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PhysicsDual Nature of Matter
A photon and an electron both have the same energy $E$. If $\lambda_{ph}$ is the wavelength of photon and $\lambda_e$ is the de Broglie wavelength of the electron, then $\lambda_{ph}/\lambda_e$ is proportional to:
Options
1
$\sqrt{E}$
2
$\frac{1}{\sqrt{E}}$
3
$\sqrt{\frac{E}{2mc^2}}$
4
$c\sqrt{\frac{2m}{E}}$
Correct Answer
$c\sqrt{\dfrac{2m}{E}}$
Solution
1

Photon: $E = hc/\lambda_{ph}$ $\Rightarrow$ $\lambda_{ph} = \dfrac{hc}{E}$

Electron: $E = \dfrac{p^2}{2m}$ $\Rightarrow$ $p = \sqrt{2mE}$ $\Rightarrow$ $\lambda_e = \dfrac{h}{\sqrt{2mE}}$

2
$$\frac{\lambda_{ph}}{\lambda_e} = \frac{hc/E}{h/\sqrt{2mE}} = \frac{c\sqrt{2mE}}{E} = c\sqrt{\frac{2m}{E}}$$
$\lambda_{ph} = hc/E$, $\lambda_e = h/\sqrt{2mE}$
Ratio $= c\sqrt{2m/E}$
Theory: Dual Nature of Matter
1. Photon Properties

Photon: massless particle of light. Energy $E = h\nu = hc/\lambda$. Momentum $p = E/c = h/\lambda$. No rest mass. Always travels at $c$. Cannot be at rest. de Broglie wavelength for photon $= h/p = hc/E = \lambda$ (same as wavelength from wave picture — consistent). Photon energy: visible light ($\lambda=500$ nm): $E = 2.5$ eV. X-ray ($\lambda=0.1$ nm): $E = 12.4$ keV. Gamma ray ($E = 1$ MeV): $\lambda = 1.24\times10^{-3}$ nm.

2. de Broglie Wavelength for Particles

$\lambda = h/p = h/(mv)$ for non-relativistic particles. For particle with kinetic energy $KE$: $p = \sqrt{2m\cdot KE}$, so $\lambda = h/\sqrt{2m\cdot KE}$. For electron accelerated through $V$ volts: $KE = eV$, $\lambda = h/\sqrt{2meV}$. Numerically: $\lambda(\text{Å}) = \sqrt{150/V}$ for electrons. For proton (mass $1836m_e$): $\lambda$ is $\sqrt{1836}$ times smaller than electron at same $KE$. Thermal de Broglie wavelength: $\lambda_{th} = h/\sqrt{2\pi m k_B T}$.

3. Photoelectric Effect

$KE_{max} = h\nu - \phi = h\nu - h\nu_0$. Stopping potential $V_0$: $eV_0 = KE_{max}$. Threshold frequency $\nu_0 = \phi/h$. Key results: $KE_{max}$ depends on frequency, NOT intensity. Intensity determines number of photoelectrons, not their energy. No time delay in emission. Einstein (1905): light consists of photons. Photoelectric effect proved quantum nature of light. Millikan confirmed Einstein equation (1916) — Nobel Prize.

4. Compton Scattering

X-ray photon scatters off electron: wavelength increases. $\Delta\lambda = \lambda_C(1-\cos\phi)$ where $\lambda_C = h/(m_e c) = 0.00243$ nm (Compton wavelength), $\phi$ = scattering angle. Proved photons have momentum $p = h/\lambda$. Maximum shift at $\phi = 180°$: $\Delta\lambda = 2\lambda_C = 0.00486$ nm. Compton (1923) — Nobel Prize 1927. Combined with photoelectric effect: definitive proof of wave-particle duality of light.

5. Wave-Particle Duality

Light: wave (interference, diffraction, polarization) AND particle (photoelectric, Compton). Electrons: particle (mass, charge) AND wave (diffraction, interference). Principle: nature of observed phenomenon depends on type of experiment. Double-slit with electrons: interference pattern builds up electron by electron (each is both wave and particle). Quantum eraser: restoring wave behavior by erasing which-path information. Bohr complementarity: wave and particle aspects cannot be simultaneously observed.

6. Heisenberg Uncertainty Principle

$\Delta x \cdot \Delta p_x \geq \hbar/2$. $\Delta E \cdot \Delta t \geq \hbar/2$. Fundamental limits — not instrument limitations. Origin: wave nature of particles (localizing a wave requires superposition of many wavelengths → momentum uncertainty). Consequences: electrons cannot exist in nucleus (confinement gives huge $\Delta p$ → kinetic energy too large). Zero-point energy: ground state has $E > 0$ (cannot have $E = 0$ as that would require $\Delta p = 0$ → infinite $\Delta x$).

