1. If a photon has 100 eV energy then its frequency is[TOM 2009]
- 2.5 × 1015 Hz
- 2.5 × 1010 Hz
- 2.5 × 1024 Hz
- 2.5 × 1032 Hz
(a) E = hν ⇒ ν = E/h = (100 × 1.6×10-19)/(6.63×10-34) ≈ 2.5 × 1015 Hz 2. Planck's constant has the dimension of[MOE 2013]
- energy
- mass
- frequency
- angular momentum
(d) [h] = [E]/[ν] = ML2T-2/T-1 = ML2T-1 (same as angular momentum) 3. If a pd of 1V is applied across an electron, the energy gained by it will be[MOE 2012]
(b) Energy = qV = e × 1V = 1 eV 4. Electron, proton, neutron and alpha particle have the same K.E. Which has the highest de-Broglie wavelength?[MOE 2012 & 2068]
- electron
- proton
- neutron
- alpha particle
(a) λ = h/√(2mK) ⇒ λ ∝ 1/√m. Electron has smallest mass, hence longest wavelength. 5. A metal (work function 3.31 eV) is illuminated by light of wavelength 5×10-7 m. The threshold frequency is:[MOE 2010]
- 8 × 1014 Hz
- 1.6 × 1015 Hz
- 2.4 × 1015 Hz
- 3.2 × 1015 Hz
(a) ν0 = φ/h = (3.31 × 1.6×10-19)/(6.63×10-34) ≈ 8 × 1014 Hz 6. Light (λ=400nm) on metal (threshold λ=600nm) produces current I. If λ is doubled, photoelectric current will be:[MOE 2068]
(d) Doubled λ (800nm) > threshold λ (600nm) ⇒ No photoemission. 7. If K.E. of a particle increases by four times, de-Broglie wavelength becomes:[MOE 2010]
- 2 times
- 1/2 times
- 1/4 times
- √2 times
(b) λ = h/√(2mK) ⇒ λ ∝ 1/√K. 4×K ⇒ λ/2. 8. The de-Broglie wavelength of electron is 1.224 Å. The energy of electron in eV is:[MOE 2010]
(c) λ = h/√(2mE) ⇒ E ≈ 100 eV for λ = 1.224 Å. 9. An electron (mass 'm', charge 'e') is accelerated from rest through pd. 'V' volts. Its speed will be:
(b) K.E. = eV = ½mv² ⇒ v = √(2eV/m) 10. In photoelectric effect, the number of ejected electrons per second depends on:[KU 2012]
- frequency of incident radiation
- intensity of incident radiation
- wavelength of incident radiation
- time of exposure
(b) Number of electrons ∝ intensity (number of photons). 11. The wavelength associated with an electron (mass m, velocity v) is:[KU 2011]
(a) de-Broglie relation: λ = h/p = h/mv 12. Work function = 2 eV. Velocity of emitted electron when λ=230 nm light is incident?[KU 2010]
- 1.1 × 106 m/s
- 2 × 106 m/s
- 1 × 106 m/s
- 7 × 106 m/s
(a) K.E. = hc/λ - φ ≈ 1.1 × 106 m/s 13. Which wavelength falls under visible light?
