Question

Find the de-broglie wavelength associated with an electron accelerated through a potential difference of 30 kV.

Answer 1

Kinetic energy of the electron

K = qV = e(30 kV) = 30 keV = 30 × 10 ³ × 1.6 × 10^–19 J = 4.8 × 10^–15 J

Now, momentum of the electron

∴ De-broglie wavelength, λ = h / Momentum