Dual Nature of Radiation and Matter MCQ Class XII.pptx

Physics Multiple choice questions and answers with explanation. (Class XII Physics TN State board) Unit - 9

1. Dual Nature of Radiation and Matter MCQ Class-XII (TN State Board) (With explanation)

2. By Dr M. Arunachalam Head, Department of Physics ( Rtd .) Sri SRNM College, Sattur

3. 1. The wavelength λe of an electron and λp of a photon of same energy E are related by λ p e b) ) λ p e c) λ p e d ) λ p e For photon: E = h ν = hc / λ p ie. λ p = hc/E ie λ p 1/E For electron: e = h/ E ie e 1/ e 1/E ie. λ p e Ans: d  

4. 2. In an electron microscope, the electrons are accelerated by a voltage of 14 kV. If the voltage is changed to 224 kV , then the de Broglie wavelength associated with the electrons would a) increase by 2 times b) decrease by 2 times c) decrease by 4 times d) increase by 4 times The wavelength of an electron is given by the equation λ = 12.27/ λ – de Broglie wavelength of electron V – Accelerating voltage So we can write λ 1 = 12.27/ ie λ 1 = 12.27/ ----- 1  

5. Contd … λ 2 = 12.27/ ie λ 1 = 12.27/ ----- 2 Dividing equation 1 by equation λ 2/ λ 1 = 12.27 x = 11.8/47.3 = 0.25 [App. 12/50 = 0.24] λ 2 = λ 1x0.25 = λ 2 = λ 1/4 de Broglie wavelength of the electrons would decrease by 4 times Ans: c  

6. 3. The wave associated with a moving particle of mass 3 × 10 –6 g has the same wavelength as an electron moving with a velocity 6x 10 6 m s -1 . The velocity of the particle is a) 1.82 x10 -18 m s -1 b) 9 6x10 -2 m s -1 c) 3 x 10 -31 m s -1 d) 1.82 x10 -15 m s -1 wavelength of the particle = wavelength the electron = λ Mass of the particle = 3×10 –6 g = 3×10 –9 kg. h – Planck’s constant v - V elocity of the particle = ?

7. Contd … For the particle: λ = h/mv = h/ 3×10 –9 xv ---- 1 For electron: Mass of electron = 9.1 ×10 –31 kg. v - V elocity of electron = 6x10 6 m s -1 λ = h/mv = h/ 9.1 ×10 –31 x 6x10 6 ---- 2 From equations 1 and 2 h / 3×10 –9 xv = h / 9.1 ×10 –31 x 6x10 6 v = 9.1 ×10 –31 x 2 6 x10 6 / 3 ×10 –9 = 18.2x10 16 = 1.82x10 15 m s -1 Ans: d

8. 4. When a metallic surface is illuminated with radiation of wavelength λ , the stopping potential is V . If the same surface is illuminated with radiation of wavelength 2 λ , the stopping potential is V/4 . The threshold wavelength for the metallic surface is 4 λ b) 5 λ c) 5/2 d) 3 λ By Einstein’s Photo electric equation: h ν = h ν + mv 2 /2 At stopping potential the equation becomes, h ν = h ν + eV ie . eV = h ν - h ν V – Stopping potential ν – Frequency of radiation = hc / λ

9. Contd … So we can write, . eV = hc / λ - hc / λ λ - Threshold wavelength of the metal surface For case I, we have, eV = hc / λ - hc / λ - --- 1 For case II, stopping potential is V/4 and wavelength is 2 λ . So we have eV/4 = hc /2 λ - hc / λ - --- 2

10. Dividing equation 1 by equation 2 we get eV/ (eV/4) = hc [1/ λ - 1/ λ ] ie . 4 = [1/ λ - 1/ λ ] hc [1/2 λ - 1/ λ ] [1/2 λ - 1/ λ ] ie . 4 [1/2 λ - 1/ λ ] = [1/ λ - 1/ λ ] ie . 4 /2 λ - 4/ λ = 1/ λ - 1/ λ

11. Contd … ie . 2/ λ - 1/ λ = 4/ λ - 1/ λ (2-1)/ λ = (4 – 1)/ λ ie . 1/ λ = 3/ λ λ = 3 λ Ans: d

12. 5. If a light of wavelength 330 nm is incident on a metal with work function 3.55 eV, the electrons are emitted. Then the wavelength of the wave associated with the emitted electron is (Take h = 6.6 × 10 –34 Js ) a) < 75x10 -9 .m b) 75x10 -9 .m c) 75x10 -12 .m d) < 2.5x10 -10 .m Energy of incident light = E = h ν = hc / λ c – Speed of light = 3x10 8 ms -1 h – Planck’s constant = 6.6 × 10 –34 Js λ – Wavelength of light = 330 nm = 330 x10 -9 .m= 3.3x10 -7 m  

