Calculate the
de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.

Explanation

Given, `V = 56 V, e = 1.6xx10^(-19)c, m = 9xx10^(-31)kg`
`lambda = (12.27)/(sqrt(V)) Å = (12.37)/(sqrt(56))= 0.164xx10^(-9)m therefore lambda = 0.164 nm`.