Q: Electrons are accelerated through a potential difference of 150 V. Calculate the de-Broglie wavelength.

Sol: V = 150 V ; h = 6.62 × 10^{-34} Js, m = 9.1 x 10^{-31} kg,

e = 1.6 x 10^{-19} C

$\large \lambda = \frac{h}{\sqrt{2 m e V}} $

$\large \lambda = \frac{6.62 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19}\times 150}}$

= 1 A°