Axiom Physics | Class 12 - de Broglie Hypothesis & Matter Waves

⚡ 01. The Spark (Analogy)

Socho ek two-sided coin ko: agar nature ne light wave ko particle (photon) jaisa behave karne ki aazadi di hai (Photoelectric Effect & Compton Effect), toh kya nature particle (jaise electron ya cricket ball) ko wave jaisa behave karne ki aazadi nahi degi?
1924 mein French physicist Louis de Broglie ne ek daring philosophical hypothesis rakha: "Nature loves symmetry!" Pure universe do hi cheezon se milkar bana hai—Matter aur Radiation. Agar radiation ke paas dual nature hai, toh moving matter ke paas bhi wave nature hona hi chahiye! Jab aap daudte hain, toh aapke sath bhi ek wave chalti hai, par aapka mass itna vishal ($m \approx 60\text{ kg}$) hota hai ki wavelength Planck's constant ($10^{-34}$) ki wajah se sub-atomic size se bhi choti ($10^{-36}\text{ m}$) hokar gayab ho jaati hai. Lekin ek nanhe electron ($m \sim 10^{-30}\text{ kg}$) ke liye yahi matter wave X-ray jitni badi ($\sim 1\text{ \AA}$) hokar physical crystal planes se diffract hone lagti hai!

📐 02. The Core Law & Wavelength Formulae

1. de Broglie Relation in Different Forms

Wavelength associated with a moving particle of mass $m$ and velocity $v$:

$$\lambda = \frac{h}{p} = \frac{h}{mv} = \frac{h}{\sqrt{2mK}}$$

$$\lambda = \frac{h}{\sqrt{2mqV}} \quad (\text{For charge } q \text{ accelerated through } V)$$

$\lambda \propto 1/v$ and $\lambda \propto 1/m$. Matter waves are NOT electromagnetic waves; they are not generated by accelerated charges and carry zero electric/magnetic vector oscillations.

2. Special Particles & Thermal Matter Waves

Direct numerical evaluation constants for competitive exams:

$$\text{Electron: } \lambda_e = \frac{1.227}{\sqrt{V}}\text{ nm} = \frac{12.27}{\sqrt{V}}\text{ \AA}$$

$$\text{Proton: } \lambda_p = \frac{0.286}{\sqrt{V}}\text{ \AA}, \quad \text{Alpha: } \lambda_\alpha = \frac{0.101}{\sqrt{V}}\text{ \AA}$$

$$\text{Thermal Neutron: } \lambda_n = \frac{h}{\sqrt{3mkT}} \quad (K = \frac{3}{2}kT)$$

Where $k = 1.38 \times 10^{-23}\text{ J/K}$ is Boltzmann's constant, and $T$ is absolute temperature in Kelvin.

03. Interactive Visualizer: Real-Time Matter Wave Packet & de Broglie Simulator

PARTICLE: ELECTRON (q = e, m = m_e)

Accelerating Potential ($V$) 100 V

Select Quantum Particle Electron

Animation Controls Pilot Wave

de Broglie Wavelength ($\lambda$):

1.227 Å (0.123 nm)

Kinetic Energy ($K = qV$):

100 eV

Momentum ($p = \sqrt{2mqV}$):

5.40 × 10⁻²⁴ kg·m/s

Classical Velocity ($v$):

5.93 × 10⁶ m/s

📝 04. Comprehensive Stepwise Derivations Blueprint

Primary Board Derivation

Deduction of de Broglie Wavelength of an Electron Accelerated Through Potential $V$

Fig 11.3A: Synthesis of matter-wave duality; localized quantum particle guided by de Broglie pilot wave packet of wavelength $\lambda = h/p$.

1. Consider an electron of rest mass $m$, charge $e$, initially at rest, accelerated through an electric potential difference of $V$ volts.