13. Contd … Therefore, E = 6.6 × 10 –34 x 3x10 8 / 3.3x10 -7 E = 6.6 2 × 10 –34 x 3x10 8 / 3.3 x10 -7 = 6x 10 –34 x10 8 x10 7 ie . E = 6x 10 –19 J = 6x 10 –19 /1.6 x 10 –19 = 3.75eV Energy for work function = φ = h ν = 3.355eV KE of electron = h ν - h ν = 3.75 – 3.55 = 0.2eV = 0.2x1. 6x 10 –19 = 0.32 x 10 -19 J= E K

14. Contd … λ = h/ K - wavelength of light m – Mass of electron = 9x 10 -31 kg ie . λ = 6.6×10 –34 / 9x10 -31 x0.32 x 10 -19 = 6.6×10 –34 / 9x10 -50 = = 6.6×10 –34 / x10 -25 = (6.6/2.4) 10 -34 x 10 +25 = 2.75x10 -9 m ie λ 2.75x10 -9 m Ans: b  

15. 6. A photoelectric surface is illuminated successively by monochromatic light of wavelength λ and λ /2 . If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the material is a ) hc / λ b) 2hc/ λ c) hc /3 λ d) hc /2 λ KE of electrons in the II case is 3 times to that of KE in the I case ie . KE 2 = 3KE 1 ----- 1 Now we can write KE 1 = h ν 1 - φ = hc / λ - φ and

16. Contd … KE 2 = h ν 2 - φ = hc /( λ /2) - φ ie . KE 2 = 2hc/ λ - φ Substituting the above values in equation 1 we get 2hc/ λ - φ = 3[ hc / λ - φ ] 2hc/ λ - φ = 3hc/ λ - 3 φ

17. Contd … Rearranging the above equation, we get, 3hc/ λ - 2hc/ λ = 3 φ - φ ie . hc / λ = 2 φ ie . φ = hc /2 λ Ans : d

18. 7. In photoelectric emission, a radiation whose frequency is 4 times threshold frequency of a certain metal is incident on the metal. Then the maximum possible velocity of the emitted electron will be a) (hv /m) b) (6hv /m) c) 2 ( hv /m) d) ( hv /2m) By Photo electric equation, E = φ + KE = hv + KE v - Threshold frequency . During the photo electric emission, the frequency of radiation is 4v So we can write 4hv = hv + KE = hv + (1/2)mv 2  

19. Contd … ie. (1/2)mv 2 = 4hv - hv = 3 hv ie. v 2 = 2x 3 hv / m ie. v = 6 hv / m Ans: b  

20. 8. Two radiations with photon energies 0.9 eV and 3.3 eV respectively are falling on a metallic surface successively. If the work function of the metal is 0.6 eV, then the ratio of maximum speeds of emitted electrons in the two cases will be a) 1:4 b) 1:3 c) 1:1 d)1:9 By Photo electric equation, hv = hv + (1/2)mv 2 ie. (1/2)mv 2 = hv - hv hv – Photo electric work function of the metal = 0.6eV m – Mass of electron v – Velocity of electron

21. Contd … For case I: (1/2)mv 1 2 = 0.9 – 0.6 = 0.3eV v 1 2 = 2x 0.3/m ----- 1 For case II: (1/2)mv 2 2 = 3 .3 – 0.6 = 2.7eV v 1 2 = 2x 2.7/m ----- 2 Dividing equation 1 by equation 2 we get v 1 2 / v 2 2 = ( 2 x 0.3 / m )x( m / 2 x 2.7 ) 9 = 1/9 ie . v 1 : v 2 = 1:3 Ans: b

22. 9. A light source of wavelength 520 nm emits 1.04 × 10 15 photons per second while the second source of 460 nm produces 1.38 × 10 15 photons per second. Then the ratio of power of second source to that of first source is a) 1.00 b) 1.02 c) 1.5 d) 0.98 Power of source P = N hv = N hc/ λ N – No. of photons emitted per second h – Planck’s constant c – Speed of light λ – Wavelength of light Case I: P 1 = N 1 hc/ λ 1 Case II: P 2 = N 2 hc/ λ 2

23. Contd … ie . P 2 / P 1 = N 2 hc λ 1 / N 1 hc λ 2 No. of photons emitted/s in case I = 1.04 × 10 15 No. of photons emitted/s in case II = 1.38 × 10 15 Wavelength of light in case I = 520nm Wavelength of light in case II = 460nm P 2 / P 1 = 1.38× 10 15 hc x520/ 1.04× 10 15 hc x460 = 138x52/104x46 = 1.5 ie . P 2 / P 1 = 1.5 Ans: c

24. 10. If the mean wavelength of light from sun is taken as 550 nm and its mean power as 3.8 × 10 26 W, then the number of photons emitted per second from the sun is of the order of a ) 10 45 b) 10 42 c) 10 54 d) 10 51 Power of source P = N hv = Nhc/ λ ie . N = P λ / hc N – No. of photons emitted per second c – Speed of light = 3x10 8 m/s h – Planck’s constant = 6.6x10 -34 J-s λ – Wavelength of light = 550nm Power P = 3.8×10 26 W

