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`
(ii) Therefore, the speed v of the electron is `usqrt((2eV)/m)`
Hence, the de Broglie wavelength of the electron is
`lamda = h/(mv) = h/sqrt(2 emV)`
(iii) Substituting the known values in the above equation, we get
`lamda = (6.626xx10^(-34))/sqrt(2Vxx1.6xx10^(-19)xx9.11xx10^(-31))`
` = (12.27xx10^(-10))/sqrtV`meter(or)
`lamda = (12.27)/sqrtV Å`
(iv) Since the kinetic energy of the electron, K = eV, then the de Broglie wavelength associated with electron can be also written as `lamda = h/sqrt(2mK)`