Dual Nature of Radiation and Matter NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Dual Nature of Radiation and Matter question from the NEET & AIPMT papers on this site (2005–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Dual Nature of Radiation and Matter previous year questions
44 questions. Tap an option to check your answer.
Q1NEET 2021
An electromagnetic wave of wavelength 'λ' is incident on a photosensitive surface of negligible work function. If 'm' mass is of photoelectron emitted from the surface has de-Broglie wavelength λd, then :
Not quite — the answer is C.
With negligible work function, hc/λ = K = p²/2m and p = h/λd. So hc/λ = h²/(2mλd²), giving λ = (2mc/h)λd². Check dimensions: mc/h has units of 1/length, so (mc/h)λd² is a length.
Q2NEET 2021
The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of 3.3 × 10⁻³ watt will be : (h = 6.6 × 10⁻³⁴ Js)
Not quite — the answer is C.
Photon energy E = hc/λ = (6.6 × 10⁻³⁴ × 3 × 10⁸)/(600 × 10⁻⁹) = 3.3 × 10⁻¹⁹ J. Number per second = P/E = (3.3 × 10⁻³)/(3.3 × 10⁻¹⁹) = 10¹⁶.
Q3NEET 2020 · Q157
Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material. What will be the photoelectric current if the frequency is halved and intensity is doubled ?
Not quite — the answer is D.
Halving 1.5ν₀ gives 0.75ν₀, which is below the threshold frequency, so no photoelectrons are emitted whatever the intensity and the current is zero. Intensity controls the current only when ν exceeds ν₀.
Q4NEET 2020 · Q159
The energy required to break one bond in DNA is 10⁻²⁰ J. This value in eV is nearly :
Not quite — the answer is C.
1 eV = 1.6×10⁻¹⁹ J, so 10⁻²⁰ J = 10⁻²⁰/1.6×10⁻¹⁹ ≈ 0.0625 eV ≈ 0.06 eV. Multiplying instead of dividing by the conversion factor gives meaningless magnitudes.
Q5NEET 2020 · Q175
An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is 1.227×10⁻² nm, the potential difference is :
Not quite — the answer is D.
For an electron λ = 1.227/√V nm, so √V = 1.227/(1.227×10⁻²) = 100 and V = 10⁴ V. Forgetting to square √V gives 10² V.
Q6NEET 2019 · Q13
An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is, (nearly): (mₑ = 9 × 10⁻³¹ kg)
Not quite — the answer is B.
λ = h/√(2meV) ≈ 1.227/√V nm = 1.227/100 nm = 1.227 × 10⁻¹¹ m = 12.2 × 10⁻¹² m. 12.2 nm would correspond to an accelerating voltage of only about 0.01 V.
Q7NEET 2018 · Q12
An electron of mass m with an initial velocity V⃗ = V₀î (V₀ > 0) enters an electric field E⃗ = −E₀î (E₀ = constant > 0) at t = 0. If λ₀ is its de-Broglie wavelength initially, then its de-Broglie wavelength at time t is
Not quite — the answer is C.
Force on the electron F = −eE⃗ = +eE₀î, along its velocity, so v = V₀ + eE₀t/m. Since λ = h/mv, λ = λ₀/(1 + eE₀t/(mV₀)); the wavelength decreases, so the expression that multiplies λ₀ by the growing factor is wrong.
Q8NEET 2018 · Q14
When the light of frequency 2ν₀ (where ν₀ is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is v₁. When the frequency of the incident radiation is increased to 5ν₀, the maximum velocity of electrons emitted from the same plate is v₂. The ratio of v₁ to v₂ is
Not quite — the answer is C.
½mv₁² = h(2ν₀ − ν₀) = hν₀ and ½mv₂² = h(5ν₀ − ν₀) = 4hν₀. So v₁²/v₂² = 1/4 and v₁ : v₂ = 1 : 2; 1 : 4 is the ratio of kinetic energies, not speeds.
Q9NEET 2017 (Gujarati paper) · Q138
In an experiment of photoelectric effect the stopping potential was measured to be V₁ and V₂ with incident light of wavelength λ and λ/2, respectively. The relation between V₁ and V₂ is:
Not quite — the answer is D.
eV₁ = hc/λ − φ and eV₂ = 2hc/λ − φ = 2(eV₁ + φ) − φ = 2eV₁ + φ. Since the work function φ > 0, V₂ > 2V₁. V₂ = 2V₁ would hold only if the work function were zero.
