Question

Derive an expression for de Broglie wavelength of electrons.

Obtain an expression for the de-Broglie wavelength associated with an electron accelerated from rest through a potential difference of ‘V’ volts.

Answer 1

An electron of mass m is accelerated through a potential difference of V volts. 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((2 eV)/m)`

Hence, the de-Broglie wavelength of the electron is:

λ = `h/(mv)`

= `h/sqrt (2 e m V)`

Substituting the known values in the above equation, we get,

λ = `(6.626 xx 10^-34)/(sqrt (2 V xx 1.6 xx 10^-19 xx 9.11 xx 10^-31))`

= `(12.27 xx 10^-10)/sqrt V` meter

= `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 the electron can also be written as:

λ = `h/(sqrt (2 m K))`