Derive an expression for De Broglie wavelength.

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(2 emV))`
Substituting the known values in the above equation, we get
`lambda = (6.626 xx 10^(-34))/(sqrt(2V xx 1.6 xx 10^(-19) xx 9.11 xx 10^(-31)))`
`lambda = (12.27 xx 10^(-10))/(sqrt(V))"meter" (or) lambda = (12.27)/(sqrt(V)) Å`
For example, if an electron is accelerated through a potential difference of 100 V, then its de Broglie wavelength is 1.227 `Å`. Since the kinetic energy of the electron, K = eV, then the de Broglie wavelength associated with electron can be also written as
`lambda = (h)/(sqrt(2mK))`