Let v be the velocity gained by the given charged particle when it is accelerated through a potential difference of V volts.
Kinetic energy of the particle = 12mv2.
Kinetic energy of the particle = Work done on the particle by electric field.
∴ 12mv2=qV
or P22m=qV
p= √2mqV
de-Broglie wavelength (λ) associated with the particle,
λ=hp=h√2mqV