Calculate the de-Broglie wavelength of an α-particle accelerated through a potential difference of 200 volts

Solution:

 Given- potential difference V= 200Volts

Where, q = charge on α-particle which is double of electronic charge = 2e= 2 x 1.6 x 10^-19C = 3.2 x 10^-19 C, h = Plank constant = 6.626x10^-34Js

and m= mass of α-particle = 2(mass of proton)+ 2(mass of neutron)

 m= 2x(1.67x10^-27Kg) + 2x(1.67x10^-27Kg)

 m= 6.68x10^-27Kg