(b) Visible range: ~400-700 nm. 640 nm = red light. 14. Which has frequency 6 × 1015 Hz?[BP 2012]
- Radio waves
- UV-rays
- Microwaves
- X-rays
(b) UV range: ~8×1014-3×1016 Hz. 15. Two photons traveling opposite directions have relative velocity:[I.E. 2009]
(b) Relative velocity = c - (-c) = 2c (special relativity doesn't apply to photons). 16. Photoelectric effect is explained by:[I.E. 2009]
- Wave theory only
- EM theory only
- Quantum theory only
- None
(c) Quantum theory (photon concept) explains photoelectric effect. 17. An α-particle and proton accelerated through same potential. Ratio of final velocities:[I.E. 2011]
(b) v = √(2qV/m). For α (q=2e, m=4u) and p (q=e, m=1u), ratio is 1 : √2. 18. Particle nature of light is shown by:[TOM 2014]
- Photoelectric effect
- Velocity of light
- Diffraction
- Refraction
(a) Photoelectric effect demonstrates light's particle (photon) nature. 19. Light (1.5× threshold frequency) on material. If frequency is halved and intensity doubled, photocurrent becomes:[MOE 2014]
- Quadrupled
- Doubled
- Halved
- Zero
(d) Halved frequency < threshold frequency ⇒ no emission. 20. What will be the maximum velocity of photoelectrons ejected from a metal (work function 1eV) when light of wavelength 3000Å falls on it?[MOE 2014]
- 103 m/s
- 104 m/s
- 105 m/s
- 106 m/s
(d) K.E. = hc/λ - φ ≈ 3.14 eV ⇒ v = √(2K.E./me) ≈ 106 m/s 21. Which phenomenon does NOT support the wave theory of light?[BP 2014]
- Polarization
- Interference
- Diffraction
- Photoelectric effect
(d) Photoelectric effect requires particle (quantum) theory. 22. If threshold frequency increases, what happens to the K.E. of photoelectrons?[BP 2014]
- Increases
- Decreases
- Remains constant
- First increases then decreases
(b) K.E. = h(ν - ν0). Higher ν0 ⇒ lower K.E. for same ν. 23. Value of Planck’s constant (h) is:[BP 2014]
- 6.625 × 10-34 Js
- 6.625 × 10-27 Js
- 6.625 × 10-20 Js
- 6.625 × 10-10 Js
(a) Standard value: h ≈ 6.626 × 10-34 Js. 24. Uncertainty in proton position is 6×10-8 m. Minimum uncertainty in speed is:[BP 2014]
(b) Δp ≈ ħ/Δx ⇒ Δv ≈ 1 m/s for a proton. 25. Ratio of de Broglie wavelengths of proton and α-particle with same K.E.:[BP 2014]
(a) λ = h/√(2mK) ⇒ λp/λα = √(mα/mp) = 2:1 (mα = 4mp). 26. Work function = 3.3 eV. Threshold frequency is:[MOE 066]
- 8 × 1014 Hz
- 8 × 1015 Hz
- 8 × 1016 Hz
- 5 × 1020 Hz
(a) ν0 = φ/h ≈ 8 × 1014 Hz. 27. UV photon (work function = 2 eV) produces photoelectron with 2 eV energy. Photon wavelength is:[MOE 2065]
(d) E = hc/λ = φ + K.E. ⇒ λ ≈ 3100 Å. 28. Photoelectric effect conserves:[MOE 2063]
- Energy
- Momentum
- Angular momentum
- Power
(a) Energy conservation: Ephoton = φ + K.E.electron. 29. Light (frequency 3ν0) on material. If frequency is halved and intensity doubled, photocurrent becomes:
- Zero
- Halved
- Quadrupled
- Doubled
(a) 1.5ν0 > ν0, but halved (1.5ν0 → 0.75ν0) < ν0 ⇒ no emission. 30. Behind cutoff voltage, photoelectron emission is proportional to:[I.E. 06]
- Voltage
- K.E.
- Number of photons
- Field
(c) Emission rate ∝ photon flux (intensity). 31. Photoelectrons from a monochromatic beam have:[BPKIHS-08]
- Energy spread with no sharp limits
- Energy spread with a lower limit
- Definite energy only
- Energy spread with an upper limit
(d) Maximum K.E. = hν - φ (upper limit). 32. Increasing light intensity affects:[BPKIHS 02]
- Photocurrent increases
- Frequency increases
- Maximum K.E. increases
- Photocurrent decreases
(a) Higher intensity ⇒ more photons ⇒ more photoelectrons. 33. When frequency increases:[BPKIHS 02]
- Photocurrent increases
- Maximum K.E. increases
- Stopping potential decreases
- Photocurrent decreases
(b) K.E.max = hν - φ ⇒ increases with ν. 34. Photocell current is:[BPKIHS-04]
- Directly proportional to intensity
- Directly proportional to square of intensity
- Inversely proportional to square of intensity
- Inversely proportional to intensity
(a) Current ∝ number of photons ∝ intensity. 35. UV radiation (6.2 eV) on aluminum (φ = 4.2 eV). K.E.max is:[BPKIHS 05]
- 4.2 × 10-17 J
- 3 × 10-17 J
- 1.4 × 10-17 J
- 2.65 × 10-17 J
(c) K.E.max = 6.2 - 4.2 = 2 eV ≈ 3.2 × 10-19 J (options may need adjustment). 36. Photoelectric effect occurs if incident light frequency is:[BPKIHS 05]
- Greater than threshold frequency
- Less than threshold frequency
- Equal to threshold frequency
- Equal to heat radiation frequency
37. Photoelectric effect proves the existence of:
- Electrons
- EM fields
- Protons
- Photons
(d) Demonstrates light’s particle nature (photons). 38. Number of photons for fixed energy varies:
- Inversely with wavelength
- Inversely with frequency
- Directly with frequency
- Independent of frequency
(a) E = nhc/λ ⇒ n ∝ λ for fixed E. 39. X-ray photon (λ = 0.01 Å) momentum is:[MOE 2008]
- 6.6 × 10-22 kg m/s
- 6.6 × 10-32 kg m/s
- 3.3 × 10-32 kg m/s
- 0
(a) p = h/λ ≈ 6.6 × 10-22 kg m/s. 40. Energy of green light photon (λ = 5000 Å) is:
- 3.459 × 10-19 J
- 4.132 × 10-19 J
- 3.973 × 10-19 J
- 8.453 × 10-19 J
(c) E = hc/λ ≈ 3.973 × 10-19 J. 41. Energy of photon (λ = 6600 Å) in eV is:
- 0.1875 eV
- 1.875 eV
- 18.75 eV
- 198 eV
(b) E(eV) = 12400/λ(Å) ≈ 1.875 eV. 42. Radio transmitter (1000 kHz, 66W) emits how many photons/second?