2. The work done by the electric field on the electron equals its acquired kinetic energy $K$:

$$K = e V$$

3. If $v$ is the velocity acquired by the electron, its kinetic energy in terms of momentum $p$ is:

$$K = \frac{1}{2}m v^2 = \frac{p^2}{2m} \implies p = \sqrt{2mK} = \sqrt{2m e V}$$

4. According to de Broglie's fundamental hypothesis, the wavelength $\lambda$ associated with this electron is:

$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m e V}}$$

5. Substituting standard fundamental physical constants:

• Planck's constant $h = 6.626 \times 10^{-34}\text{ J}\cdot\text{s}$

• Mass of electron $m = 9.109 \times 10^{-31}\text{ kg}$

• Electronic charge $e = 1.602 \times 10^{-19}\text{ C}$

$$\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times (9.109 \times 10^{-31}) \times (1.602 \times 10^{-19}) \times V}} = \frac{6.626 \times 10^{-34}}{\sqrt{2.918 \times 10^{-49} \times V}} = \frac{6.626 \times 10^{-34}}{5.402 \times 10^{-25}\sqrt{V}}$$

$$\lambda = \frac{1.227 \times 10^{-9}}{\sqrt{V}}\text{ m} = \frac{1.227}{\sqrt{V}}\text{ nm} = \frac{12.27}{\sqrt{V}}\text{ \AA}$$

Physical Consequence: For an electron accelerated through typical laboratory potential $V = 120\text{ V}$, $\lambda = \frac{1.227}{\sqrt{120}} \approx 0.112\text{ nm} = 1.12\text{ \AA}$, which is identical to the interatomic spacing in crystals. Hence, electron waves can be directly diffracted by crystals, as experimentally verified by Davisson and Germer!

Fundamental Limit

Why Wave Nature is Not Observed in Everyday Macroscopic Bodies

1. Macroscopic Body (Cricket Ball)

Mass $m = 0.15\text{ kg}$, Velocity $v = 30\text{ m/s}$ ($108\text{ km/h}$):

$$\lambda = \frac{6.63 \times 10^{-34}}{0.15 \times 30} \approx 1.47 \times 10^{-34}\text{ m}$$

This wavelength is $\approx 10^{20}$ times smaller than the diameter of an atomic nucleus ($10^{-14}\text{ m}$). No physical aperture or grating exists in the universe that small; hence, wave nature cannot be observed.

2. Microscopic Body (Thermal Electron)

Mass $m = 9.1 \times 10^{-31}\text{ kg}$, Voltage $V = 100\text{ V}$:

$$\lambda = \frac{1.227}{\sqrt{100}} = 0.123\text{ nm} = 1.23\text{ \AA}$$

This wavelength lies exactly in the X-ray regime and matches natural nickel crystal lattice spacing ($d \approx 0.91\text{ \AA}$), allowing distinct diffraction peaks.

⚠️ 05. Red Alert Exam Trap Zone

Trap 1: Proton vs Alpha Particle Accelerated Through Same Potential $V$

*Board Favorite:* A proton and an $\alpha$-particle are accelerated from rest through the same potential $V$. Find the ratio of their de Broglie wavelengths $\lambda_p / \lambda_\alpha$.
• Formula: $\lambda = \frac{h}{\sqrt{2mqV}} \implies \lambda \propto \frac{1}{\sqrt{m q}}$.
• For proton: $m_p = m, q_p = e$.
• For alpha: $m_\alpha = 4m, q_\alpha = 2e$. $$\mathbf{\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}} = \sqrt{\frac{4m \times 2e}{m \times e}} = \sqrt{8} = 2\sqrt{2} \approx 2.828}$$

Trap 2: Proton vs Alpha Particle with Equal Kinetic Energy $K$

*Watch Out for the Wording:* If the question states "Same Kinetic Energy" instead of "Same Potential": $$\lambda = \frac{h}{\sqrt{2mK}} \implies \lambda \propto \frac{1}{\sqrt{m}} \quad (\text{Charge } q \text{ is completely IRRELEVANT!})$$ $$\mathbf{\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha}{m_p}} = \sqrt{\frac{4m}{m}} = \sqrt{4} = 2 : 1}$$

Trap 3: Are Matter Waves Electromagnetic Waves?