Q10NEET 2016 Phase 2 · Q39
Electrons of mass m with de-Broglie wavelength λ fall on the target in an X-ray tube. The cutoff wavelength (λ₀) of the emitted X-ray is
Not quite — the answer is A.
The electron's kinetic energy is K = p²/(2m) = h²/(2mλ²). The shortest X-ray wavelength carries all of this energy: hc/λ₀ = h²/(2mλ²), so λ₀ = 2mcλ²/h. The cutoff wavelength is not equal to the electron's de Broglie wavelength.
Q11NEET 2016 Phase 2 · Q40
Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV. When photons of energy 6 eV are incident on C, no photoelectrons will reach the anode A, if the stopping potential of A relative to C is
Not quite — the answer is D.
Work function φ = 5 − 2 = 3 eV. For 6 eV photons the maximum kinetic energy is 6 − 3 = 3 eV, so the anode must be at least 3 V negative with respect to the cathode: −3 V. A positive potential would attract the electrons rather than stop them.
Q12NEET 2016 Phase 1 · Q20
An electron of mass m and a photon have same energy E. The ratio of de-Broglie wavelengths associated with them is: (c being velocity of light)
Not quite — the answer is A.
For the electron λe = h/√(2mE); for the photon λp = hc/E. Ratio λe/λp = E/(c√(2mE)) = (1/c)√(E/2m). The expression without 1/c is not dimensionless, so it cannot be a ratio of wavelengths.
Q13NEET 2016 Phase 1 · Q29
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 2λ, the stopping potential is V/4. The threshold wavelength for the metallic surface is:
Not quite — the answer is D.
hc/λ − φ = eV and hc/2λ − φ = eV/4. Subtracting: hc/2λ = 3eV/4, so eV = 2hc/3λ, and φ = hc/λ − 2hc/3λ = hc/3λ. Threshold wavelength = hc/φ = 3λ.
Q14AIPMT 2015
A certain metallic surface is illuminated with monochromatic light of wavelength λ. The stopping potential for photo-electric current for this light is 3V₀. If the same surface is illuminated with light of wavelength 2λ, the stopping potential is V₀. The threshold wavelength for this surface for photo-electric effect is:
Not quite — the answer is C.
hc/λ = φ + 3eV₀ and hc/2λ = φ + eV₀. Subtracting, hc/2λ = 2eV₀, so eV₀ = hc/4λ and φ = hc/2λ − hc/4λ = hc/4λ. Threshold wavelength = hc/φ = 4λ.
Q15AIPMT 2014
If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is:
Not quite — the answer is A.
λ = h/√(2mK), so λ ∝ 1/√K. Increasing K 16 times makes λ one quarter of its value, a 75% decrease; 25% is the remaining fraction, not the change.
Q16AIPMT 2014
When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal surface increased from 0.5 eV to 0.8 eV. The work function of the metal is:
Not quite — the answer is A.
E − φ = 0.5 eV and 1.2E − φ = 0.8 eV. Subtracting, 0.2E = 0.3 eV, so E = 1.5 eV and φ = 1.0 eV; 1.5 eV is the photon energy, not the work function.
Q17NEET 2013
For photoelectric emission from certain metal the cutoff frequency is ν. If radiation of frequency 2ν impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass)
Not quite — the answer is C.
Work function = hν (cutoff). KEmax = h(2ν) − hν = hν = (1/2)mv², so v = √(2hν/m). Using √(hν/m) drops the factor 2 from the kinetic energy formula.
Q18NEET 2013
The wavelength λe of an electron and λp of a photon of same energy E are related by
Not quite — the answer is A.
Electron: λe = h/√(2mE), so λe² ∝ 1/E. Photon: λp = hc/E ∝ 1/E. Therefore λp ∝ λe². A direct proportion would be true only if both depended on E in the same way.
Q19AIPMT 2012 Mains
If the momentum of electron is changed by P, then the de Broglie wavelength associated with it changes by 0.5%. The initial momentum of electron will be
Not quite — the answer is A.
λ = h/p gives |Δλ/λ| = Δp/p for small changes. So 0.005 = P/p₀ and p₀ = P/0.005 = 200P. P/200 inverts the relation.