(c) n = P/(hν) ≈ 1029 photons/s. 43. Light (1.5× threshold frequency). If frequency is halved, photoelectrons:
- Four times
- Doubled
- Halved
- Zero
(d) 0.75ν0 < ν0 ⇒ no emission. 44. Threshold frequency = ν0. If incident frequency doubles, K.E. becomes:
- Doubled
- Halved
- More than doubled
- Less than doubled
(c) K.E. = h(2ν0 - ν0) = hν0 → 2hν0 (more than double initial K.E.). 45. If light frequency doubles in photoelectric experiment, stopping potential:
- Doubles
- Halves
- More than doubles
- Less than doubles
(c) Vs ∝ (ν - ν0). Doubling ν more than doubles Vs if ν0 > 0. 46. Threshold wavelength = 5000 Å. Photoemission occurs with:
- 50W infrared lamp
- 1W infrared lamp
- 0.5W infrared lamp
- 50W ultraviolet lamp
(d) UV light (λ < 5000 Å) can cause emission; IR cannot. 47. Threshold wavelength = 5200 Å. Photoemission occurs with:
- 50W infrared lamp
- 25W red light lamp
- 50W green light lamp
- All of above
(c) Green light (~5000 Å) may exceed threshold; IR and red light likely do not. 48. Photon energy = 6 eV, maximum K.E. = 4 eV. Stopping potential is:
(b) e Vs = K.E.max ⇒ Vs = 4 V. 49. Green light ejects electrons; yellow does not. Red light will:
- Emit more energetic electrons
- Emit less energetic electrons
- Depend on intensity
- Not emit electrons
(d) Red light (λ > yellow) has ν < ν0 ⇒ no emission. 50. Stopping potential is V when λ = λ. For λ = 2λ, stopping potential = V/3. Threshold λ is:
(d) Solve hc/λ = φ + eV and hc/(2λ) = φ + eV/3 ⇒ φ = hc/(4λ). 51. Photoelectrons have K.E. ratio 1:K for frequencies ν1 and ν2 (ν1 > ν2). Threshold frequency is:
- (ν1 - Kν2)/(1 - K)
- (Kν2 - ν1)/(K - 1)
- (ν1 + Kν2)/(1 + K)
- (Kν1 - ν2)/(K - 1)
(b) Solve hν1 - φ = K(hν2 - φ) ⇒ φ = (Kν2 - ν1)/(K - 1). 52. Work function = 2.2 eV. Maximum wavelength for photoemission is:[MOE Bangladesh 2009]
(b) λmax = hc/φ ≈ 567 nm. 53. Radiation with photon energies 2φ and 10φ incident on metal. Ratio of maximum photoelectron velocities:
(d) v ∝ √(K.E.) ⇒ √(2φ - φ) : √(10φ - φ) = 1 : 3. 54. Photon energy = 5 eV, threshold frequency = 1.6 × 1015 Hz. K.E. of photoelectron (in eV):[KU 2014]
(c) φ = hν0 ≈ 6.63 × 10-34 × 1.6 × 1015 ≈ 1.06 eV ⇒ K.E. = 5 - 1.06 ≈ 3.94 eV (closest to 1.2 eV may be typo). 55. Cutoff potential = V0 at 1m. If source is moved to 2m, cutoff potential becomes:
(c) Cutoff potential depends on photon energy (frequency), not distance. 56. Speed of electron with λ = 10-10 m is:
- 7.25 × 106 m/s
- 5.25 × 106 m/s
- 6.25 × 106 m/s
- 4.24 × 106 m/s
(a) λ = h/(mev) ⇒ v ≈ 7.25 × 106 m/s. 57. K1 and K2 are maximum K.E. for wavelengths λ1 and λ2. If λ1 = 3λ2, then:[BP 2014]
- K1 > K2/3
- K1 < K2/3
- K1 = 3K2
- K2 = 3K1
(b) K.E. ∝ 1/λ ⇒ K1 ≈ K2/3 (but φ reduces it further). 58. Work functions: A = 1.92 eV, B = 2.0 eV, C = 5 eV. Which emit photoelectrons for λ = 4100 Å?
