68 free MCQs on Dual Nature of Radiation and Matter with worked answers and explanations. Photoelectric effect, de Broglie waves, Bohr's model, atomic spectra.
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In the photoelectric setup, light striking the metal plate ejects electrons that cross the evacuated tube to the collector, producing a measurable current on the ammeter.
Q1.
The photoelectric effect demonstrates that light has:
Answer: B. Particle nature
Why: Photoelectric effect shows light behaves as particles (photons). Wave theory cannot explain why electrons are only emitted above a threshold frequency.
Q2.
The energy of a photon with frequency f (h = Planck's constant) is:
Answer: B. hf
Why: E = hf. Photon energy is proportional to frequency. Higher frequency = more energetic photon.
Q3.
Work function of a metal is the minimum energy needed to:
Answer: C. Remove an electron from its surface
Why: Work function (phi) is the minimum energy required to remove a conduction electron from the metal surface.
Q4.
Threshold frequency in photoelectric effect is the frequency below which:
Answer: D. All of above
Why: Below threshold frequency: E = hf < phi. No electrons are emitted regardless of light intensity.
Q5.
Stopping potential in photoelectric effect is the voltage needed to:
Answer: B. Stop all emitted electrons
Why: Stopping potential V<sub>0</sub> halts all emitted electrons. eV<sub>0</sub> = maximum KE of emitted electrons = hf - phi.
Q6.
De Broglie wavelength of a particle of momentum p:
Answer: C. h/p
Why: De Broglie wavelength lambda = h/p = h/(mv). Every particle has an associated wave.
Q7.
In Bohr's model, electrons occupy:
Answer: B. Only specific quantized orbits
Why: Bohr's postulate: electrons can only occupy specific quantized orbits where angular momentum = n x h/(2pi).
Q8.
Planck's constant h has the value:
Answer: A. 6.626 x 10<sup>-34</sup> J s
Why: h = 6.626 x 10<sup>-34</sup> J s. It is the fundamental quantum of action (energy x time).
Q9.
Increasing intensity of light in photoelectric effect increases:
Answer: C. Number of emitted electrons (photocurrent)
Why: Higher intensity = more photons per second = more electrons emitted per second = higher photocurrent. KE is unaffected.
Q10.
Thomson's atomic model is called:
Answer: C. Plum pudding model
Why: Thomson proposed a plum pudding model: electrons (plums) embedded in a spread-out positive charge (pudding).
Q11.
Photon has mass:
Answer: B. Zero rest mass but carries energy
Why: Photon has zero rest mass but carries energy E = hf and momentum p = h/lambda.
Q12.
Compton effect shows that:
Answer: B. Photons have momentum
Why: Compton effect: X-ray photons scattered by electrons change wavelength. This proves photons carry momentum p = h/lambda.
Q13.
Davisson-Germer experiment confirmed:
Answer: A. Wave nature of electrons
Why: Davisson and Germer (1927) showed electron diffraction from a crystal: confirming de Broglie's hypothesis of electron waves.
Q14.
Light can behave both as a wave and as a:
Answer: A. particle
Why: Light shows a dual wave–particle nature.
Q15.
A quantum (packet) of light energy is called a:
Answer: A. photon
Why: Light energy comes in discrete packets called photons.
Q16.
The photoelectric effect is the emission of ___ when light strikes a metal surface:
Answer: A. electrons
Why: Incident light ejects electrons from the metal in the photoelectric effect.
Q17.
The energy of a photon is directly proportional to its:
Answer: A. frequency
Why: E = hν, so photon energy increases with frequency.
Q18.
The photoelectric effect was successfully explained by:
Answer: A. Einstein
Why: Einstein explained it using the photon (quantum) idea in 1905.
Q19.
A photon has a rest mass of:
Answer: A. zero
Why: Photons are massless; they travel at the speed of light.
Q20.
Increasing the intensity of the incident light increases the number of emitted:
Answer: A. electrons
Why: Greater intensity means more photons, so more electrons are ejected per second.
Q21.
De Broglie wavelength of a particle with momentum p is lambda =
Answer: A. h/p
Why: De Broglie: lambda = h/p where h = Planck constant and p = momentum. Applies to all matter.
Q22.
In the photoelectric effect, stopping potential depends on:
Answer: B. Frequency only (and work function)
Why: Stopping potential V₀ = (hf - phi)/e. It depends on frequency f and work function phi, not intensity.
Q23.
The work function of a metal is 2 eV. Threshold frequency (h = 6.6×10⁻³⁴ J·s) is approx:
Answer: A. 4.8 × 10¹⁴ Hz
Why: phi = hf₀. f₀ = phi/h = (2×1.6×10⁻¹⁹)/(6.6×10⁻³⁴) = 3.2×10⁻¹⁹/6.6×10⁻³⁴ ≈ 4.8×10¹⁴ Hz.
Q24.
