Question

Derive an expression for de Broglie wavelength of electrons.

Answer 1

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` mv² = 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"emV")`

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))`

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

λ = `"h"/(sqrt(2"mK"))`