7. Electron Diffraction

Davisson-Germer (1927): 54 eV electrons diffracted by Ni crystal. Peak at $\phi = 50°$ consistent with $\lambda = h/p = 1.67$ Å. First experimental proof of electron wave nature. G.P. Thomson (1927): diffraction rings from thin Al foil — electrons behave like X-rays with similar wavelengths. Both won Nobel Prize 1937. Applications: electron microscope ($\lambda \sim 0.004$ nm at 100 kV → 100,000× magnification), low-energy electron diffraction (LEED) for surface structure analysis, transmission electron microscopy (TEM) for atomic-resolution imaging.

8. Matter Waves and Schrodinger Equation

de Broglie (1924): particles have associated wavelength $\lambda = h/p$. Schrodinger (1926): wave equation for matter waves. $-\frac{\hbar^2}{2m}\nabla^2\psi + V\psi = i\hbar\frac{\partial\psi}{\partial t}$. $|\psi|^2$ = probability density. For stationary states: $-\frac{\hbar^2}{2m}\nabla^2\psi + V\psi = E\psi$ (time-independent). For hydrogen: solutions give quantized energies $E_n = -13.6/n^2$ eV — same as Bohr, but with correct wavefunctions and selection rules. Solved the problems of Bohr model (multi-electron atoms, intensities, selection rules, spin).

Frequently Asked Questions
1. Why is the electron wavelength shorter than photon wavelength at same energy?
Photon: $\lambda_{ph} = hc/E$. Electron: $\lambda_e = h/\sqrt{2mE}$. Ratio: $\lambda_{ph}/\lambda_e = c\sqrt{2m/E}$. For $E = 1$ eV: $\lambda_{ph} = 1240$ nm (infrared). $\lambda_e = h/\sqrt{2m_e \times 1.6\times10^{-19}} = 6.63\times10^{-34}/\sqrt{2\times9.11\times10^{-31}\times1.6\times10^{-19}} = 6.63\times10^{-34}/(5.41\times10^{-25}) = 1.23$ nm. Ratio $= 1240/1.23 \approx 1000$. Photon wavelength is ~1000 times larger. Electron waves are much shorter → better resolution in electron microscopes.
2. What is the energy of a photon with same wavelength as electron at 100 eV?
Electron at 100 eV: $\lambda_e = h/\sqrt{2m_e \times 100\times1.6\times10^{-19}} = 0.123$ nm = 1.23 Å. Photon with same wavelength: $E_{ph} = hc/\lambda = 1240\text{ eV·nm}/0.123\text{ nm} = 10,081$ eV $\approx 10$ keV. So a 100 eV electron has same wavelength as a 10 keV X-ray photon! Electrons are much more efficient at achieving short wavelengths (lower energy needed). This is why electron microscopes don't need X-ray sources.
3. How does de Broglie wavelength change with temperature for gas molecules?
Thermal de Broglie wavelength: $\lambda_{th} = h/\sqrt{2\pi mk_BT}$. For electron at room temperature ($T=300$ K): $\lambda_{th} = 6.63\times10^{-34}/\sqrt{2\pi\times9.11\times10^{-31}\times1.38\times10^{-23}\times300} \approx 4.3$ nm. For helium atom: $\lambda_{th} \approx 0.08$ nm at 300 K. For proton: $\lambda_{th} \approx 0.1$ nm. At low temperature ($T\to0$): $\lambda_{th}\to\infty$ → macroscopic quantum effects (Bose-Einstein condensation, superfluidity) when $\lambda_{th} \sim$ interparticle spacing.
4. What is the relativistic de Broglie wavelength?
Non-relativistic: $\lambda = h/\sqrt{2mE}$ (valid when $E \ll mc^2$). Relativistic: total energy $E_{total} = \sqrt{(pc)^2 + (mc^2)^2}$. Kinetic energy $KE = E_{total} - mc^2$. So $pc = \sqrt{(KE+mc^2)^2 - (mc^2)^2} = \sqrt{KE^2+2KE\cdot mc^2}$. $\lambda = h/p = hc/\sqrt{KE^2+2KE\cdot mc^2}$. For $KE \ll mc^2$: $\lambda \approx h/\sqrt{2mKE}$ (non-relativistic limit). For $KE \gg mc^2$: $\lambda \approx hc/KE$ (ultra-relativistic, like photon). At 100 MeV for electrons ($mc^2 = 0.511$ MeV): relativistic formula gives significantly different answer.
5. What is the significance of $c\sqrt{2m/E}$?
The ratio $\lambda_{ph}/\lambda_e = c\sqrt{2m/E}$. Substituting constants: $c\sqrt{2m_e/E} = (3\times10^8)\sqrt{2\times9.11\times10^{-31}/E}$ where $E$ in joules. This ratio is dimensionless-equivalent (if we use consistent units). For $E = 1$ eV $= 1.6\times10^{-19}$ J: ratio $\approx 1000$ (as computed above). The ratio decreases as $1/\sqrt{E}$ — at higher energies, the two wavelengths get closer (at $E = mc^2 \approx 0.511$ MeV for electrons: ratio $= c\sqrt{2/c^2} = \sqrt{2} \approx 1.41$).
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