Davisson-Germer experiment — electron diffraction

WHAT is being claimed?

The genius of the experiment: it gives two roads to $\lambda$ that must agree.

Theory road: accelerate electrons through voltage $V$ , compute $\lambda = h/p$ .
Experiment road: measure the diffraction angle, use crystal geometry to get $\lambda$ .

If both give the same number, matter waves are real.

DERIVATION 1 — The "theory road" wavelength

Formula Wavelength from accelerating voltage — derived

Step 1 — Energy: the work done by the field equals kinetic energy gained. $eV = \tfrac12 m v^2$ Why this step? The field does work $qV = eV$ on charge $e$ ; starting from rest, all of it becomes KE.

Step 2 — Momentum in terms of KE: $p = mv, \qquad KE = \frac{p^2}{2m} \ \Rightarrow\ p = \sqrt{2m\,KE} = \sqrt{2meV}$ Why this step? We want $\lambda = h/p$ , so we re-express KE as momentum.

Step 3 — de Broglie: $\boxed{\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}}$

Step 4 — Plug numbers ( $h=6.626\times10^{-34}$ , $m=9.11\times10^{-31}$ , $e=1.6\times10^{-19}$ ): $\lambda = \frac{12.27}{\sqrt{V}}\ \text{Å} \quad (V \text{ in volts})$ Why this step? Collapsing all constants gives a clean working formula. At $V=54$ V: $\lambda = \frac{12.27}{\sqrt{54}} = 1.67\ \text{Å}$

DERIVATION 2 — The "experiment road" wavelength

Formula Surface-grating analysis (the relation D–G actually used)

Treat the regularly spaced surface atoms as a 2-D grating with atomic row spacing $a$ . Constructive interference of waves scattered from adjacent surface atoms requires the path difference $a\sin\phi$ to equal a whole number of wavelengths: $a\sin\phi = n\lambda$ Why this step? For grazing scatter off a surface row, the extra path between neighbouring atoms is $a\sin\phi$ , measured directly from the incident-beam angle $\phi$ — no plane-angle conversion needed.

For the Ni(111) surface the atomic-row spacing is $a = 2.15$ Å, with $\phi = 50^\circ$ , $n=1$ : $\lambda = a\sin\phi = (2.15)\sin 50^\circ = 1.65\ \text{Å}$

Formula Bragg cross-check (using the CORRECT Ni spacing)

Bragg condition for constructive interference from atomic planes: $2d\sin\theta = n\lambda$ where $d$ = spacing of atomic planes, $\theta$ = glancing angle (from the plane).

Geometry link: the scattering angle $\phi$ (between incident and scattered beam) relates to the glancing angle by $\theta = 90^\circ - \frac{\phi}{2}$ Why? The incident and diffracted rays are symmetric about the plane normal; bisecting the $\phi$ angle and subtracting from $90^\circ$ gives the glancing angle.

For the Ni(111) planes the correct interplanar spacing is $d = 2.03$ Å. With $\phi = 50^\circ \Rightarrow \theta = 65^\circ$ , first order $n=1$ : $\lambda = 2d\sin\theta = 2(2.03)\sin 65^\circ = 3.68\ \text{Å}$ This corresponds to a higher-order reflection ( $n=2$ : $\lambda = 1.84$ Å) — the point is that the grating analysis above, using the surface row spacing $a=2.15$ Å, is the one that directly yields $1.65$ Å. (Older textbooks quote $d=0.91$ Å; that value is not the Ni(111) plane spacing and should not be used.)

Forecast-then-Verify

Recall Forecast before reading the answer

Q: If you increase the accelerating voltage $V$ , does the diffraction peak move to a larger or smaller detector angle $\phi$ ?

Verify: Higher $V$ → larger $p$ → smaller $\lambda$ (since $\lambda=h/p$ ). Surface-grating view ( $a\sin\phi = n\lambda$ ): smaller $\lambda$ → smaller $\sin\phi$ → the peak moves to a smaller $\phi$ . Bragg view ( $2d\sin\theta = n\lambda$ ): smaller $\lambda$ → smaller $\theta$ ; but since $\phi = 180^\circ - 2\theta$ , a smaller $\theta$ gives a larger $\phi$ . ⚠️ These look opposite because $\phi$ is defined differently in the two pictures (grating $\phi$ is from the incident beam to the outgoing ray along the surface convention; the Bragg $\phi=180^\circ-2\theta$ is the full beam-to-beam deflection). The physically observed result, using the standard surface-grating relation D–G applied, is that the peak shifts toward smaller $\phi$ as $V$ increases. The peak can even disappear if $\lambda$ becomes too small for that order. Lesson: always state which geometric convention your $\phi$ refers to before predicting the shift direction.

Common mistakes (Steel-manned)

Worked Example (full numbers)

Flashcards

What did the Davisson–Germer experiment prove?

That electrons (matter) exhibit wave behaviour — direct confirmation of de Broglie matter waves.

What target was used and why?

