How did Davisson–Germer actually confirm de Broglie's wavelength at 54 V, and why can't we see a cricket ball diffract?

7 answers & replies

• [+][–]Kavya NaikDropperSample· 20d ago

  The number first. $\lambda = \dfrac{h}{\sqrt{2meV}}$, and putting the constants in gives $\lambda = \dfrac{1.227}{\sqrt V}$ nm. With $V = 54$: $\sqrt{54} \approx 7.35$, so $\lambda = 1.227/7.35 \approx 0.167$ nm. And I use the shorter form $12.27/\sqrt{V}$ Å for numericals, since the unit is Angstrom: 1 Å = 0.1 nm. Then $\lambda$ for 100 V is 1.227 Å. Easy.

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• [+][–]Ishaan Bose11thSample· 20d ago

  Quick mental maths for MCQs: $\lambda\,(\text{Å}) \approx 12.3/\sqrt{V}$. So 100 V gives about 1.23 Å, 144 V gives $12.27/12 \approx 1.02$ Å, and 400 V gives about 0.61 Å. Also note the scaling: if you make the voltage 4 times, $\lambda$ halves ($\lambda \propto 1/\sqrt V$). These come up a lot as "ratio of wavelengths" questions.

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• [+][–]Zoya Khan12thSample· 20d ago

  On what was actually measured: it showed that electrons behave like waves. The set-up: a beam of electrons accelerated through a voltage hits a nickel crystal, and a detector measures how many electrons scatter at each angle. If electrons were only particles, you'd get a smooth, boring distribution. Instead, at 54 V there's a strong peak at a scattering angle of 50°. That's diffraction: the electron waves reflected from the regular rows of atoms interfere constructively only at particular angles, just like X-rays in Bragg diffraction.

  From the crystal's geometry you can compute the wavelength that would produce a peak there and it comes to about 0.165 nm, very close to the de Broglie value 0.167 nm. That agreement confirmed the matter-wave idea.

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  • [+][–]Ishita GoyalDropperSample· 20d agoAuthor

    ahh so the peak at one particular angle = constructive interference, so it must be waves. thank you! also the 10^19 times smaller than a nucleus line is dramatic 😂 putting it in my formula sheet

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  • [+][–]Ishaan Bose11thSample· 20d ago

    ahh so the peak at one angle is the evidence for waves. noted, and the 4x voltage halves wavelength thing is useful

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• [+][–]Vivek Mishra12thSample· 20d ago

  Cricket ball. $\lambda = \dfrac{h}{mv} = \dfrac{6.6\times10^{-34}}{0.15\times20} = 2.2\times10^{-34}$ m. What to compare with: the size of things the ball goes past, like the stumps, are about centimetres; the nucleus is about $10^{-15}$ m. A wave diffracts noticeably only when the size of the opening is comparable to its wavelength. Here the wavelength is smaller than even a nucleus by a factor of $10^{19}$, so the wave nature is impossible to see. For the electron at 54 V, 0.167 nm is of the order of the spacing between atoms in a crystal, that's why the crystal works as a grating.

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• [+][–]Lakshmi TripathiDropperSample· 20d ago

  Short ones to memorise for NEET: $\lambda = h/p = h/mv$. In terms of kinetic energy, $\lambda = \dfrac{h}{\sqrt{2mK}}$. For a photon, $p = h/\lambda$. For the electron through V volts, $\lambda = 12.27/\sqrt V$ Å. For a given kinetic energy a lighter particle has a longer wavelength, that's why electron microscopes use electrons: shorter wavelength than visible light, better resolution.

  For your third question, in the standard picture the electron is detected as a single point, but where it lands follows the wave pattern (intensity = probability). I don't think NEET asks that.

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