In Bohr model, the radius of nth orbit rn ∝
Answer: B. n²
Why: Bohr radius: rn = n² × a₀ where a₀ = 0.529 Angstrom (Bohr radius). rn is proportional to n².
Q25.
The energy of the electron in nth orbit of hydrogen: En =
Answer: A. -13.6/n² eV
Why: En = -13.6/n² eV. Negative means bound state. Ground state (n=1): -13.6 eV.
Q26.
Heisenberg uncertainty principle: Delta_x × Delta_p ≥
Answer: A. h/4π
Why: Uncertainty principle: Delta_x × Delta_px >= h/(4π) = hbar/2. Simultaneous precise position and momentum is impossible.
Q27.
X-rays are produced when:
Answer: B. High-energy electrons hit a metal target
Why: X-rays: produced when fast electrons decelerate suddenly in metal target (bremsstrahlung) or when inner electrons are knocked out.
Q28.
If wavelength of electron = wavelength of photon, which has more energy?
Answer: B. Photon
Why: Photon energy E = hc/lambda. Electron KE = p²/2m = h²/(2m lambda²). Since hc/lambda >> h²/(2m lambda²) for typical wavelengths, photon has more energy.
Q29.
The first emission line in Balmer series corresponds to transition:
Answer: B. n=3 to n=2
Why: Balmer series: transitions to n=2. First line: n=3 to n=2 (lowest energy, red H-alpha line at 656 nm).
Q30.
Kinetic energy of emitted photoelectron when light of frequency f hits metal:
Answer: B. hf - phi (work function)
Why: Photoelectric equation: KE_max = hf - phi. Energy of photon minus work function gives kinetic energy of emitted electron.
Q31.
Compton scattering: a photon scatters from an electron. The scattered photon has:
Answer: C. Lower energy (longer wavelength)
Why: Compton effect: photon transfers momentum to electron and loses energy. Scattered photon has longer wavelength (lower energy).
Q32.
Ionization energy of hydrogen from ground state:
Answer: C. 13.6 eV
Why: Energy to remove electron from n=1: E = 0 - (-13.6) = 13.6 eV. This is the ionization energy.
Q33.
The wavelength of matter waves decreases when:
Answer: B. Momentum increases
Why: lambda = h/p. If momentum p increases (higher speed or mass), wavelength decreases.
Q34.
Bohr model explains only the spectrum of:
Answer: C. Hydrogen-like (one-electron) atoms
Why: Bohr model: works for hydrogen and hydrogen-like ions (He+, Li²+). Fails for multi-electron atoms.
Q35.
The energy of a photon of frequency ν is given by:
Answer: A. hν
Why: Photon energy E = hν, where h is Planck’s constant.
Q36.
The minimum frequency of light needed to eject electrons from a metal is the ___ frequency:
Answer: A. threshold
Why: Below the threshold frequency, no photoelectrons are emitted.
Q37.
The minimum energy required to free an electron from a metal surface is called the:
Answer: A. work function
Why: The work function is the least energy needed to release an electron from the metal.
Q38.
The de Broglie wavelength of a particle of momentum p is:
Answer: A. h/p
Why: The matter wavelength is λ = h/p.
Q39.
The stopping potential in the photoelectric effect depends on the ___ of the incident light:
Answer: A. frequency
Why: The maximum kinetic energy, and hence the stopping potential, depends on the light’s frequency.
Q40.
Below the threshold frequency, no electrons are emitted no matter how great the light’s:
Answer: A. intensity
Why: Intensity cannot compensate for too low a frequency - the effect depends on frequency.
Q41.
Moseley law for characteristic X-ray frequency: sqrt(f) ∝
Answer: B. Z-1
Why: Moseley law: sqrt(f) = a(Z - b) where b is approximately 1 for K-alpha lines. So sqrt(f) proportional to (Z-1).
Q42.
The quantum numbers (n, l, ml, ms) for the outermost electron of Na (Z=11):
Answer: A. 3, 0, 0, +1/2
Why: Na configuration: 1s²2s²2p⁶3s¹. Outermost electron in 3s: n=3, l=0, ml=0, ms=±1/2.
Q43.
The minimum wavelength of X-rays produced by electrons accelerated through potential V is:
Answer: A. hc/(eV)
Why: Duane-Hunt law: lambda_min = hc/(eV). All electron KE converts to single X-ray photon.
Q44.
In Compton scattering, the Compton wavelength shift is given by:
Answer: A. Delta_lambda = (h/m<sub>e</sub> c)(1 - cos theta)
Why: Compton shift: Delta_lambda = (h/m<sub>e</sub> c)(1-cos theta) where theta is scattering angle. At theta=90°: Delta = h/m<sub>e</sub> c = 2.43 pm.
Q45.
The angular momentum of an electron in s orbital is:
Answer: C. 0
Why: For l=0 (s orbital): L = sqrt(l(l+1)) hbar = 0. S orbitals have zero orbital angular momentum.
Q46.