Q20AIPMT 2012 Mains
Two radiations of photons energies 1 eV and 2.5 eV, successively illuminate a photosensitive metallic surface of work function 0.5 eV. The ratio of the maximum speeds of the emitted electrons is :
Not quite — the answer is B.
Maximum kinetic energies are 1 − 0.5 = 0.5 eV and 2.5 − 0.5 = 2 eV, a ratio of 1 : 4. Speed goes as √KE, so the speeds are in the ratio 1 : 2; 1 : 4 is the energy ratio, not the speed ratio.
Q21AIPMT 2012 Prelims
An α-particle moves in a circular path of radius 0.83 cm in the presence of a magnetic field of 0.25 Wb/m². The de Broglie wavelength associated with the particle will be
Not quite — the answer is A.
In a magnetic field, momentum p = qBr with q = 2e for an α-particle. λ = h/(2eBr) = 6.63 × 10⁻³⁴/(2 × 1.6 × 10⁻¹⁹ × 0.25 × 0.83 × 10⁻²) ≈ 1.0 × 10⁻¹² m = 0.01 Å. Taking the charge as e instead of 2e doubles the wavelength.
Q22AIPMT 2012 Prelims
A 200 W sodium street lamp emits yellow light of wavelength 0.6 µm. Assuming it to be 25% efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is
Not quite — the answer is B.
Light power = 25% of 200 W = 50 W. Energy per photon = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(0.6 × 10⁻⁶) ≈ 3.3 × 10⁻¹⁹ J. Number per second = 50/3.3 × 10⁻¹⁹ ≈ 1.5 × 10²⁰. Using the full 200 W gives about 6 × 10²⁰, ignoring the efficiency.
Q23AIPMT 2012 Prelims
Monochromatic radiation emitted when an electron in hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured to be 3.57 V. The threshold frequency of the material is
Not quite — the answer is D.
The n = 2 → 1 transition in hydrogen emits a 10.2 eV photon. Work function = 10.2 − 3.57 = 6.63 eV = 6.63 × 1.6 × 10⁻¹⁹ J. Threshold frequency ν₀ = φ/h = (6.63 × 1.6 × 10⁻¹⁹)/(6.63 × 10⁻³⁴) = 1.6 × 10¹⁵ Hz. Using the full 10.2 eV gives about 2.5 × 10¹⁵ Hz, which is the photon frequency, not the threshold.
Q24AIPMT 2011 Mains
The threshold frequency for a photosensitive metal is 3.3 × 10¹⁴ Hz. If light of frequency 8.2 × 10¹⁴ Hz is incident on this metal, the cut-off voltage for the photoelectric emission is nearly
Not quite — the answer is C.
eV₀ = h(ν − ν₀) = 6.63 × 10⁻³⁴ × (8.2 − 3.3) × 10¹⁴ = 3.25 × 10⁻¹⁹ J, so V₀ = 3.25 × 10⁻¹⁹/1.6 × 10⁻¹⁹ ≈ 2 V. Using the incident frequency alone (ignoring the threshold) would give about 3.4 V, which is not the cut-off voltage.
Q25AIPMT 2011 Prelims
A radioactive nucleus of mass M emits a photon of frequency ν and the nucleus recoils. The recoil energy will be
Not quite — the answer is C.
The photon carries momentum p = hν/c, and the nucleus recoils with equal and opposite momentum. Its kinetic energy is p²/2M = h²ν²/(2Mc²). It is not hν, which is the photon's own energy.
Q26AIPMT 2011 Prelims
In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic electrons is 0.5 eV. The corresponding stopping potential is
Not quite — the answer is D.
The stopping potential satisfies eV₀ = K(max) = 0.5 eV, so V₀ = 0.5 V. The work function is irrelevant once K(max) is given; adding it (2.3 V) gives the photon energy, not the stopping potential.
Q27AIPMT 2011 Prelims
Electrons used in an electron microscope are accelerated by a voltage of 25 kV. If the voltage is increased to 100 kV then the de-Broglie wavelength associated with the electrons would
Not quite — the answer is C.
For electrons accelerated through V, λ = h/√(2meV), so λ ∝ 1/√V. Increasing V from 25 kV to 100 kV (4 times) reduces λ by √4 = 2 times. 'Decrease by 4 times' wrongly takes λ ∝ 1/V.
