NEET Dual Nature of Matter and Radiation PYQs | 2013–2025

2025

Q1. Derive the de Broglie wavelength of a particle moving with momentum p.

Q2. A photon of wavelength 400 nm strikes a metal surface with work function 2 eV. Calculate the maximum kinetic energy of the emitted photoelectron.

Q3. Explain the photoelectric effect and state the experimental observations that confirm the particle nature of light.


2024

Q1. Derive the expression for the energy of a photon in terms of its frequency.

Q2. Calculate the threshold frequency for a metal with work function 4.5 eV.

Q3. Explain why increasing the intensity of light does not increase the kinetic energy of photoelectrons.


2023

Q1. A particle has mass m and velocity v. Write the expression for its de Broglie wavelength and discuss the significance.

Q2. Explain Einstein’s photoelectric equation and its terms.

Q3. Discuss experimental verification of matter waves using electron diffraction.


2022

Q1. A photon of wavelength 250 nm falls on a surface with work function 3 eV. Calculate the stopping potential.

Q2. Derive the expression for de Broglie wavelength in terms of kinetic energy.

Q3. Explain the difference between wave and particle nature of light with suitable examples.


2021

Q1. Derive the expression for de Broglie wavelength for electrons accelerated through a potential difference V.

Q2. Explain the significance of the Davisson–Germer experiment in confirming the wave nature of electrons.

Q3. A photon of frequency f strikes a metal surface. Write the expression for the maximum velocity of emitted electrons.


2020

Q1. Explain the concept of matter waves and derive the de Broglie relation.

Q2. A metal surface has work function 2.5 eV. Light of wavelength 300 nm is incident. Find the maximum kinetic energy of photoelectrons.

Q3. Discuss the limitations of classical wave theory in explaining the photoelectric effect.


2019

Q1. Derive the relation between wavelength and momentum for a particle.

Q2. Explain the photoelectric effect with a suitable diagram.

Q3. An electron is accelerated through 100 V. Calculate its de Broglie wavelength.


2018

Q1. Derive the expression for photon energy in terms of wavelength.

Q2. Explain the stopping potential and its relation with kinetic energy of photoelectrons.

Q3. Explain how electron diffraction confirms the wave nature of matter.


2017

Q1. A photon of wavelength 500 nm falls on a metal surface (work function 2 eV). Find kinetic energy of photoelectron.

Q2. Write the expression for de Broglie wavelength of electron in terms of accelerating voltage.

Q3. Describe an experiment to demonstrate wave nature of electrons.


2016

Q1. Derive the energy–momentum relation for a photon.

Q2. A particle of mass m moves with velocity v. Find the corresponding de Broglie wavelength.

Q3. Explain why the classical theory fails to explain the photoelectric effect.


2015

Q1. Derive the de Broglie wavelength of a particle moving under a potential difference V.

Q2. Explain Einstein’s photoelectric equation.

Q3. Electron accelerated through 200 V → calculate its de Broglie wavelength.


2014

Q1. Derive the expression for the kinetic energy of photoelectrons in the photoelectric effect.

Q2. Explain the significance of the de Broglie hypothesis.

Q3. Describe an experiment that confirms wave nature of electrons.


2013

Q1. Photon of frequency f strikes a metal with work function φ. Derive expression for stopping potential.

Q2. Derive the de Broglie wavelength for a particle of mass m and kinetic energy K.

Q3. Discuss the dual nature of matter and radiation with examples.

Dual Nature of Matter and Radiation — Solutions (2025 → 2013)


2025

Q1. de Broglie wavelength:λ=hpwhere p=momentum of particle, h=Planck’s constant\lambda = \frac{h}{p} \quad \text{where } p = \text{momentum of particle, } h = \text{Planck’s constant}λ=ph​where p=momentum of particle, h=Planck’s constant


Q2. Photon λ = 400 nm, work function φ = 2 eV.Ephoton=hcλ=6.626×10343×108400×1094.97×1019J3.1eVE_{\text{photon}} = \frac{hc}{\lambda} = \frac{6.626\times10^{-34} \cdot 3\times10^8}{400\times10^{-9}} \approx 4.97\times10^{-19}\,\text{J} \approx 3.1\,\text{eV}Ephoton​=λhc​=400×10−96.626×10−34⋅3×108​≈4.97×10−19J≈3.1eV

Maximum kinetic energy:Kmax=Ephotonϕ=3.12=1.1eVK_{\text{max}} = E_{\text{photon}} – \phi = 3.1 – 2 = 1.1\,\text{eV}Kmax​=Ephoton​−ϕ=3.1−2=1.1eV


Q3. Photoelectric effect observations:

  • Instantaneous emission of electrons.
  • KE depends on frequency, not intensity.
  • Number of electrons depends on intensity.

Confirms particle nature of light (photons).


2024

Q1. Photon energy: E=hνE = h \nuE=hν, ν=\nu =ν= frequency

Q2. Threshold frequency: f0=ϕ/h=4.5/4.136×10151.088×1015Hzf_0 = \phi/h = 4.5 / 4.136\times10^{-15} \approx 1.088\times10^{15}\,\text{Hz}f0​=ϕ/h=4.5/4.136×10−15≈1.088×1015Hz

Q3. Increasing intensity increases number of electrons but not their kinetic energy, since KE depends on frequency.


