Write the expression for the de Broglie wavelength associated with a charged particle of charge q and mass m, when it is accelerated through a potential V.

Explanation

An electron of mass m is accelerated through a potential difference of V volt. The kinetic energy acquired by the electron is given by `1/2 mv^2 = eV`
therefore , the speed v of the electron is `v=sqrt((2eV)/(m))`
Hence, the de Broglie wavelength of the electron is `lambda=(h)/(mv) = (h)/(sqrt(2emV))`