25. Contd … ie . N = 3.8×10 26 x550x 10 -9 / 6.6x10 -34 x 3x10 8 N = 3.8×10 26 x 5.5 x 10 -7 / 6.6 x10 -34 x 3x10 8 ie . = 3.8×10 26 x5x 10 -7 /6 x 3x10 -26 = (3.8x5/18) x10 45 = 19/18x10 45 ie . N 1x10 45 No. of photons emitted /sec from the sun N = 10 45 Ans: a  

26. 11. The threshold wavelength for a metal surface whose photoelectric work function is 3.313 eV is a) 4125Å b) 3750Å c) 6000Å d) 2062.5Å Work function φ = h ν = hc / λ ie . λ = hc / φ ----- 1 c – Speed of light = 3x10 8 m/s h – Planck’s constant = 6.626x10 -34 J-s λ – Threshold Wavelength of the metal surface = ? φ - photoelectric work function = 3.313 eV = 3.313 x1.6x10 -19 J

27. Contd … Substituting the values in equation 1 we get, λ = 6.626x10 -34 x 3x10 8 / 3.313 x1.6x10 -19 ie . = 6.626 2 x10 -34 x 3x10 8 / 3.313 x1.6x10 -19 = (6/1.6) x10 -7 ie . λ = (60/16) x10 -7 ie . λ = 3.75x10 -7 m = 3750 x10 -10 m ie . λ = 3750 Å Ans: b

28. 12. A light of wavelength 500 nm is incident on a sensitive metal plate of photoelectric work function 1.235 eV. The kinetic energy of the photo electrons emitted is (Take h = 6.6 × 10 –34 Js ) a) 0.58 eV b) 2.48 eV c) 1.24 eV d) 1.16 eV By Photo electric equation, hv = hc / λ = φ + (1/2)mv 2 ie. (1/2)mv 2 = hv - φ : KE of the photo electron λ – Wavelength of light = 500 nm = 500 x10 -9 m= 5x10 -7 m φ – Photo electric work function of the metal = 1.235eV

29. Contd … h – Planck’s constant = 6.6 × 10 –34 Js ) c – Velocity of light = 3x10 8 m/s hc / λ = Energy of photon = 6.6×10 –34 x 3x10 8 /5x10 -7 J hc / λ = 6.6×10 –34 x 3x10 8 /5x10 -7 x1.6x 10 -19 eV hc / λ = 19.8×10 –26 /8x10 -26 =2.475eV KE of electron = (1/2)mv 2 = 2.475 – 1.235 = 1.24eV Ans: c

30. 13. Photons of wavelength λ are incident on a metal. The most energetic electrons ejected from the metal are bent into a circular arc of radius R by a perpendicular magnetic field having magnitude B . The work function of the metal is ( hc / λ ) - m e + e 2 B 2 R 2 /2m e ` b) ( hc / λ ) + 2m e [ eBR /2m e ] 2 c) ( hc / λ ) - m e c 2 - e 2 B 2 R 2 /2m e d) ( hc / λ ) - 2m e [ eBR /2m e ] 2 By Photo electric equation, hv = hc / λ = φ + KE Max hc / λ – Energy of the photon and φ = hc / λ - KE Max - ---- 1

31. Contd … To find KE Max The ejected electrons are bent in to a circular arc by a perpendicular magneti field B. Radius of the circular arc R = m e v/ Bq = m e v / Bq therefore v = BeR / m e m e - Mass of electron v – velocity of electron B – Magnetic field q = e – charge of electron

32. Contd … KE Max = (1/2) m e v 2 Substituting the value of v we get, KE Max = (1/2) m e [ BeR / m e ] 2 Multiplying the numerator and denominator of RHS by 4 we get, KE Max = 4x(1/4x 2 ) m e [ BeR / m e ] 2 ie . KE Max = 2 x(1/4x 2 ) m e [ BeR / m e ] 2

33. Contd … ie . KE Max = 2 m e [ BeR /2m e ] 2 Substituting this in equation 1 we get φ = hc / λ - 2 m e [ BeR /2m e ] 2 Ans: d

34. 14. The work functions for metals A , B and C are 1.92 eV, 2.0 eV and 5.0 eV respectively. The metal/metals which will emit photoelectrons for a radiation of wavelength 4100Å is/are a) A only b) both A and B c) all these metals d) none Energy of radiation E = hv = hc / λ joules h – Planck’s constant = 6.6 × 10 –34 Js c – Velocity of light = 3x10 8 m/s λ – Wavelength of radiation = 4100Å = 4100x10 –10 m= 4.1x10 –7 m

35. Contd … E = 6.6 ×10 –34 x 3x10 8 /4.1 x10 -7 joules E = 6.6 ×10 –34 x 3x10 8 /4.1 x10 -7 x 1.6x 10 -19 eV = 19.8x 10 -26 /6.56 x 10 -26 E 3eV The value of E is greater than the work functions of A and B. But less than that of C. So only A and B will emit photo electrons Ans: b  

36. 15. Emission of electrons by the absorption of heat energy is called………emission. a) photoelectric b) field c) thermionic d) secondary Thermionic emission Ans: c

37. Thank You