*Direct Viva & Board Trap:* Are matter waves electromagnetic in nature?
*Correct Answer:* STRICTLY NO!
1. EM waves are produced exclusively by accelerated electric charges; matter waves are associated with ANY moving material body (even neutral neutrons or uncharged atoms).
2. EM waves travel at speed of light $c$ in vacuum; matter waves travel at velocity $v < c$ of the particle itself.

Trap 4: de Broglie Wavelength of a Photon vs Electron

For a photon of energy $E$: $E = pc \implies \mathbf{\lambda_{\text{photon}} = \frac{hc}{E}}$.
For an electron with the same kinetic energy $E$: $\mathbf{\lambda_{\text{electron}} = \frac{h}{\sqrt{2mE}}}$. $$\frac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}} = \frac{hc/E}{h/\sqrt{2mE}} = c\sqrt{\frac{2m}{E}} \gg 1$$ For the same energy, a photon always has a vastly larger wavelength than an electron!

💡 06. Handcrafted Exemplars

Board Classic Level

Problem: An electron is accelerated through a potential difference of $100\text{ V}$. What is its de Broglie wavelength? What is the momentum associated with this electron?

Solution: Accelerating potential $V = 100\text{ V}$.

1. de Broglie wavelength of electron:

$$\lambda = \frac{1.227}{\sqrt{V}}\text{ nm} = \frac{1.227}{\sqrt{100}} = \frac{1.227}{10} = 0.1227\text{ nm} = 1.227\text{ \AA}$$

2. Momentum $p = h/\lambda$:

$$p = \frac{6.626 \times 10^{-34}\text{ J}\cdot\text{s}}{0.1227 \times 10^{-9}\text{ m}} \approx 5.40 \times 10^{-24}\text{ kg}\cdot\text{m/s}$$

JEE/NEET Competitive Trap

Problem: What is the de Broglie wavelength of a nitrogen molecule in air at $300\text{ K}$? Assume that the molecule moves with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen $= 14.0076\text{ u}, k = 1.38 \times 10^{-23}\text{ J/K}$).

Solution: Nitrogen molecule $N_2$ has molecular mass:

$$m = 2 \times 14.0076 \times (1.66 \times 10^{-27}\text{ kg}) \approx 4.65 \times 10^{-26}\text{ kg}$$

At temperature $T = 300\text{ K}$, average translational kinetic energy of a molecule is $K = \frac{3}{2}kT$. Therefore, thermal de Broglie wavelength:

$$\lambda = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2m\left(\frac{3}{2}kT\right)}} = \frac{h}{\sqrt{3mkT}}$$

$$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{3 \times (4.65 \times 10^{-26}) \times (1.38 \times 10^{-23}) \times 300}} = \frac{6.63 \times 10^{-34}}{\sqrt{5.775 \times 10^{-46}}} = \frac{6.63 \times 10^{-34}}{2.403 \times 10^{-23}}$$

$$\lambda \approx 2.76 \times 10^{-11}\text{ m} = 0.0276\text{ nm} = 0.276\text{ \AA}$$

Significance: Even a neutral gas molecule at room temperature exhibits a de Broglie wavelength comparable to atomic dimensions!

🗂️ 07. Quick Revision Flashcard Core Relations

de Broglie Relation

$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}$

Electron Constant

$\lambda_e = \frac{1.227}{\sqrt{V}}\text{ nm}$

Thermal Neutrons

$\lambda = \frac{h}{\sqrt{3mkT}}$

Proton / Alpha Ratio

$\frac{\lambda_p}{\lambda_\alpha} = 2\sqrt{2}$ (Same $V$)

de Broglie Hypothesis, Matter Waves & Electron Wavelength