In the Davisson-Germer experiment, an electron emitted from the filament is accelerated through a voltage V. The de Broglie wavelength of that electron will be m. Here m and e are the mass and charge of the electron.

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(4) h/(√2Vem). Idea: accelerating through V gives the electron kinetic energy K = eV. Momentum from kinetic energy: K = (p²)/(2m), so p = √2mK = √2meV. De Broglie wavelength: λ = h/p = h/(√2meV) Check: for V = 100 V this gives about 1.23×10⁻¹⁰ m, the familiar value. So λ = h/(√2Vem), the same product 2meV written in a different order.

In the Davisson-Germer experiment, an electron emitted from the filament is accelerated through a voltage V. The de Broglie wavelength of that electron will be ______ m. Here m and e are the mass and charge of the electron.

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