Answer 1

Potential difference, V = 56 V

Planck’s constant, h = 6.6 × 10⁻³⁴ Js

Mass of an electron, m = 9.1 × 10⁻³¹ kg

Charge on an electron, e = 1.6 × 10⁻¹⁹ C

De Broglie wavelength of an electron accelerating through a potential V, is given by the relation:

`lambda = 12.27/ sqrt(V)` Å

`= 12.27/sqrt56 xx 10^(-10) m`

= 0.1639 nm

Therefore, the de Broglie wavelength of each electron is 0.1639 nm.