The cutoff frequency (threshold) in photoelectric effect for a metal with work function phi:
Answer: A. f₀ = phi/h
Why: Threshold: hf₀ = phi. f₀ = phi/h. Below this frequency, no photoelectrons regardless of intensity.
Q47.
The ground state wavefunction of hydrogen: psi ∝ e<sup>-r/a₀</sup>. Probability density peaks at:
Answer: B. r = a₀
Why: Radial probability P(r) = |psi|² × 4pi r² peaks at a₀ (Bohr radius), even though |psi|² is maximum at r=0.
Q48.
Electron affinity differs from work function because:
Answer: B. Electron affinity: energy to ADD electron to neutral atom; work function: energy to REMOVE from solid
Why: Work function: energy to remove electron from metal surface. Electron affinity: energy released when electron is added to neutral atom. Different quantities.
Q49.
Photo current in photoelectric effect is proportional to:
Answer: C. Intensity (number of photons)
Why: Photocurrent proportional to number of emitted electrons proportional to intensity (number of incident photons per second).
Q50.
Electron spin quantum number ms can only be:
Answer: B. ±1/2
Why: Spin quantum number ms = ±1/2 (spin-up or spin-down). This is a fundamental quantum property with no classical analogue.
Q51.
Pauli exclusion principle states that:
Answer: B. No two electrons can have all four identical quantum numbers
Why: Pauli exclusion: no two electrons in an atom can have the same set of four quantum numbers (n, l, ml, ms).
Q52.
Einstein’s photoelectric equation is KE_max = hν minus the:
Answer: A. work function φ
Why: KE_max = hν − φ, where φ is the work function of the metal.
Q53.
The de Broglie (matter) wavelength is appreciable only for particles of very ___ mass:
Answer: A. small
Why: Because λ = h/mv, only very light particles have a measurable wavelength.
Q54.
The photoelectric effect provides direct evidence for the ___ nature of light:
Answer: A. particle
Why: The instantaneous, frequency-dependent emission supports the particle (photon) model.
Q55.
A photon of wavelength λ carries a momentum of:
Answer: A. h/λ
Why: Photon momentum p = h/λ.
Q56.
If the frequency of the incident light is increased above threshold, the maximum kinetic energy of the photoelectrons:
Answer: A. increases
Why: Since KE_max = hν − φ, a higher frequency raises the maximum kinetic energy.
Q57.
The work function of a metal determines its ___ frequency:
Answer: A. threshold
Why: The threshold frequency ν₀ = φ/h is fixed by the work function.
Q58.
An electron accelerated through a potential difference V has a de Broglie wavelength proportional to:
Answer: A. 1/√V
Why: λ = h/√(2meV), so λ ∝ 1/√V.
Q59.
The Davisson–Germer experiment confirmed the ___ nature of electrons:
Answer: A. wave
Why: Electron diffraction in the experiment demonstrated the wave nature of electrons.
Q60.
Above the threshold frequency, the photoelectric current is proportional to the ___ of the incident light:
Answer: A. intensity
Why: The number of photoelectrons, and hence the current, is proportional to the light intensity.
Q61.
Light of photon energy 5 eV falls on a metal of work function 2 eV. The maximum kinetic energy of the emitted photoelectrons is:
Answer: B. 3 eV
Why: K<sub>max</sub> = E<sub>photon</sub> − work function = 5 − 2 = 3 eV.
Q62.
If the incident light frequency is below the threshold frequency of a metal, then increasing the intensity results in:
Answer: A. no photoemission
Why: Below the threshold frequency no photoelectrons are emitted regardless of intensity.
Q63.
The de Broglie wavelength of a particle is inversely proportional to its momentum. If the momentum is doubled, the wavelength:
Answer: A. halves
Why: λ = h/p, so doubling p halves the wavelength.
Q64.
In the photoelectric effect, the stopping potential is:
Answer: A. independent of the light intensity
Why: Stopping potential depends on frequency, not intensity; it is independent of the light intensity.
Q65.
An electron accelerated through a potential difference of 100 V has a de Broglie wavelength of about:
Answer: B. 1.23 Å
Why: λ = 12.27/√V Å = 12.27/√100 = 1.23 Å.
Q66.
The energy of a photon of wavelength 500 nm (using hc = 1240 eV·nm) is:
Answer: B. 2.48 eV
Why: E = hc/λ = 1240/500 = 2.48 eV.
Q67.
Increasing the intensity of incident light (frequency unchanged, above threshold) in the photoelectric effect results in:
Answer: A. more photoelectrons with the same maximum KE
Why: Higher intensity means more photons per second, so more photoelectrons, but the maximum KE (set by frequency) is unchanged.
Q68.
A photon has wavelength 6.6×10⁻⁷ m (h = 6.6×10⁻³⁴ J·s). Its momentum is:
Answer: A. 1×10⁻²⁷ kg·m/s
Why: p = h/λ = 6.6×10⁻³⁴/6.6×10⁻⁷ = 1×10⁻²⁷ kg·m/s.