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Wavelength of electron accelerated by 100 V

Short answer

Using de Broglie relation λ = h/√(2meV): for V=100V, λ = 1.23×10⁻¹⁰ m = 0.123 nm (1.23 Å). Alternatively, λ = 1.227/√V nm.

Find the wavelength of an electron accelerated by 100V.

An electron accelerated from rest through a potential difference VV gains kinetic energy eVeV. Its momentum is p=2meVp = \sqrt{2meV} and its de Broglie wavelength is λ=h/p\lambda = h/p.

Given: V=100V = 100 V, m=9.1×10−31m = 9.1\times10^{-31} kg, e=1.6×10−19e = 1.6\times10^{-19} C, h=6.626×10−34h = 6.626\times10^{-34} J s.

Kinetic energy and momentum

KE=eV=1.6×10−19×100=1.6×10−17 Jp=2m eV=2×9.1×10−31×1.6×10−17=5.396×10−24 kg m s−1\begin{aligned} KE &= eV = 1.6\times10^{-19}\times100 = 1.6\times10^{-17}\ \text{J} \\ p &= \sqrt{2m\,eV} = \sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-17}} \\ &= 5.396\times10^{-24}\ \text{kg m s}^{-1} \end{aligned}

Wavelength

λ=hp=6.626×10−345.396×10−24=1.228×10−10 m\lambda = \frac{h}{p} = \frac{6.626\times10^{-34}}{5.396\times10^{-24}} = 1.228\times10^{-10}\ \text{m}

Answer: λ=1.23×10−10\lambda = 1.23\times10^{-10} m =0.123= 0.123 nm =1.23= 1.23 Å.

Shortcut: λ=1.227V\lambda = \dfrac{1.227}{\sqrt{V}} nm =1.22710=0.1227= \dfrac{1.227}{10} = 0.1227 nm. This wavelength is of the same order as the spacing of atoms in a crystal, so such electrons are diffracted by crystals (Davisson–Germer experiment).

✓ Answer checked by Claude Opus 5.5 (high effort)

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