Q28AIPMT 2011 Prelims
Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5 eV successively. Ratio of maximum speeds of emitted electrons will be
Not quite — the answer is C.
K(max) = hν − φ gives 1 − 0.5 = 0.5 eV and 2.5 − 0.5 = 2.0 eV. Since v ∝ √K, v₁ : v₂ = √0.5 : √2 = 1 : 2. Taking the ratio of photon energies (1 : 2.5) or of kinetic energies (1 : 4) are the traps.
Q29AIPMT 2011 Prelims
Photoelectric emission occurs only when the incident light has more than a certain minimum
Not quite — the answer is A.
Photoelectric emission requires the frequency to exceed the threshold frequency of the metal, whatever the intensity. Intensity or power only changes the number of electrons once emission occurs; a minimum wavelength is the wrong sense since there is a maximum wavelength.
Q30AIPMT 2010 Mains · Q66
When monochromatic radiation of intensity I falls on a metal surface, the number of photoelectron and their maximum kinetic energy are N and T respectively. If the intensity of radiation is 2I, the number of emitted electrons and their maximum kinetic energy are respectively
Not quite — the answer is B.
Doubling intensity doubles the number of photons per second, so the number of photoelectrons doubles to 2N. The maximum kinetic energy depends only on frequency and work function (hν − φ), so it stays T. Doubling the kinetic energy would need a change in frequency, not intensity.
Q31AIPMT 2010 Mains · Q79
The electron in the hydrogen atom jumps from excited state (n = 3) to its ground state (n = 1) and the photons thus emitted irradiate a photosensitive material. If the work function of the material is 5.1 eV, the stopping potential is estimated to be (the energy of the electron in nth state En = −13.6/n² eV)
Not quite — the answer is D.
Photon energy for n = 3 → 1 is 13.6(1 − 1/9) = 12.09 eV. Maximum kinetic energy = 12.09 − 5.1 ≈ 7.0 eV, so the stopping potential is about 7 V. 12.1 V forgets to subtract the work function.
Q32AIPMT 2010 Prelims · Q39
A source S₁ is producing 10¹⁵ photons per second of wavelength 5000 Å. Another source S₂ is producing 1.02 × 10¹⁵ photons per second of wavelength 5100 Å. Then (power of S₂)/(power of S₁) is equal to
Not quite — the answer is B.
Power = n·hc/λ. P₂/P₁ = (n₂/n₁)(λ₁/λ₂) = 1.02 × 5000/5100 = 1.00. 1.02 counts only the photon numbers and ignores the lower energy of longer-wavelength photons.
Q33AIPMT 2009 Prelims · Q4
The number of photo electrons emitted for light of a frequency ν (higher than the threshold frequency ν₀) is proportional to:
Not quite — the answer is B.
Above threshold, each photon ejects at most one electron, so the number of photoelectrons depends on the number of photons arriving per second, i.e. the intensity. The frequency (and ν − ν₀) decides the kinetic energy of the electrons, not their number.
Q34AIPMT 2009 Prelims · Q30
Monochromatic light of wavelength 667 nm is produced by a helium neon laser. The power emitted is 9 mW. The number of photons arriving per sec on the average at a target irradiated by this beam is:
Not quite — the answer is A.
Energy per photon = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(667 × 10⁻⁹) ≈ 2.98 × 10⁻¹⁹ J. Photons per second = 9 × 10⁻³/2.98 × 10⁻¹⁹ ≈ 3 × 10¹⁶. Forgetting the milli prefix gives 3 × 10¹⁹.
Q35AIPMT 2008 Prelims · Q41
The work function of a surface of a photosensitive material is 6.2 eV. The wavelength of the incident radiation for which the stopping potential is 5 V lies in the
Not quite — the answer is B.
Photon energy = φ + eV₀ = 6.2 + 5 = 11.2 eV, so λ = 12400/11.2 eV·Å ≈ 1100 Å (about 110 nm). This lies in the ultraviolet; visible light is only about 1.8 to 3.1 eV, and X-rays need wavelengths below about 100 Å.
Q36AIPMT 2008 Prelims · Q42
A particle of mass 1 mg has the same wavelength as an electron moving with a velocity of 3 × 10⁶ m s⁻¹. The velocity of the particle is (mass of electron = 9.1 × 10⁻³¹ kg)
Not quite — the answer is B.