2023

Q1. de Broglie wavelength: λ=h/p=h/(mv)\lambda = h / p = h / (mv)λ=h/p=h/(mv)

Q2. Einstein’s photoelectric equation: Kmax=hνϕK_{\text{max}} = h\nu – \phiKmax​=hν−ϕ

Q3. Electron diffraction: Electrons diffracted by crystal lattice → interference pattern → confirms wave nature.


2022

Q1. Photon λ = 250 nm, φ = 3 eV:E=hcλ=6.626×10343×108250×1097.95×1019J4.97eVE = \frac{hc}{\lambda} = \frac{6.626\times10^{-34}\cdot3\times10^8}{250\times10^{-9}} \approx 7.95\times10^{-19}\,\text{J} \approx 4.97\,\text{eV}E=λhc​=250×10−96.626×10−34⋅3×108​≈7.95×10−19J≈4.97eV

Stopping potential:Kmax=eVs=Eϕ=4.973=1.97eV    Vs=1.97VK_{\text{max}} = eV_s = E – \phi = 4.97 – 3 = 1.97\,\text{eV} \implies V_s = 1.97\,\text{V}Kmax​=eVs​=E−ϕ=4.97−3=1.97eV⟹Vs​=1.97V

Q2. de Broglie wavelength in terms of kinetic energy:λ=hp=h2mK\lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m K}}λ=ph​=2mK​h​

Q3. Wave nature: diffraction/interference, particle nature: photoelectric effect.


2021

Q1. Electron accelerated through V:λ=hp=h2meV\lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m e V}}λ=ph​=2meV​h​

Q2. Davisson–Germer experiment: electrons diffracted by crystal → wave behavior confirmed.

Q3. Photon incident → max velocity of photoelectron:Kmax=12mvmax2=hνϕ    vmax=2(hνϕ)mK_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2 = h\nu – \phi \implies v_{\text{max}} = \sqrt{\frac{2(h\nu – \phi)}{m}}Kmax​=21​mvmax2​=hν−ϕ⟹vmax​=m2(hν−ϕ)​​


2020

Q1. Matter waves: λ=h/p\lambda = h/pλ=h/p, significance: particles show wave behavior at atomic scale.

Q2. λ = 300 nm, φ = 2.5 eV → E = 4.14 eV → Kmax = 4.14 – 2.5 = 1.64 eV

Q3. Classical wave theory predicts KE ∝ intensity, cannot explain threshold frequency → quantum theory needed.


2019

Q1. Wavelength-momentum relation: λ=h/p\lambda = h/pλ=h/p

Q2. Photoelectric effect diagram: light → metal → emitted electrons → stopping potential measurement.

Q3. Electron accelerated through V = 100 V:p=2meV,λ=hp=h2meV1.23×1010mp = \sqrt{2 m e V}, \quad \lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m e V}} \approx 1.23 \times 10^{-10}\, \text{m}p=2meV​,λ=ph​=2meV​h​≈1.23×10−10m


2018

Q1. Photon energy in terms of wavelength: E=hc/λE = hc/\lambdaE=hc/λ

Q2. Stopping potential: Vs=Kmax/e=(hνϕ)/eV_s = K_{\text{max}}/e = (h\nu – \phi)/eVs​=Kmax​/e=(hν−ϕ)/e

Q3. Electron diffraction → wave nature confirmed via interference pattern.


2017

Q1. λ = 500 nm photon, φ = 2 eV → E = 2.48 eV → Kmax = 0.48 eV

Q2. de Broglie wavelength for electron accelerated through V:λ=h2meV\lambda = \frac{h}{\sqrt{2 m e V}}λ=2meV​h​

Q3. Experiment: electron diffraction by crystal → observed interference → wave nature.


2016

Q1. Photon energy-momentum relation: E=pcE = pcE=pc

Q2. Particle mass m, velocity v → λ=h/(mv)\lambda = h / (mv)λ=h/(mv)

Q3. Classical theory fails: cannot explain threshold frequency, instantaneous emission → quantum theory needed.


2015

Q1. Electron accelerated through V → λ=h/2meV\lambda = h / \sqrt{2 m e V}λ=h/2meV​

Q2. Einstein’s photoelectric equation: Kmax=hνϕK_{\text{max}} = h\nu – \phiKmax​=hν−ϕ

Q3. V = 200 V → λ=h/2meV8.6×1012m\lambda = h / \sqrt{2 m e V} \approx 8.6 \times 10^{-12}\, \text{m}λ=h/2meV​≈8.6×10−12m


2014

Q1. Photoelectron KE: Kmax=hνϕK_{\text{max}} = h\nu – \phiKmax​=hν−ϕ

Q2. de Broglie hypothesis: particles can show wave behavior → λ=h/p\lambda = h/pλ=h/p

Q3. Electron diffraction confirms wave behavior of matter.


2013

Q1. Stopping potential:eVs=Kmax=hνϕ    Vs=hνϕee V_s = K_{\text{max}} = h\nu – \phi \implies V_s = \frac{h\nu – \phi}{e}eVs​=Kmax​=hν−ϕ⟹Vs​=ehν−ϕ​

Q2. de Broglie wavelength:λ=h2mK\lambda = \frac{h}{\sqrt{2 m K}}λ=2mK​h​

Q3. Dual nature:

  • Light → wave: interference, diffraction; particle: photoelectric effect
  • Matter → particle: mass, momentum; wave: electron diffraction, de Broglie wavelength