Dual Nature of Matter and Radiation is a Class 12 Physics chapter in the NEET (UG) syllabus. NEET720 has 497 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.
102
easy
304
medium
91
hard
Topics covered
Photoelectric effect · Photons · de Broglie waves · Davisson-Germer experiment · X-ray production · de Broglie hypothesis · Photon and particle momentum · Electron emission · X-rays · Wave Nature of Matter · Photon · Wave-Particle Duality · Cathode rays · Photoelectric Effect and Matter Waves · Dual nature of radiation · de Broglie wavelength · Dual nature of radiation and matter · Photoelectric effect vs X-rays · Compton effect · Matter waves · Dual Nature · Photoelectric Effect · De Broglie Wavelength · Photon
8 free Dual Nature of Matter and Radiation practice questions with answers
Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.
Question 1 · medium · de Broglie waves
A student claims: "Just as sound needs air and water waves need water, de Broglie matter waves must need some physical medium to oscillate in — otherwise what exactly is 'waving' as the electron moves?" This claim is
- A.correct, matter waves are mechanical waves that require a supporting medium like a very thin gas of ether particles
- B.correct, but the medium is the electron's own electric field oscillating in space
- D.incorrect; de Broglie waves are not mechanical (or electromagnetic) waves needing a medium — they are associated with the probability of finding the particle, and propagate without any material medium
- C.incorrect; de Broglie waves do need a medium, but only inside solids, not in free space
Show answer and explanation
Answer: D. incorrect; de Broglie waves are not mechanical (or electromagnetic) waves needing a medium — they are associated with the probability of finding the particle, and propagate without any material medium
Matter waves are not mechanical or electromagnetic waves; they need no physical medium — the wave describes the probability amplitude associated with the particle.
de Broglie/matter waves are fundamentally different from sound (mechanical) or light (electromagnetic) waves. They do not require any oscillating medium; instead, they are best understood (in the later quantum-mechanical framework) as related to the probability of finding a particle at a given place and time. The wave nature manifests through interference/diffraction of particle beams (as in the Davisson-Germer experiment), not through mechanical oscillation of any substance.
Common mistake: Assuming all waves must have a physical medium, by analogy with sound
Question 2 · medium · Photoelectric effect
Light of wavelength 300 nm falls on a metal surface with work function 2.0 eV. Taking hc = 1240 eV·nm, the maximum kinetic energy of emitted photoelectrons is closest to:
- A.2.14 eV
- B.4.14 eV
- C.0.14 eV
- D.1.86 eV
Show answer and explanation
Answer: A. 2.14 eV
KE_max = hc/λ - W = 1240/300 - 2.0 = 4.14 - 2.0 = 2.14 eV.
By Einstein's photoelectric equation, KE_max = (hc/λ) - W. Photon energy = 1240/300 ≈ 4.14 eV. Subtracting the work function: KE_max = 4.14 - 2.0 = 2.14 eV.
Common mistake: Forgetting to subtract the work function from the photon energy
Question 3 · medium · Photoelectric effect
A metal has a work function of 2.0 eV. Taking hc = 1240 eV·nm, the threshold wavelength for photoemission from this metal is approximately:
- A.621 nm
- B.2480 nm
- C.310 nm
- D.1240 nm
Show answer and explanation
Answer: A. 621 nm
λ₀ = hc/W = 1240/2.0 = 620 nm ≈ 621 nm.
The threshold wavelength corresponds to photon energy exactly equal to the work function: W = hc/λ₀, so λ₀ = hc/W = 1240 eV·nm / 2.0 eV = 620 nm ≈ 621 nm.
Common mistake: Inverting the formula and multiplying hc by the work function instead of dividing
Question 4 · medium · de Broglie wavelength
An electron is accelerated from rest through a potential difference of 100 V. Its de Broglie wavelength is approximately:
- A.0.123 nm
- B.1.23 nm
- C.12.3 nm
- D.0.0123 nm
Show answer and explanation
Answer: A. 0.123 nm
Using λ=h/√(2meV), for V=100V, λ≈0.123nm — the well-known result λ(nm)≈1.227/√V.
For an electron accelerated through potential V, λ = h/√(2meV). Using the standard result λ(nm) ≈ 1.227/√V, for V = 100 V: λ ≈ 1.227/10 ≈ 0.1227 nm ≈ 0.123 nm.
Common mistake: Misplacing a decimal point when using the shortcut formula λ(nm)=1.227/√V
Question 5 · easy · Photoelectric effect
If the maximum kinetic energy of photoelectrons emitted from a metal is 2.14 eV, the stopping potential required to just stop these electrons is:
- A.2.14 V
- B.1.07 V
- C.4.28 V
- D.0 V, since stopping potential is unrelated to kinetic energy
Show answer and explanation
Answer: A. 2.14 V
eV₀ = KE_max, so numerically the stopping potential (in volts) equals the maximum KE (in eV).