Equal de Broglie wavelengths (λ = h/mv) mean equal momenta. 1 mg = 10⁻⁶ kg, so v = (9.1 × 10⁻³¹ × 3 × 10⁶)/10⁻⁶ ≈ 2.7 × 10⁻¹⁸ m s⁻¹. Taking 1 mg as 10⁻³ kg gives 2.7 × 10⁻²¹ m s⁻¹.
Q37AIPMT 2007 Prelims · Q29
Monochromatic light of frequency 6.0 × 10¹⁴ Hz is produced by a laser. The power emitted is 2 × 10⁻³ W. The number of photons emitted, on the average, by the sources per second is
Not quite — the answer is D.
n = P/(hf) = 2 × 10⁻³/(6.63 × 10⁻³⁴ × 6.0 × 10¹⁴) = 2 × 10⁻³/3.98 × 10⁻¹⁹ ≈ 5 × 10¹⁵ per second. 5 × 10¹⁶ is a power-of-ten slip.
Q38AIPMT 2007 Prelims · Q36
A 5 watt source emits monochromatic light of wavelength 5000 Å. When placed 0.5 m away, it liberates photoelectrons from a photosensitive metallic surface. When the source is moved to a distance of 1.0 m, the number of photoelectrons liberated will be reduced by a factor of
Not quite — the answer is D.
The number of photoelectrons is proportional to intensity, which falls as 1/r². Doubling the distance from 0.5 m to 1.0 m reduces it by 2² = 4. A factor of 2 assumes intensity ∝ 1/r.
Q39AIPMT 2006 Prelims · Q22
A photo-cell employs photoelectric effect to convert :
Not quite — the answer is B.
Above the threshold, the photoelectric current is proportional to the intensity of the incident light, so a photocell turns changes in illumination into changes in photocurrent. Frequency changes alter the kinetic energy of the electrons, not the current, and the work function is a fixed property of the cathode.
Q40AIPMT 2006 Prelims · Q25
In a discharge tube ionization of enclosed gas is produced due to collisions between :
Not quite — the answer is B.
In a discharge tube the strong electric field accelerates the free electrons, which knock electrons off neutral gas atoms or molecules and ionise them. Positive ions are heavy and gain much less speed between collisions, so they are not the main cause of ionisation.
Q41AIPMT 2006 Prelims · Q26
When photons of energy hν fall on an aluminium plate (of work function E₀), photoelectrons of maximum kinetic energy K are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be :
Not quite — the answer is D.
Einstein's equation gives K = hν − E₀. At frequency 2ν, K' = 2hν − E₀ = (hν − E₀) + hν = K + hν. The kinetic energy does not simply double, because the work function is subtracted only once.
Q42AIPMT 2006 Prelims · Q31
The momentum of a photon of energy 1 MeV in kg m/s, will be :
Not quite — the answer is D.
E = 1 MeV = 1.6 × 10⁻¹³ J and p = E/c = 1.6 × 10⁻¹³/(3 × 10⁸) ≈ 5.3 × 10⁻²² kg m/s. Skipping the MeV-to-joule conversion gives a meaningless number, and 5 × 10⁻²³ is off by a factor of ten.
Q43AIPMT 2005 Prelims · Q12
The work functions for metals A, B and C are respectively 1.92 eV, 2.0 eV and 5 eV. According to Einstein's equation, the metals which will emit photoelectrons for a radiation of wavelength 4100 Å is/are :
Not quite — the answer is C.
E = 12400/4100 ≈ 3.0 eV. This exceeds the work functions of A (1.92 eV) and B (2.0 eV) but not C (5 eV), so only A and B emit. 'All three' ignores that 5 eV > 3 eV.
Q44AIPMT 2005 Prelims · Q22
A photosensitive metallic surface has work function, hν₀. If photons of energy 2hν₀ fall on this surface, the electrons come out with a maximum velocity of 4 × 10⁶ m/s. When the photon energy is increased to 5hν₀, then maximum velocity of photoelectrons will be :
Not quite — the answer is D.
KE_max = E − hν₀. With 2hν₀ it is hν₀; with 5hν₀ it is 4hν₀. Since v ∝ √KE, the speed doubles to 8 × 10⁶ m/s. 16 × 10⁶ m/s wrongly takes v ∝ KE.
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Questions from this chapter in each paper on this site.
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