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A photon and an electron have the same kinetic energy $E = 20.2$ eV. The ratio of the momentum of the electron to that of the photon is: (given $m_e c^2 = 0.511$ MeV)

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Solution written and verified by Roshan, science educator with 5 years of experience teaching NEET and JEE aspirants. Last reviewed September 2026.
Options
1
225
2
150
3
100
4
450
Correct Answer
225
Solution
1

$p_{ph} = E/c$; $p_e = \sqrt{2m_eE}$

Ratio $= \sqrt{2m_ec^2/E}$

2

$= \sqrt{2\times511000/20.2} = \sqrt{50594} \approx \mathbf{225}$

Answer: 225

$p_e/p_{ph} = \sqrt{2m_ec^2/E}$; $E=20.2$ eV, $m_ec^2=511000$ eV → ratio = 225
Theory: Modern Physics / Dual Nature
1. Wave-Particle Duality

Historical development: Newton (light as particles) → Huygens (wave theory) → Young (interference proved wave nature) → Maxwell (EM waves) → Planck/Einstein (photons, particle nature) → de Broglie (matter waves). Bohr complementarity principle: wave and particle aspects are complementary; cannot observe both simultaneously in same experiment.

2. Photoelectric Effect Revisited

Einstein\'s photon model (1905, Nobel 1921): $E = hf = hc/\lambda$. Photoelectric: $hf = \phi + KE_{max}$; stopping potential $V_0 = (hf-\phi)/e$; threshold $f_0 = \phi/h$. Applications: photomultiplier tubes, photodiodes, solar cells, CCD cameras. Photovoltaic effect (solar cells): similar to photoelectric but internal — electron-hole pairs created in semiconductor.

3. Matter Waves and Quantum Mechanics

de Broglie wavelength $\lambda = h/p = h/mv$. Large macroscopic objects have extremely small $\lambda$ (unobservable). Electron microscope: uses electron waves ($\lambda \sim 0.001$ nm at 100 keV) for much higher resolution than light microscopes ($\lambda \sim 400-700$ nm). Electron diffraction: electrons diffracting through crystal lattice confirms wave nature. Quantum confinement: when object size approaches $\lambda$, quantum effects dominate (quantum dots, nanoscale devices).

4. Bohr Model Limitations and Quantum Mechanics

Bohr model works only for one-electron systems (H, He+, Li2+). Fails for: multi-electron atoms, cannot explain intensity of spectral lines, cannot explain fine structure, cannot explain chemical bonding. Schrodinger equation (1926): $H\psi = E\psi$ (wave equation for matter). Wavefunction $\psi$: $|\psi|^2$ gives probability density of finding particle. Quantum numbers: $n$ (principal), $l$ (angular momentum), $m_l$ (magnetic), $m_s$ (spin). Pauli exclusion principle: no two electrons can have same set of all 4 quantum numbers.

5. de Broglie Wavelength and Where It Matters

Every moving particle has an associated wavelength $\lambda = h/p = h/mv$. For an electron accelerated through a potential V, this becomes the convenient form $\lambda = \frac{12.27}{\sqrt{V}}$ Å. The wavelength is inversely proportional to mass, which is why wave behaviour is observable for electrons but not for a cricket ball — a 150 g ball at 30 m/s has a wavelength around $10^{-34}$ m, far too small to detect.

6. The Davisson-Germer Experiment

Electrons accelerated through 54 V were scattered off a nickel crystal and showed a diffraction maximum at 50°, exactly where the Bragg condition predicted for a wavelength of 1.65 Å. The de Broglie formula gives 1.67 Å for that voltage. The agreement was the first direct experimental proof that matter has wave properties, and it is the standard example asked for when a question requests evidence of the dual nature of matter.

7. Photon Momentum and Radiation Pressure

A photon has no rest mass but does carry momentum, $p = h/\lambda = E/c$. When light is absorbed by a surface the momentum transferred produces a pressure $I/c$; when it is perfectly reflected the momentum change doubles and so does the pressure, giving $2I/c$. That single factor of 2 is the most frequently dropped mark in this topic, and it is also the principle behind solar sails.

Where students lose the mark

Using $\lambda = h/mv$ for a photon. A photon has no mass. Use $\lambda = h/p$ with $p = E/c$ instead.

Assuming brighter light gives faster photoelectrons. Electron energy depends only on frequency. Intensity changes how many electrons are emitted, not how fast.

Frequently Asked Questions
1. How do you compare photon and electron momenta at same KE? ⌄
Photon: $E = pc$ → $p_{ph} = E/c$. Electron (non-relativistic): $E = p^2/2m$ → $p_e = \sqrt{2mE}$. Ratio: $p_e/p_{ph} = c\sqrt{2m_e/E} = \sqrt{2m_ec^2/E}$.
2. What is de Broglie wavelength? ⌄
$\lambda = h/p$. For electron: $\lambda_e = h/\sqrt{2m_eE} = h/\sqrt{2meV}$ (after acceleration through $V$ volts). For $V=100$ V: $\lambda_e \approx 1.23/\sqrt{100}$ Angstrom $= 0.123$ nm. X-rays have similar wavelength → electron diffraction experiments.
3. At what energy does relativistic treatment become necessary? ⌄
When $E_{kinetic} \geq 0.1 m_ec^2 = 51.1$ keV, relativistic treatment needed. For $E \ll m_ec^2$: $p_e = \sqrt{2m_eE}$ (non-relativistic). For $E \gg m_ec^2$: $p_e \approx E/c$ (ultra-relativistic, like photon). The 20.2 eV in this problem is far below 511 keV → non-relativistic approximation valid.
4. What is the Compton wavelength of electron? ⌄
$\lambda_c = h/m_ec = 2.43\times10^{-12}$ m. When photon wavelength $\sim \lambda_c$, quantum effects dominate. Compton scattering: X-ray photon hits electron, wavelength increases by $\Delta\lambda = \lambda_c(1-\cos\phi)$. Proved photon has momentum.
5. What is wave-particle duality? ⌄
Both light and matter exhibit wave and particle properties. Light: wave (interference, diffraction) and particle (photoelectric, Compton). Electrons: particle (discrete charge, mass) and wave (electron diffraction - Davisson-Germer 1927). de Broglie (1924): all matter has associated wavelength. Heisenberg uncertainty principle: $\Delta x\Delta p \geq h/4\pi$ (fundamentally, not instrumental limitation).
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