The stopping potential V₀ is defined by the condition that the work done against it equals the maximum kinetic energy of the photoelectrons: eV₀ = KE_max. Since KE_max = 2.14 eV, we directly get V₀ = 2.14 V.
Common mistake: Applying an unnecessary scaling factor instead of the direct numerical equivalence
Question 6 · medium · Photoelectric effect
If the intensity of monochromatic light (frequency held fixed, above threshold) falling on a photosensitive surface is increased, then:
- A.The photoelectric current increases, but the maximum kinetic energy of photoelectrons remains unchanged
- B.Both the photoelectric current and the maximum kinetic energy increase
- C.The photoelectric current remains the same, but the maximum kinetic energy increases
- D.Both the current and maximum kinetic energy decrease
Show answer and explanation
Answer: A. The photoelectric current increases, but the maximum kinetic energy of photoelectrons remains unchanged
More intensity means more photons per second (more electrons emitted, higher current), but each photon's energy (set by frequency) is unchanged, so KE_max stays the same.
Intensity is related to the number of photons striking the surface per second, not to the energy of each individual photon. Increasing intensity at fixed frequency increases the number of photoelectrons emitted per second, raising the photoelectric current, but since each photon still carries the same energy hf, the maximum kinetic energy of individual photoelectrons (given by hf - W) remains unaffected.
Common mistake: Believing that increasing intensity increases the energy of each photon
Question 7 · medium · Photoelectric effect
If the frequency of incident light on a photosensitive surface is increased (kept above the threshold frequency) while intensity is held constant, then:
- A.The maximum kinetic energy of photoelectrons increases, but the saturation photocurrent remains essentially unchanged
- B.The saturation photocurrent increases, but the maximum kinetic energy remains unchanged
- C.Both maximum kinetic energy and saturation current increase proportionally
- D.Neither changes, since only intensity affects photoemission
Show answer and explanation
Answer: A. The maximum kinetic energy of photoelectrons increases, but the saturation photocurrent remains essentially unchanged
KE_max = hf - W increases directly with frequency; at fixed intensity, roughly the same number of photons arrive per second, keeping saturation current about the same.
According to Einstein's photoelectric equation, KE_max = hf - W, so increasing frequency directly increases KE_max. Since intensity (which determines the number of photons per second, and hence saturation current) is held constant, the saturation photocurrent stays essentially the same even as the individual photoelectrons now carry more kinetic energy.
Common mistake: Mixing up which photoelectric quantity (current vs KE_max) depends on frequency versus intensity
Question 8 · easy · Photoelectric effect
Einstein's photoelectric equation, hf = W + KE_max, is best interpreted as:
- A.Each absorbed photon transfers its entire energy to a single electron, part of which overcomes the work function and the rest becomes the electron's kinetic energy
- B.The total energy of all photons striking the surface per second equals the work function
- C.Electrons absorb energy continuously over time until they accumulate enough to escape
- D.Photon energy is shared equally among all free electrons in the metal
Show answer and explanation
Answer: A. Each absorbed photon transfers its entire energy to a single electron, part of which overcomes the work function and the rest becomes the electron's kinetic energy
Photoelectric effect is a one-photon, one-electron quantum process: absorbed photon energy = work function + electron's kinetic energy.
Einstein explained the photoelectric effect using the quantum (photon) picture: a single photon of energy hf is absorbed entirely by a single electron. Part of this energy (equal to the work function W) is used to free the electron from the metal surface, and the remaining energy appears as the electron's kinetic energy, with the maximum possible kinetic energy given by KE_max = hf - W.
Common mistake: Treating the photoelectric equation as describing a continuous or collective absorption process instead of a discrete one-photon event
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Questions about Dual Nature of Matter and Radiation for NEET
How many NEET questions does NEET720 have on Dual Nature of Matter and Radiation?＋
NEET720 has 497 reviewed practice questions on Dual Nature of Matter and Radiation (Physics): 102 easy, 304 medium and 91 hard. 8 of them are free on this page with full explanations; the rest are available in the app.
Is Dual Nature of Matter and Radiation a Class 11 or Class 12 chapter for NEET?＋
Dual Nature of Matter and Radiation is a Class 12 Physics chapter in the NEET (UG) syllabus. Read the NCERT chapter first, then practise chapter-wise MCQs and previous-year questions.
How should I practise Dual Nature of Matter and Radiation for NEET?＋
Attempt the questions below without looking at the options for more than a few seconds, mark your answer, then read the explanation even when you were right. Record every mistake and revisit it after a gap. On NEET720 this happens automatically: wrong answers go to your Mistake Book and are scheduled for spaced revision.
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