A single crystal of nickel, because its atomic spacing (~0.1 nm) matches electron wavelengths, enabling diffraction.

Wavelength of electron accelerated through voltage V (derive)?

\lambda = h/\sqrt{2meV} = 12.27/\sqrt{V}

Å.

At what voltage and scattering angle was the famous peak observed?

V = 54

V, scattering angle

\phi = 50^\circ

Which relation directly gives

\lambda=1.65

Å from the data?

The surface-grating relation

a\sin\phi=n\lambda

with

a\approx2.15

Å,

\phi=50^\circ

n=1

What is the actual Ni(111) interplanar spacing (and what NOT to use)?

d\approx2.03

Å (0.203 nm); do not use the old

0.91

Å value.

Relation between detector angle

\phi

and Bragg glancing angle

\theta

\theta = 90^\circ - \phi/2

, equivalently

\phi=180^\circ-2\theta

Why is non-relativistic treatment valid at 54 V?

KE = 54 eV

\ll

511 keV rest energy, so

v\ll c

Using the surface-grating convention, how does the peak

\phi

shift as

V

increases?

\lambda

decreases,

\sin\phi=n\lambda/a

decreases, so the peak moves to smaller $\phi$ .

Recall Feynman: explain to a 12-year-old

Imagine throwing tiny balls at a comb full of evenly spaced teeth. If the balls were just balls, they'd scatter everywhere randomly. But these special balls (electrons) act like ripples on a pond — and ripples bouncing off evenly spaced teeth line up and reinforce in one special direction, making a bright "loud" spot. Davisson and Germer saw that bright spot. Balls don't do that; waves do. So electrons are secretly wavy. And the spacing of the comb teeth told them exactly how "wavy" the electron was — and it matched de Broglie's formula perfectly.

Connections

de Broglie hypothesis — the prediction this experiment confirmed.
Bragg's law — geometry used to cross-check $\lambda$ from the angle.
Wave-particle duality — D–G is the matter-wave half; photoelectric effect is the wave→particle half.
Photoelectric effect — complementary evidence.
Electron microscope — practical application of electron waves (short $\lambda$ → high resolution).
Heisenberg uncertainty principle — follows from treating particles as wave packets.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Davisson–Germer ka pura point yeh hai: pehle log sochte the ki electron sirf ek tiny particle hai. Agar aisa hota, to electron beam ko nickel crystal pe maaroge to woh randomly har taraf bikhar jaata — koi special direction nahi. Lekin experiment me unhe ek sharp intensity peak mila ek khaas angle ( $\phi = 50^\circ$ ) par, jab voltage $54$ V tha. Aisa peak sirf waves bana sakti hain (interference/diffraction se), particles nahi. Matlab electron me wave nature bhi hai — yahi de Broglie ne predict kiya tha.

Ab proof kaise pakka hua? Do raaste se same $\lambda$ nikaala. Pehla theory road: voltage se energy ( $eV = \tfrac12 mv^2$ ), usse momentum $p=\sqrt{2meV}$ , phir $\lambda = h/p = 12.27/\sqrt{V}$ Å = $1.67$ Å. Dusra experiment road: crystal ke surface atoms ko grating maano aur relation $a\sin\phi = n\lambda$ use karo, jaha $a\approx2.15$ Å Ni(111) surface ka row spacing hai — isse $\lambda \approx 1.65$ Å. Dono almost barabar! Dhyan rakho: purane notes me jo $d=0.91$ Å likha hota hai woh Ni(111) ka sahi plane spacing nahi hai (sahi $\approx2.03$ Å); clean $1.65$ Å surface-grating relation se aata hai, fudged Bragg se nahi.

Ek important trap yaad rakhna: detector ka angle $\phi$ aur Bragg ka angle $\theta$ same nahi hote. $\theta = 90^\circ - \phi/2$ , yaani $\phi=180^\circ-2\theta$ . Jis convention me kaam kar rahe ho, pehle woh fix karo — warna " $V$ badhao to peak kidhar jaayega" me ulta jawab aa jaayega. Surface-grating convention me $V\uparrow$ se $\lambda\downarrow$ , $\sin\phi\downarrow$ , to peak chhote $\phi$ ki taraf jaata hai. Aur relativity ki tension mat lo — $54$ eV bahut kam hai $511$ keV rest energy ke saamne, to simple non-relativistic formula chalega.

Yeh experiment isliye important hai kyunki yahi se electron microscope ka idea aaya — chhota wavelength matlab high resolution. Aur ye wave–particle duality ka asli foundation hai: photoelectric effect light ko particle dikhata hai, Davisson–Germer matter ko wave dikhata hai. Dono milke quantum mechanics ki neev banate hain.

Davisson-Germer experiment — electron diffraction

WHAT is being claimed?

HOW the apparatus works

DERIVATION 1 — The "theory road" wavelength

DERIVATION 2 — The "experiment road" wavelength

Forecast-then-Verify

Common mistakes (Steel-manned)

Worked Example (full numbers)

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Modern Physics

Connections