number of nuclei, at t = 0 4N0 N0 Half-life 1 min 2 min Number of nuclei after NP NQ time t Let after t min the number of nuclei of P and Q are equal. ∴ NP = 4N0 ( 1 2 ) t/1 and NQ = N0 ( 1 2 ) t/2 As NP = NQ ∴ 4N0 ( 1 2 ) t/1 = N0 ( 1 2 ) t/2 4 2 t/1 = 1 2 t/2 or 4 = 2 t 2 t/2 or 4 = 2 t/2 or 2 2 = 2 t/2 or t 2 = 2 or t = 4 min After 4 minutes, both P and Q have equal number of nuclei. ∴ Number of nuclei of R = (4N0 − N0 4 ) + (N0 − N0 4 ) = 15N0 4 + 3N0 4 = 9N0 2 40. (A) Formation of covalent bonds due to the wave nature of particle is done in compounds. 41. (D) According to activity law R = R0e −λt Where, R0 = initial activity at t = 0 R = activity at time t λ = decay constant According to given problem, R0 = N0 counts per minute R = N0 e counts per minute t = 5 minutes Substituting these values in equation (i), we get N0 e = N0e −5λ e −1 = e −5λ 5λ = 1 or λ = 1 5 per minute At t = T1/2, the activity R reduces to R0 2 . Where T1/2 = half life of a radioactive sample From equation (i), we get : R0 2 = R0e −λT1/2 e λT1/2 = 2 Taking natural logarithms of both sides of above equation, we get λT1/2 = log 2 or T1/2 = loge 2 λ = loge 2 ( 1 5 ) = 5loge 2 minutes 42. (D) In emission spectrum, number of bright lines is given by n(n−1) 2 = 4(4−1) 2 = 6 43. (C) At the distance of closest approach d, Kinetic energy = potential energy 1 2 mv 2 = 1 4πε0 (2e)(Ze) d Where, Ze = charge of target nucleus 2e = charge of alpha nucleus