Question

Show that the de Broglie wavelength of an electron accelerated from rest through potential difference of V volts is 1.226/squareroot V nm.

Answer 1

The first argument: λ∝1Vλ∝1V. From the Planck relation, we have:

E=hf=hc/λ

But after acceleration through the electric potential difference, the energy of the electron is eVeV, where ee is the charge of the electron. As such,

eV=hcλeV=hc/λ

λ=hc/e x 1/V

⟹λ ∝ 1 / V

The second argument: λ ∝ 1 / V From de Brogile's equation, we have

p=h / λ

mv= h / λ

m²V² = h² /  λ²

m²v² / 2m = h² / 2mλ²

mv² / 2 = h² / 2m x 1 / λ

But since the kinetic energy of the electron is equal to the energy gained from accelerating through the electric potential,

eV=h² / 2m x 1 / λ²

λ² = h² / 2meV

λ=h / √2me x 1 / √V

h = Planck's Constant (6.634*10^-34 Js),
m = mass of electron (9.1*10^-31kg),
e = Electron charge (1.602*10^-19 C)
V = Accelerating voltage.

put these valusesin above equation

lemda = 1.226 / root(V) Ans