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
`1/2mv^(2)=eV`
Therefore,the speed v of the electron is `v=sqrt((2eV)/m)`
hence,the de Broglie wavelength of the electrons is `lambda=h/(mv)=h/(sqrt(2emV))`
Subsituting the known values in the above equation,we get
`lambda=(6.626xx10^(-34))/(sqrt(2Vxx1.6xx10^(-19)xx9.11xx10^(-31)))implies lambda=(12.27xx10^(-10))/(sqrtV)meter(Or)lambda=(12.27)/(sqrtV)` Å
For example ,if an electron is accelerated through a potential difference of 100V,then its de Broglie wavelength is `1.277`Å.Since the kinetic energy of the electron, K=eV then the Broglie wavelength associated with electron can be also written as
`lambda=h/(sqrt2mK)`