De Broglie hypothesis — matter waves λ = h - p

1. What is a matter wave?

WHAT it is not: it is not the particle physically smeared into a wave in ordinary space. It is a wave of probability — where the wave is intense, the particle is likely to be found. But for getting $\lambda$ , treat it as a plain wavelength.

2. Deriving $\lambda = h/p$ from first principles

We build it by steel-manning the photon and demanding symmetry.

Useful rewrites (derive, don't memorize)

A particle of mass $m$ with kinetic energy $K$ : $K = \tfrac12 m v^2 = \frac{p^2}{2m}\;\Rightarrow\; p = \sqrt{2mK}.$ Why? $p = mv$ so $p^2 = m^2v^2 = 2m(\tfrac12 mv^2)=2mK$ . Hence $\lambda = \frac{h}{\sqrt{2mK}}.$

A charge $q$ accelerated through potential $V$ gains $K = qV$ : $\lambda = \frac{h}{\sqrt{2mqV}}.$ For an electron, plug numbers (do it once, then reuse): $\lambda = \frac{12.27}{\sqrt{V}}\ \text{Å}\quad(V\text{ in volts}).$ Why this constant? $\dfrac{h}{\sqrt{2 m_e e}} = \dfrac{6.626\times10^{-34}}{\sqrt{2(9.11\times10^{-31})(1.6\times10^{-19})}}\approx 1.227\times10^{-9}\ \text{m·V}^{1/2}=12.27\ \text{Å·V}^{1/2}.$

Figure — De Broglie hypothesis — matter waves λ = h - p

3. Worked examples

4. Common mistakes (Steel-man + fix)

Recall Feynman: explain to a 12-year-old

Imagine throwing a ball. De Broglie said the ball secretly comes with a tiny ripple, like a wave. How big is the ripple? It depends on how much "push" (momentum) the ball has — more push, smaller ripple. For a real ball the ripple is so unbelievably tiny ( $10^{-34}$ m!) you'd never notice it. But for a teeny electron, the ripple is about the size of an atom — big enough that electrons can spread out and make wave patterns, just like water waves through gaps. So matter is secretly a little bit wave-y; we just only see it when things are super small.

5. Forecast-then-Verify

Flashcards

De Broglie relation (state it)

\lambda = h/p

, wavelength of a particle with momentum

p

;

h

= Planck's constant.

Why does the wave nature of a cricket ball never show up?

Its momentum is huge, so

\lambda=h/p\sim10^{-34}

m — far smaller than any slit/object, so no diffraction.

Express

\lambda

in terms of kinetic energy

K

and mass

m

\lambda = h/\sqrt{2mK}

(since

p=\sqrt{2mK}

Electron accelerated through

V

volts: quick formula for

\lambda

\lambda = 12.27/\sqrt{V}

Å.

At equal kinetic energy, which is wavier, electron or proton, and by what factor?

Electron;

\lambda\propto1/\sqrt m

, factor

\sqrt{m_p/m_e}=\sqrt{1836}\approx 42.8

Which experiment confirmed matter waves?

Davisson–Germer (1927), electron diffraction off a nickel crystal.

\lambda=h/p

a derivation or a hypothesis for matter?

A hypothesis (postulate by symmetry with light); confirmed by experiment.

Doubling the speed of a non-relativistic particle does what to

\lambda

Halves it (

\lambda\propto1/p\propto1/v

What does the matter wave actually represent (Born)?

A probability amplitude — where the particle is likely to be found, not a physical smear.

Connections

Photoelectric effect — gives photons momentum $p=E/c$ , the seed of $\lambda=h/p$ .
Davisson–Germer experiment — experimental proof of electron waves.
Heisenberg Uncertainty Principle — follows from wave nature ( $\Delta x\,\Delta p\gtrsim\hbar$ ).
Bohr model — quantized orbits = standing matter waves ( $2\pi r = n\lambda$ ).
Wave–particle duality — the umbrella concept.
Compton effect — photon momentum confirmed.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, idea bilkul simple hai: jaise light ek wave bhi hai aur particle (photon) bhi, waise hi de Broglie ne 1924 mein bola ki har chalti hui cheez — electron, proton, yahan tak ki cricket ball — ke saath bhi ek "matter wave" judi hoti hai. Uss wave ki wavelength milti hai $\lambda = h/p$ se, jahan $p = mv$ momentum hai aur $h$ Planck's constant. Matlab jitna zyada momentum, utni chhoti wavelength.

Yeh formula nikalta kaise hai? Photon ke liye $E=h\nu$ aur $E=pc$ — dono ko combine karo, $c=\nu\lambda$ daalo, toh $\lambda=h/p$ aa jata hai. De Broglie ka jugaad yeh tha ki is formula mein kahin bhi "light-only" cheez nahi hai (na $c$ , na massless), toh yeh sab matter pe bhi lagega. Yeh ek hypothesis thi, aur Davisson–Germer ne electron diffraction dekh ke prove kar diya.

Ab samajhne wali baat: cricket ball ka $\lambda$ nikalo toh $10^{-34}$ m aata hai — itna chhota ki kabhi wave nature dikhegi hi nahi, isiliye badi cheezein "classical" lagti hain. Lekin electron ko 100 V se accelerate karo toh $\lambda = 12.27/\sqrt{V} = 1.23$ Å — yeh atom ke size jaisa hai, isiliye electron crystal se diffract karta hai. Yahi reason hai ki wave-particle duality sirf chhoti (microscopic) duniya mein clearly dikhti hai.

Exam tip: ratta mat maaro $h/mv$ — hamesha $\lambda = h/p$ se start karo, phir $p$ ko situation ke hisaab se nikaalo ( $p=mv$ , ya $p=\sqrt{2mK}$ , ya $p=\sqrt{2mqV}$ ). Galat jagah mass-form use karoge toh photon jaise massless case mein phass jaoge.

De Broglie hypothesis — matter waves λ = h - p

1. What is a matter wave?

2. Deriving $\lambda = h/p$ from first principles

Useful rewrites (derive, don't memorize)

3. Worked examples

4. Common mistakes (Steel-man + fix)

5. Forecast-then-Verify

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Modern Physics

Connections