- None
- A only
- A and B only
- All three
(c) E = hc/λ ≈ 3.0 eV ⇒ A and B (φ < 3.0 eV) emit; C does not. 59. For a metal, ν = 2ν0 gives vmax = 4 × 106 m/s. If ν = 5ν0, vmax will be:
- 2 × 106 m/s
- 8 × 106 m/s
- 2 × 107 m/s
- 8 × 107 m/s
(b) v ∝ √(ν - ν0) ⇒ √(5ν0 - ν0)/√(2ν0 - ν0) = 2 ⇒ v = 8 × 106 m/s. 60. X-rays on sodium and copper surfaces. Stopping potential is:
- Equal for both
- Greater for sodium
- Greater for copper
- Infinite for both
(b) Sodium has lower work function ⇒ higher K.E. ⇒ higher stopping potential. 61. Monochromatic source at distance 'd' emits n electrons/s with K.E. = E. At distance d/2:
- 2n and 2E
- 4n and 4E
- 4n and E
- n and 4E
(c) Intensity ∝ 1/d² ⇒ 4× photons ⇒ 4n electrons. K.E. depends on ν (unchanged). 62. UV light (λ = 300 nm, I = 1.0 W/m²) on photosensitive material (1% efficiency). Photoelectrons emitted from 1.0 cm² area:
- 9.61 × 1013 s-1
- 4.12 × 1011 s-1
- 1.51 × 1012 s-1
- 2.13 × 1010 s-1
(c) n = (IA/hν) × 1% ≈ 1.51 × 1012 s-1. 63. Particle (mass 5m) decays into 2m and 3m. Ratio of de Broglie wavelengths:
(a) Momentum conservation ⇒ p1 = p2 ⇒ λ1/λ2 = 1. 64. Proton and α-particle accelerated through same V. Ratio of de Broglie wavelengths:
(b) λ = h/√(2mqV) ⇒ λp/λα = √(mαqα/mpqp) = 2√2 : 1. 65. Proton (λ = λ) accelerated through V volt. To get same λ for α-particle, potential needed is:
(a) λ ∝ 1/√(qV) ⇒ Vα = Vp/8 (qα = 2qp). 66. Particles with same velocity. Maximum de Broglie wavelength is for:
- Proton
- α-particle
- Nucleus
- β-particle
(d) λ = h/(mv). β-particle (electron) has smallest mass ⇒ longest λ. 67. Electron and photon with same λ have same:
- Energy
- Linear momentum
- Angular momentum
- Speed
(b) p = h/λ (same for both). Energy differs (Ephoton = pc, Ee = p²/2m). 68. Energy added to electron to reduce λ from 10-10 m to 0.5 × 10-10 m:
- Twice initial energy
- Thrice initial energy
- Four times initial energy
- Equal to initial energy
(b) λ ∝ 1/√E ⇒ 4× energy ⇒ ΔE = 3× initial energy. 69. Electron and photon have same K.E. = 10-20 J. Their wavelengths relate as:
- λe = λph
- λe > λph
- λe < λph
- λe = 0
(c) λph = hc/E, λe = h/√(2mE) ⇒ λe < λph. 70. Electron accelerated from 20V to 40V. de Broglie wavelength at 40V:
(a) λ ∝ 1/√V ⇒ λ40V = λ20V/√2 ≈ 0.75 Å (if λ20V ≈ 1.06 Å). 71. de Broglie wavelength order for e-, p, n, α with same K.E.:
- α < n < p < e-
- e- < p < n < α
- p < n < α < e-
- Cannot compare
(a) λ = h/√(2mK) ⇒ λ ∝ 1/√m. Mass order: α > n ≈ p > e-. 72. Initial momentum of electron if momentum change Δp = pm causes 0.5% λ change:
(a) Δλ/λ = Δp/p ⇒ p = Δp/(Δλ/λ) = pm/0.005 = 200 pm. 73. Ratio of momentum of e- and α accelerated through 100V:
- √(me/mα)
- √(2me/mα)
- √(mα/me)
- √(mα/2me)
(a) p = √(2mqV) ⇒ pe/pα = √(me/mα). 74. de Broglie wavelength of particle (rest mass m0) moving at c:
(b) Relativistic mass → ∞ ⇒ λ = h/p → 0. 75. Photon momentum = 3.3 × 10-27 kg m/s. Frequency is:
- 1.5 × 1015 Hz
- 7.5 × 1015 Hz
- 6.0 × 1015 Hz
- 3.0 × 1015 Hz
(a) E = pc = hν ⇒ ν ≈ 1.5 × 1015 Hz. 76. Ratio of de Broglie wavelengths of proton and α with same energy:[IOM/BPKIHS]
(a) λ ∝ 1/√(mq) ⇒ λp/λα = √(mαqα/mpqp) = 2:1. 77. de Broglie wavelength of electron in first Bohr orbit:
- Equal to circumference
- Equal to twice circumference
- Equal to half circumference
- Equal to one-fourth circumference
(a) 2πr = nλ ⇒ λ = 2πr for n=1. 78. Particle (v = 2.25 × 108 m/s) has same λ as photon. Ratio of K.E./Ephoton:
(a) K.E. = ½mv², Ephoton = pc = hc/λ ⇒ ratio = v/(2c) = 3/8. 79. If K.E. of free electron doubles, de Broglie wavelength changes by factor:
80. Uncertainty in electron position = 10-10 m. Minimum uncertainty in momentum:
- 3.3 × 10-24 kg m/s
- 1.03 × 10-24 kg m/s
- 6.6 × 10-24 kg m/s
- 6.6 × 10-20 kg m/s
(c) Δp ≈ ħ/Δx ≈ 6.6 × 10-24 kg m/s. 81. Laser pulse period = 0.25 μs. Uncertainty in energy:
- 4.2 × 10-9 J
- 4.2 × 10-12 J
- 4.2 × 10-15 J
- 4.2 × 10-30 J
(a) ΔE ≈ ħ/Δt ≈ 4.2 × 10-9 J. 82. Uncertainty in proton speed (Δx = 6 × 10-8 m):
- 1 cm/s
- 1 m/s
- 1 mm/s
- 100 mm/s
(b) Δv ≈ ħ/(mΔx) ≈ 1 m/s. 83. Electron velocity accuracy = 0.005%. Position measurement accuracy:
- 4.634 × 10-4 m
- 4.634 × 10-5 m
- 4.634 × 10-6 m
- 4.634 × 10-7 m
(a) Δx ≈ ħ/(mΔv) ≈ 4.634 × 10-4 m. 84. Electron excitation time = 10-6 μs. Uncertainty in photon frequency:
- 1.6 × 108 Hz
- 1.6 × 109 Hz
- 1.6 × 1010 Hz
- 1.6 × 1011 Hz
(b) Δν ≈ 1/Δt ≈ 1.6 × 109 Hz. 85. de Broglie wavelength of neutron at 927°C is λ. At 27°C, it is:
(a) λ ∝ 1/√T ⇒ λ27°C/λ927°C = √(1200/300) = 2 ⇒ λ → λ/2. 86. Wavelength of 10 keV electron:
(a) λ ≈ 12.3/√V(Å) ≈ 0.12 Å for 10 keV. 87. Particles A (+q) and B (+4q) with same mass 'm' fall through same V. Ratio vA/vB:
(b) v = √(2qV/m) ⇒ vA/vB = √(qA/qB) = 1/2. 88. de Broglie wavelength of body (mass m, energy E):[IOM 2015]
- h/√(2mE)
- √(h/2mE)
- h/(2mE)
- √(2mE)/h
(a) Non-relativistic: λ = h/√(2mE). 89. Quantum theory is explained by:[KU 2016]
- Photon
- Positron
- Electron
- None
(a) Quantum theory introduced photons. 90. Radioactive particle decays into two pieces. Ratio of de Broglie wavelengths λ1/λ2:[IOM 2016]
(c) Momentum conservation ⇒ p1 = p2 ⇒ λ1/λ2 = 1.